calculus success in 20 minutes a day2ndedition[1]
DESCRIPTION
This is a very helpful self paced Calculus course. Includes 20 different chapters with topics typically covered in a college level Calculus I course. There is a pre-test, a post-test, examples and exercises throughout the book.TRANSCRIPT
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CALCU
LUS
Successin
20M
inutesa
Day
MASTER CALCULUS IN JUST 20 MINUTES A DAY!
STUDY GUIDES/Mathematics
SKILL BUILDERS
� Packed with key calculus concepts including ratesof change, optimization, antidifferentiation,techniques of integration, and much more
� Includes hundreds of practice questionswith detailed answer explanations
� Measure your progress with pre–and posttests
� Build essential calculus skills for successon the AP exams!
SKILL BUILDERS
CALCULUS ESSENTIALS INSIDE:
• Functions • Trigonometry • Graphs • Limits • Rates of change • Derivatives• Basic rules • Derivatives of sin(x) and cos(x) • Product and quotient rules • Chain
rule • Implicit differentiation • Related rates • Graph sketching • Optimization• Antidifferentiation • Areas between curves • The fundamental theorem of
calculus • Techniques of integration • and more!
CALCULUS SUCCESS PRACTICE
A good knowledge of calculus is essential for success on many tests and applicable fora wide range of careers. Calculus Success in 20 Minutes a Day helps students refresh andacquire important calculus skills. This guide provides a thorough review that fits into anybusy schedule. Each step takes just 20 minutes a day!
Pretest—Pinpoint your strengths and weaknesses
Lessons—Master calculus essentials with hundreds of exercises
Posttest—Evaluate the progress you’ve made
BONUS! Additional resources for preparing for important standardized tests ADDEDVALU
E—Access
toonline
practicew
ithInstantScoring
2ND EDITIONCompletely Revised and Updated!
FREE Calculus Practice!
Visit LearningExpress’s Online Practice Center to:
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� Receive immediate scoring and detailed answer explanations
� Focus your study with our customized diagnostic report,and boost your overall score to guarantee success
Mark A. McKibben
CALCULUSSUCCESS
in 20 Minutes a Day
Calc2e_00_i-x_FM.qxd 11/18/11 12:32 AM Page i
Calc2e_00_i-x_FM.qxd 11/18/11 12:32 AM Page ii
CALCULUSSUCCESS
in 20 Minutes a Day
N E W Y O R K
®
Second Edition
Mark A. McKibbenChristopher Thomas
Calc2e_00_i-x_FM.qxd 11/18/11 12:32 AM Page iii
Copyright © 2012 LearningExpress, LLC.
All rights reserved under International and Pan-American Copyright Conventions.
Published in the United States by LearningExpress, LLC, New York.
Library of Congress Cataloging-in-Publication Data:
McKibben, Mark A.
Calculus success in 20 minutes a day / Mark A. McKibben.—2nd ed.
p. cm.
Previous ed.: Calculus success in 20 minutes a day / Thomas, Christopher. © 2006.
ISBN 978-1-57685-889-9
1. Calculus—Problems, exercises, etc. I. Thomas, Christopher, 1973– Calculus success
in 20 minutes a day. II. Title. III. Title: Calculus success in twenty minutes a day.
QA303.2.T47 2012
515—dc23
2011030506
Printed in the United States of America
9 8 7 6 5 4 3 2 1
ISBN 978-1-57685-889-9
For information or to place an order, contact LearningExpress at:
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Calc2e_00_i-x_FM.qxd 11/18/11 12:32 AM Page iv
Mark A. McKibben is a professor of mathematics and computer science at Goucher College in Baltimore,
Maryland. During his 12 years at this institution, he has taught more than 30 different courses spanning the
mathematics curriculum, and has published two graduate-level books with CRC Press, more than two dozen
journal articles on differential equations, and more than 20 supplements for undergraduate texts on algebra,
trigonometry, statistics, and calculus.
Christopher Thomas is a professor of mathematics at the Massachusetts College of Liberal Arts. He has
taught at Tufts University as a graduate student, Texas A&M University as a postdoctorate professor, and the
Senior Secondary School of Mozano, Ghana, as a Peace Corps volunteer. His classroom assistant is a small
teddy bear named ex.
ABOUT THE AUTHOR
v
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INTRODUCTION ix
PRETEST 1
LESSON 1 Functions 15
LESSON 2 Graphs 23
LESSON 3 Exponents and Logarithms 31
LESSON 4 Trigonometry 37
LESSON 5 Limits and Continuity 47
LESSON 6 Derivatives 55
LESSON 7 Basic Rules of Differentiation 61
LESSON 8 Rates of Change 67
LESSON 9 The Product and Quotient Rules 75
LESSON 10 Chain Rule 81
LESSON 11 Implicit Differentiation 85
LESSON 12 Related Rates 91
LESSON 13 Limits at Infinity 97
LESSON 14 Using Calculus to Graph 107
LESSON 15 Optimization 115
CONTENTS
vii
Calc2e_00_i-x_FM.qxd 11/18/11 12:32 AM Page vii
LESSON 16 The Integral and Areas under Curves 121
LESSON 17 The Fundamental Theorem of Calculus 127
LESSON 18 Antidifferentiation 133
LESSON 19 Integration by Substitution 139
LESSON 20 Integration by Parts 145
POSTTEST 151
SOLUTION KEY 165
GLOSSARY 193
ADDITIONAL ONLINE PRACTICE 197
–CONTENTS–
vii i
Calc2e_00_i-x_FM.qxd 11/18/11 12:32 AM Page viii
INTRODUCTION
ix
If you have never taken a calculus course, and now find that you need to know calculus—this is the book
for you. If you have already taken a calculus course, but felt like you never understood what the teacher
was trying to tell you—this book can teach you what you need to know. If it has been a while since you
have taken a calculus course, and you need to refresh your skills—this book will review the basics and reteach
you the skills you may have forgotten. Whatever your reason for needing to know calculus, Calculus Success
in 20 Minutes a Day will teach you what you need to know.
Overcoming Math Anxiety
Do you like math or do you find math an unpleasant experience? It is human nature for people to like what
they are good at. Generally, people who dislike math have not had much success with math.
If you have struggles with math, ask yourself why. Was it because the class went too fast? Did you have
a chance to fully understand a concept before you went on to a new one? One of the comments students fre-
quently make is, “I was just starting to understand, and then the teacher went on to something new.” That is
why Calculus Success is self-paced. You work at your own pace. You go on to a new concept only when you
are ready.
When you study the lessons in this book, the only person you have to answer to is you. You don’t have
to pretend you know something when you don’t truly understand. You get to take the time you need to under-
stand everything before you go on to the next lesson. You have truly learned something only when you
thoroughly understand it. Take as much time as you need to understand examples. Check your work with the
answers and if you don’t feel confident that you fully understand the lesson, do it again. You might think you
don’t want to take the time to go back over something again; however, making sure you understand a lesson
Calc2e_00_i-x_FM.qxd 11/18/11 12:32 AM Page ix
completely may save you time in the future lessons.
Rework problems you missed to make sure you don’t
make the same mistakes again.
How to Use This Book
Calculus Success teaches basic calculus concepts in 20
self-paced lessons. The book includes a pretest, a
posttest, 20 lessons, each covering a new topic, and a
glossary. Before you begin Lesson 1, take the pretest.
The pretest will assess your current calculus abilities.
You’ll find the answer key at the end of the pretest.
Each answer includes the lesson number that the
problem is testing. This will be helpful in determin-
ing your strengths and weaknesses. After taking the
pretest, move on to Lesson 1, Functions.
Each lesson offers detailed explanations of a
new concept. There are numerous examples with
step-by-step solutions. As you proceed through a les-
son, you will find tips and shortcuts that will help
you learn a concept. Each new concept is followed by
a practice set of problems. The answers to the prac-
tice problems are in an answer key located at the end
of the book.
When you have completed all 20 lessons, take
the posttest. The posttest has the same format as the
pretest, but the questions are different. Compare the
results of the posttest with the results of the pretest
you took before you began Lesson 1. What are your
strengths? Do you still have weak areas? Do you need
to spend more time on some concepts, or are you
ready to go to the next level?
Make a Commitment
Success does not come without effort. If you truly
want to be successful, make a commitment to spend
the time you need to improve your calculus skills.
So sharpen your pencil and get ready to begin
the pretest!
–INTRODUCTION–
x
Calc2e_00_i-x_FM.qxd 11/18/11 12:32 AM Page x
Before you begin Lesson 1, you may want to get an idea of what you know and what you need to learn.
The pretest will answer some of these questions for you. The pretest consists of 50 multiple-choice
questions covering the topics in this book. While 50 questions can’t cover every concept or skill taught
in this book, your performance on the pretest will give you a good indication of your strengths and
weaknesses.
If you score high on the pretest, you have a good foundation and should be able to work through the
book quickly. If you score low on the pretest, don’t despair. This book will explain the key calculus concepts,
step by step. If you get a low score, you may need to take more than 20 minutes a day to work through a les-
son. However, this is a self-paced program, so you can spend as much time on a lesson as you need. You decide
when you fully comprehend the lesson and are ready to go on to the next one.
Take as much time as you need to complete the pretest. When you are finished, check your answers with
the answer key at the end of the pretest. Along with each answer is a number that tells you which lesson of
this book teaches you about the calculus skills needed to answer that question. You will find the level of dif-
ficulty increases as you work your way through the pretest.
PRETEST
1
Calc2e_00_1-14_Pre.qxd 11/18/11 12:34 AM Page 1
Calc2e_00_1-14_Pre.qxd 11/18/11 12:34 AM Page 2
3
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
1.2.3.4.5.6.7.8.9.
10.11.12.13.14.15.16.17.
18.19.20.21.22.23.24.25.26.27.28.29.30.31.32.33.34.
35.36.37.38.39.40.41.42.43.44.45.46.47.48.49.50.
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
–LEARNINGEXPRESS ANSWER SHEET–
Calc2e_00_1-14_Pre.qxd 11/18/11 12:34 AM Page 3
Calc2e_00_1-14_Pre.qxd 11/18/11 12:34 AM Page 4
1. What is the value of f(4) when f(x) = 3x2 – ?
a. 22
b. 46
c. 140
d. 142
2. Simplify g(x + 3) when g(x) = x2 – 2x + 1.
a. x2 + 2x + 4
b. x2 – 2x + 4
c. x2 + 4x + 16
d. x2 + 4x + 13
3. What is (f ° g)(x) when f(x) = x – and g(x) =
x + 3?
a.
b.
c.
d.
4. What is the domain of ?
a. all real numbers except x = 1
b. all real numbers except x = 0
c. all real numbers except x = –1 and x = 1
d. all real numbers except x = –1, x = 0, and
x = 1
Use the following figure for questions 5 and 6.
5. On what interval(s) is f(x) increasing?
a. (�∞,1) and (5,∞)
b. (1,5)
c. (1,6)
d. (5,∞)
6. Which of the following is a point of inflection
for f(x)?
a. (0,5.5)
b. (1,6)
c. (3,3)
d. (5,1)
7. What is the equation of the straight line pass-
ing through (2,5) and (�1,�1)?
a.
b.
c.
d. y � �2x � 3
y � �2x � 9
y � 2x � 1
y � 2x � 5
h1x 2 �x
x2 � 1
x � 3 �2
x � 3
x2 � 2 � 3x �6x
2x �2x
� 3
x �2x
� 3
2x
x
–PRETEST–
5
1
2
3
4
5
6
–1–2–3 1 2 3 4 5 6–1
y
x
y = f(x)
Calc2e_00_1-14_Pre.qxd 11/18/11 12:34 AM Page 5
6
8. Simplify .
a. 4
b. 8
c. 32
d. 4,096
9. Simplify .
a.
b. 8
c. �8
d. �6
10. Solve for x when 3x = 15.
a. 5
b.
c.
d.
11. Evaluate sin .
a.
b.
c.
d.
12. Evaluate .
a. �1
b. 1
c.
d.
13. Evaluate .
a. �1
b.
c.
d.
14. Evaluate .
a. 0
b. 1
c.
d. undefined
15. Evaluate .
a. ∞b. �∞
c.
d.−3
2
�14
limxS2�
x � 3x � 2
12
limxS1
x � 1x2 � 1
79
1517
35
limxS4
x2 � 1x2 � 1
2
2
2
tan3
4
π
3
2
2
2
12
�12
π3
ln112 2
ln115 2
ln13 2
ln15 2
18
2�3
6412
–PRETEST–
Calc2e_00_1-14_Pre.qxd 11/18/11 12:34 AM Page 6
16. What is the slope of at x � 5?
a. 2
b. 17
c. 3x
d. 3
17. What is the slope of at
x � 3?
a. 2
b. 8
c. 14
d.
18. Differentiate .
a.
b.
c.
d.
19. The height of a certain plant is H(t) =
inches after week. How fast is
it growing after two weeks?
a. 5 inches per week
b. 10 inches per week
c. 21 inches per week
d. 31 inches per week
20. What is the derivative of ?
a.
b.
c.
d.
21. Differentiate .
a.
b.
c.
d.
22. Differentiate
a.
b.
c.
d. 2xsin1x 2cos1x 2′ =g x( )
2xsin1x 2 � x2cos1x 2′ =g x( )
2x � cos1x 2′ =g x( )
2xcos1x 2′ =g x( )
g1x 2 � x2sin1x 2.
f ¿ 1x 2 �1x
� ex
f ¿ 1x 2 �1x
� ex
f ¿ 1x 2 � ln1x 2 � ex
f ¿ 1x 2 � ln1x 2 � ex
f 1x 2 � ln1x 2 � ex � 2
dy
dx� 2x � 3tan1x 2
dy
dx� 2x � 3cos11 2
dy
dx� 2x � 3sin1x 2
dy
dx� 2x � 3sin1x 2
y � x2 � 3cos1x 2
t � 1
41 �40t
12x2 � 5x �1x
′ =h x( )
12x2 � 5x′ =h x( )
12x2 � 5′ =h x( )
12x2′ =h x( )
h1x 2 � 4x3 � 5x � 1
2x � 2
g1x 2 � x2 � 2x � 1
f 1x 2 � 3x � 2
–PRETEST–
7
Calc2e_00_1-14_Pre.qxd 11/18/11 12:34 AM Page 7
23. Differentiate .
a. 0
b.
c.
d.
24. Differentiate .
a.
b. –cot(x)
c.
d.
25. Differentiate .
a.
b.
c.
d.
26. Differentiate m(x) = .
a.
b.
c.
d.
27. Compute if .
a.
b.
c.
d.
28. Compute if .
a.
b.
c.
d.dy
dx� 8xsec1y 2
dy
dx� cos1y 2 � 8x
dy
dx� 8xcos1y 2
dy
dx� 8x � cos1y 2
sin1y 2 � 4x2dy
dx
dy
dx
x x
y= −3
2
2
dy
dx�
3x2
1 � 2y
dy
dx�
3x2 � y
2y � x
dy
dx� x2
y2 � xy � x3 � 5dy
dx
10x1x2 � 1 24′ =m x( )
51x2 � 1 24′ =m x( )
12x 25′ =m x( )
10x′ =m x( )
1x2 � 1 25
14x2 � 7 2e4x2�8′ =f x( )
8xe4x2�7′ =f x( )
e4x2�7′ =f x( )
e8x′ =f x( )
f 1x 2 � e4x2�7
sin1x 2cos1x 2dy
dx=
cos21x 2 � sin21x 2
cos21x 2dy
dx=
dy
dx=
sec21x 2dy
dx=
y � tan1x 2
ln1x 2 � 1
x2′ =j x( )
1 � ln1x 2
x2′ =j x( )
1x
′ =j x( )
′ =j x( )
ln1x 2
xj x( ) =
8
–PRETEST–
Calc2e_00_1-14_Pre.qxd 11/18/11 12:34 AM Page 8
29. What is the slope of at ?
a. �1
b. 1
c. –
d.
30. If the radius of a circle is increasing at 4 feet
per second, how fast is the area increasing
when the radius is 10 feet?
a. 20p square feet per second
b. 80p square feet per second
c. 100p square feet per second
d. 400p square feet per second
31. The height of a triangle increases by 3 inches
every minute while its base decreases by 1 inch
every minute. How fast is the area changing
when the triangle has a height of 10 inches and
a base of 100 inches?
a. It is increasing at 145 square inches
per minute.
b. It is increasing at 500 square inches
per minute.
c. It is decreasing at 1,500 square inches
per minute.
d. It is decreasing at 3,000 square inches
per minute.
32. Evaluate .
a. 4
b. �4
c. 2
d. undefined
33. Evaluate .
a. �∞b. ∞c. �4
d. 4
34. Evaluate .
a.
b. 2
c. 3
d. 0
13
limln( )
x
x
x→−∞ +3 2
limx
x x
x x→−∞
+ ++ −
4 6 4
10 1
5
3
limx
x x
x→∞
− +−
4 5 2
1
2
2
3
3
3
3
1
2
3
2,
x2 � y2 � 1
–PRETEST–
9
Calc2e_00_1-14_Pre.qxd 11/18/11 12:34 AM Page 9
35. Which of the following is the graph of
?
a.
b.
c.
d.
y �1
x � 2
–PRETEST–
10
1
2
3
–1–2 1 2 3 4–1
–2
– 3
y
x
1
2
3
–1–2 1 2 3 4–1
–2
–3
y
x
1
2
3
–1–2 1 2 3 4–1
–2
–3
y
x
1
2
3
–1–2 1 2 3 4–1
–2
–3
y
x
Calc2e_00_1-14_Pre.qxd 11/18/11 12:34 AM Page 10
36. On what interval is con-
cave down?
a. (1,12)
b. (�6,5)
c.
d. (�1,1)
37. The surface area of a cube is increasing at a rate
of 3 square inches per minute. How fast is an
edge increasing at the instant when each side is
20 inches?
a. inch per minute
b. inch per minute
c. 80 inches per minute
d. 24,000 inches per minute
38. A box with a square bottom and no top must
contain 108 cubic inches. What dimensions
will minimize the surface area of the box?
a. 2 in. × 2 in. × 27 in.
b. 8 in. × 8 in. × 3 in.
c. 6 in. × 6 in. × 3 in.
d. 4 in. × 4 in. × 6.75 in.
39. If and , then
what is ?
a. �20
b. 1
c. 3
d. 9
40. What is ?
a. 2
b. 3
c. 10
d. 12
41. If is the area under the curve
between t � 0 and t � x, what is
?
a.
b.
c.
d. 0
42. Evaluate .
a.
b.
c.
d. x3 � 4x2 � 5x � c
x3 � 4x2 � 5x
6x � 8 � c
6x � 8
� 13x2 � 8x � 5 2 dx
14
x4 � 2x
3x2 � 4
x3 � 4x
g¿ 1x 2y � t3 � 4t
g1x 2
�4
0
f 1x 2 dx
�8
5
g1x 2 dx
�5
3
g1x 2 dx � �4�8
3
g1x 2 dx � 5
3
20
1
80
( , )− 3 3
g1x 2 � x4 � 6x2 � 5
–PRETEST–
11
1
2
3
4
1 2 3 4 5–1
y
x
(4,3)
y = f(x)
Calc2e_00_1-14_Pre.qxd 11/18/11 12:34 AM Page 11
43. Evaluate .
a. 3
b. 9
c. 18
d.
44. Evaluate
a.
b.
c.
d.
45. Evaluate
a.
b.
c.
d.
46. Evaluate
a.
b.
c.
d.
47. Evaluate .
a.
b.
c.
d.
48. Evaluate .
a.
b.
c.
d.1
1252 6( )x c+ +
xx x c
23
6
2
1
32+
+
5 22 4( )x c+ +
1
622 6( )x c+ +
x x dx+( )2 52�
43
x2sin1x3 2 � c
43
x3sin1x3 2 � c
43
sin1x3 2 � c
4sin1x3 2 � c
�4x2cos1x3 2 dx
15
e5 � c
e5 � c
e5x � c
15
e5x � c
�e5x dx˛.
1
212ln x c− +
1
212ln( )x c− +
ln x c− +1
12x2
13x3 � x
� c
� xx2 � 1
dx˛.
�sin1x 2 � c
sin1x 2 � c
�cos1x 2 � c
cos1x 2 � c
�sin1x 2 dx .
81
2
�9
0
2x dx
–PRETEST–
12
Calc2e_00_1-14_Pre.qxd 11/18/11 12:34 AM Page 12
49. Evaluate .
a.
b.
c.
d.
50. Evaluate
a.
b.
c.
d. xcos(x) – cos(x) + c
�12
x2cos1x 2 � c
12
x2cos1x 2 � c
�xcos1x 2 � sin1x 2 � c
�xsin1x 2 dx˛.
12
x2ln1x 2 �14
x2 � c
x2ln1x 2 �14
x2 � c
xln1x 2 � ln1x 2 � c
12
x2ln1x 2 � c
�xln1x 2 dx
–PRETEST–
13
Calc2e_00_1-14_Pre.qxd 11/18/11 12:34 AM Page 13
–PRETEST–
14
Answers
1. b. Lesson 1
2. a. Lesson 1
3. d. Lesson 1
4. c. Lesson 1
5. a. Lesson 2
6. c. Lesson 2
7. b. Lesson 2
8. b. Lesson 3
9. a. Lesson 3
10. c. Lesson 3
11. d. Lesson 4
12. a. Lesson 4
13. c. Lesson 5
14. c. Lesson 5
15. b. Lesson 5
16. d. Lessons 6, 7
17. b. Lessons 6, 7
18. b. Lesson 7
19. b. Lesson 8
20. a. Lesson 8
21. d. Lesson 8
22. c. Lesson 9
23. c. Lessons 8, 9
24. a. Lesson 9
25. c. Lesson 10
26. d. Lesson 10
27. b. Lesson 11
28. d. Lessons 4, 11
29. c. Lesson 11
30. b. Lesson 12
31. a. Lesson 12
32. b. Lesson 13
33. b. Lesson 13
34. d. Lesson 13
35. a. Lesson 14
36. d. Lesson 14
37. a. Lesson 12
38. c. Lesson 16
39. d. Lesson 16
40. c. Lesson 16
41. a. Lesson 17
42. d. Lesson 18
43. c. Lesson 18
44. b. Lesson 18
45. d. Lesson 19
46. a. Lesson 19
47. b. Lesson 19
48. d. Lesson 19
49. d. Lesson 20
50. a. Lesson 20
Calc2e_00_1-14_Pre.qxd 11/18/11 12:34 AM Page 14
Functions
A function is a way of matching up one set of numbers with another. The first set of numbers is called the
domain. For each of the numbers in the domain, the function assigns exactly one number from the other set,
the range.
LE
SS
ON
FUNCTIONS
Calculus is the study of change. It is often important to know when a quantity is increasing, when it
is decreasing, and when it hits a high or low point. Much of the business of finance depends on pre-
dicting the high and low points for prices. In science and engineering, it is often essential to know pre-
cisely how fast quantities such as temperature, size, and speed are changing. Calculus is the primary tool for
calculating such changes.
Numbers, which are the focus of arithmetic, do not change. The number 5 will always be 5. It never goes
up or down. Thus, we need to introduce a new sort of mathematical object, something that can change. These
objects, the centerpiece of calculus, are functions.
15
1
Calc2e_01_15-22.qxd 11/18/11 12:35 AM Page 15
For example, the domain of the function could
be the set of numbers {1, 4, 9, 25, 100}, and the range
could be {1, 2, 3, 5, 10}. Suppose the function takes 1
to 1, 4 to 2, 9 to 3, 25 to 5, and 100 to 10. This could be
illustrated by the following:
Because we sometimes use several functions in
the same discussion, it makes sense to give them
names. Let us call the function we just mentioned by
the name Eugene. Thus, we can ask, “Hey, what does
Eugene do with the number 4?” The answer is “Eugene
takes 4 to the number 2.”
Mathematicians like to write as little as possible.
Thus, instead of writing “Eugene takes 4 to the num-
ber 2,” we often write “Eugene(4) � 2” to mean the
same thing. Similarly, we like to use names that are as
short as possible, such as f (for function), g (for func-
tion when f is already being used), h, and so on. The
trigonometric functions in Lesson 4 all have three-
letter names like sin and cos, but even these are abbre-
viations. So let us save space and use f instead of
Eugene.
Because the domain is small, it is easy to write
out everything:
However, if the domain were large, this would get
very tedious. It is much easier to find a pattern and use
that pattern to describe the function. Our function f
just happens to take each number of its domain to the
square root of that number. Therefore, we can describe
f by saying:
f(a number) = the square root of that number
Of course, anyone with experience in algebra
knows that writing “a number” over and over is a waste
of time. Why not just pick a variable to represent the
number? Just as f is a typical name for a function, lit-
tle x is often used for a variable name. Using both, here
is a nice way to represent our function f :
f(x) =
This tells us that putting a number into the func-
tion f is the same as putting it into . Thus,
f(25) = = 5 and f(f) = = 2.
ExampleFind the value of g(3) if .g1x 2 � x2 � 2
425
x
f 11 2 � 1 f 14 2 � 2 f 19 2 � 3 f 125 2 � 5f 1100 2 � 10
1 S 1 4 S 2 9 S 3 25 S 5100 S 10
16
PARENTHESES HINT
It is true that in algebra, everyone is taught “parentheses mean multiplication.” This means that 5(2 + 7) = 5(9) = 45. If x is a variable, then x(2 + 7) = x(9) = 9x. However, if f is the name of a function, then f (2 + 7) = f (9) = the number to which f takes 9. The expression f (x) is pronounced “f of x” and not “f times x.” This cancertainly be confusing. But, as you gain experience, it will become second nature. Mathematicians use paren-theses to mean several different things and expect everyone to know the difference. Sorry!
Calc2e_01_15-22.qxd 11/18/11 12:35 AM Page 16
SolutionReplace each occurrence of x with 3.
g(3) � 32 � 2
Simplify.
g(3) � 9 � 2 � 11
ExampleFind the value of h(�4) if h(t) = t3 � 2t2 + 5.
SolutionReplace each occurrence of t with –4.
h(–4) = (–4)3 – 2(–4)2 + 5
Simplify.
h(–4) = –64 – 2(16) + 5 = –64 – 32 + 5 = –91
Practice
1. Find the value of when .
2. Find the value of when
.
3. Find the value of when .
4. Find the value of when .
5. Find the value of when m(t) = –5t 3.
6. Find the value of when
.
7. Suppose that after t seconds, a rock thrown off
a bridge has height
feet off the ground. What is the height above
the ground after 3 seconds?
8. Suppose that the profit on making and selling x
cookies is .
How much profit is made on selling 100 cookies?
Plugging Variables into Functions
Variables can be plugged into functions just as easily as
numbers can. Often, though, the result can’t be sim-
plified as much.
ExampleSimplify f(w) if f(x) = + 2x 2 + 2.
SolutionReplace each occurrence of x with w.
f(w) = + 2w 2 + 2
That is all we can say without knowing more about w.
ExampleSimplify if .
SolutionReplace each occurrence of t with (a � 5).
Multiply out and .
Simplify.
g1a � 5 2 � a2 � 7a � 11
g1a � 5 2 � a2 � 10a � 25 � 3a � 15 � 1
�31a � 5 21a � 5 22
g1a � 5 2 � 1a � 5 22 � 31a � 5 2 � 1
g1t 2 � t2 � 3t � 1g1a � 5 2
w
x
P xx x
( ),
dollars= − −2 10 000
102
s1t 2 � �16t2 � 20t � 100
h1x 2 � 2x � 23 x
h164 2
m −
1
5
f 1x 2 � 2f 17 2
h1t 2 � t2 �34
h a12b
g1x 2 � x3 � x2 � x � 1
g1�3 2
f 1x 2 � 2x � 1f 15 2
– FUNCTIONS–
17
(a + b)2 ≠ a2 + b2. Remember to FOIL (first, out-side, inside, last) to get (a + b)2 = a2 + 2ab + b2.
When multiplying, an even number of negativesresults in a positive number, whereas an odd num-ber of negatives results in a negative number.
Calc2e_01_15-22.qxd 11/18/11 12:35 AM Page 17
Example
Simplify if .
SolutionStart with what needs to be simplified.
Use to evaluate and .
Multiply out .
Cancel the and the .
Factor out an a.
Cancel an a from the top and bottom.
Practice
Simplify the following.
9. when
10. when
11.
12. g(x 2 + ) when
13.
14.
15. when h(x) = –2x + 1
16.
Composition of Functions
Now that we can plug anything into functions, we can
plug one function in as the input of another function.
This is called composition. The composition of func-
tion f with function g is written . This means to
plug g into f like this:
It may seem that f comes first in , read-
ing from left to right, but actually, the g is closer to the
x. This means that the function g acts on the x first.
ExampleIf f(x) = + 2x and , then what is the
composition ?
SolutionStart with the definition of composition.
(f ° g)(x) = f(g(x))
Use .
Replace each occurrence of x in f with .
Simplify.
( )( )f g x x xo = + + +4 7 8 14
( )( ) ( )f g x x xo = + + +4 7 2 4 7
4x � 7
( )( ) ( )f g x f xo = +4 7
g1x 2 � 4x � 7
( )( )f g xo
g1x 2 � 4x � 7x
( )( )f g xo
( )( ) ( ( ))f g x f g xo =
f � g
g x g xg x x
( ) ( )( )
+ − =2
23 when
h1x � a 2 � h1x 2
a
f x a f x
af x x
( ) ( )( )
+ − − + when = 2 5
g x g x g tt
t( ) ( ) ( )28
6− = − when
g1t 2 �8t
� 6tx
f x h f x
hf x
x
( ) ( )( )
+ − = when 1
2
f 1x 2 � x2 � 3x � 1f 1x � a 2
f 1x 2 � x2 � 3x � 1f 1y 2
2x � a
12x � a 2a
a
2xa � a2
a
�x2x2
x2 � 2xa � a2 � x2
a
1x � a 22
1x � a 22 � x2
a
f 1x 2f 1x � a 2f 1x 2 � x2
f 1x � a 2 � f 1x 2
a
f 1x 2 � x2f 1x � a 2 � f 1x 2
a
–FUNCTIONS–
18
Calc2e_01_15-22.qxd 11/18/11 12:35 AM Page 18
Conversely, to evaluate (g ° f )(x), we compute:
Use f(x) = + 2x.
Replace each occurrence of x in g with + 2x.
Simplify.
We can form the composition of more than two func-
tions. Just apply the functions, one at a time, working
your way from the one closest to x outward.
ExampleIf , g(x) = 2 – x, and h(x) = 4x, then
what is (f ° g ° h)(x)?
SolutionStart with the definition of composition.
(f ° g ° h)(x) = f (g(h(x)))
Use h(x) = 4x.
(f ° g ° h)(x) = f (g(4x))
Compute g(4x) by replacing each occurrence of x in g
with 4x.
g(4x) = 2 – 4x
Next, substitute this into the composition.
(f ° g ° h)(x) = f (g(4x)) = f (2 – 4x).
Replace every occurrence of x in f with 2 – 4x.
(f ° g ° h)(x) = f (2 – 4x) =
Simplify.
(f ° g ° h)(x) =
Practice
Using , , and h(x) = x
– , simplify the following compositions.
17. (f ° g)(x)
18. (g ° f )(x)
19. (f ° h)(t)
20. (f ° f )(z)
21. (h ° h)(w)
22. (g ° h)(16)
23. (h ° f ° g)(x)
24. (f ° h ° f )(2x)
Domains
When an expression is used to describe a function f (x),
it is convenient to think of the domain as the set of all
numbers that can be substituted into the expression
and get a meaningful output. This set is called the
domain. The range of the function is the set of all pos-
sible numbers produced by evaluating f at the numbers
in its domain.
In the beginning of the lesson, we considered the
function:
f(x) =
However, we left out a crucial piece of information: the
domain. The domain of this function consisted of only
the numbers 1, 4, 9, 25, and 100. Thus, we should have
written
f(x) = if x � 1, 4, 9, 25, or 100
Usually, the domain of a function is not given
explicitly like this. In such situations, it is assumed that
the domain is as large as it possibly can be, meaning that
x
x
x
g1x 2 � x3 � 2x2 � 1f 1x 2 �1x
3 41 8
−−
xx
( )( )2 4 1
2 2 4 3− +− −
xx
f xxx
( ) = +−
12 3
( )( )g f x x xo = + +4 8 7
( )( ) ( )g f x x xo = + +4 2 7
x
( )( ) ( )g f x g x xo = + 2
x
( )( ) ( ( ))g f x g f xo =
–FUNCTIONS–
19
This shows that the order in which you com-pute a composition matters! In general, ( f ° g)(x)≠ (g ° f )(x).
Calc2e_01_15-22.qxd 11/18/11 12:35 AM Page 19
it contains all real numbers that, when plugged into the
function, produce another real number. Specifically,
including a number in the domain cannot violate one
of the following two fundamental prohibitions:� Never divide by zero.� Never take an even root of a negative number.
ExampleWhat is the domain of ?
SolutionWe must never let the denominator be zero, so
x cannot be 2. Therefore, the domain of this function
consists of all real numbers except 2.
The prohibition against even roots (like square
roots) of negative numbers is less severe. An even root
of a negative number is an imaginary number. Useful
mathematics can be done with imaginary numbers.
However, for the sake of simplicity, we will avoid them
in this book.
ExampleWhat is the domain of g(x) = ?
SolutionThe numbers in the square root must not be negative,
so , thus . The domain consists
of all numbers greater than or equal to .
Do note that it is perfectly okay to take the square
root of zero, since = 0. It is only when numbers are
less than zero that even roots become imaginary.
ExampleFind the domain of k(x) = .
SolutionTo avoid dividing by zero, we need ,
so , thus and .
To avoid an even root of a negative number,
, so . Thus, the domain of k is
, , .
A nice way of representing certain collections of
real numbers is interval notation, as follows:
COLLECTION OF INTERVAL REAL NUMBERS NOTATION
a < x < b (a,b)
a ≤ x < b [a,b)
a < x ≤ b (a,b]
a ≤ x ≤ b [a,b]
x > a (a,∞)
x ≥ a [a,∞)
x < b (–∞,b)
x ≤ b (–∞,b]
All real numbers (–∞,∞)
Note: A parenthesis is used when we intend to NOT
include a point, whereas a square bracket is used when
we intend TO include a point.
The domain of the previous example would be
written as follows:
(–∞,–3), (–3,–2), and (–2,4]
Practice
Find the domain of each of the following functions.
Express your answers using interval notation.
25.
26. h(x) = x + 1
f xx x
( )( )( )
= −+ −
1
3 5 2
x � �2x � �3x � 4
x � 44 � x � 0
x � �2x � �31x � 3 2 1x � 2 2 � 0
x2 � 5x � 6 � 0
4
5 62
−+ +
x
x x
0
�23
x � �23
3x � 2 � 0
3 2x +
x � 2
f 1x 2 �3
x � 2
–FUNCTIONS–
20
Calc2e_01_15-22.qxd 11/18/11 12:35 AM Page 20
27.
28.
29.
30.
31.
32. g uu
u u( )
( )=
+ +8
3 4 3
k1x 2 �24 2 � x
x � 8
h1x 2 � 23 x
j zz z
z( )
( )( )= − ++
1 2
12
g1x 2 � x2 � 5x � 6
k1t 2 �12t � 5
–FUNCTIONS–
21
Calc2e_01_15-22.qxd 11/18/11 12:35 AM Page 21
Calc2e_01_15-22.qxd 11/18/11 12:35 AM Page 22
LE
SS
ON
GRAPHS
Afunction can be fully described by showing explicitly what happens at each number in its domain
(for example, 4 S 2) or by giving its formula (for example, f(x) = ). However, neither of these
provides a clear visual picture of the function.
Fortunately, René Descartes came up with the idea of a graph, a visual picture of a function. Rather than
say 4 S 2 or f(4) = 2, we plot the point (4,2) on the Cartesian plane, as in Figure 2.1.
x
2
23
2 up
4 over
x
y
1
2
3
4
(4,2)
1 2 3 4 5
Figure 2.1
Calc2e_02_23-30.qxd 11/18/11 12:37 AM Page 23
Practice
Plot the following points on a Cartesian plane.
1. (3,5)
2. (�3,4)
3. (2,�6)
4. (�1,�5)
5. (0,3)
6. (�5,0)
7. (0,0)
8.
For the function , plot the point
(x,f (x)) for the following values of x.
9. x � 3
10. x � 1
11. x � 0
12. x � �2
If we plotted the points (x,f (x)) for all x in the
domain of f(x) = (not just the whole numbers, but
all the fractions and decimals, too), then the points
would be so close together that they would form a con-
tinuous curve as in Figure 2.2.
The graph shows us several interesting charac-
teristics of the function f(x) = .
We can see that the function f(x) = is
increasing (the graph is going up from left to right)
and not decreasing (the graph is going down from left
to right).
The function f(x) = is concave down
because it bows downward (see Figure 2.3) like a frown
and not concave up like a smile (see Figure 2.4). We
report the input intervals in each case. So, we say that
f is increasing on (0,∞) and concave down on (0,∞).
ExampleAssume the domain of the function graphed in Figure
2.5 is all real numbers. Determine where the function
is increasing and decreasing, and where the function is
concave up and concave down.
x
x
x
x
f 1x 2 � x2 � 2x � 5
9
2
1
4,
24
NOTE ON FINDING COORDINATES
We put the y into the formula y = f (x) = to imply that the y-coordinates of our points are the numbers weget by plugging the x-coordinates into the function f .
x
1
2
1 2 3 4
y
x
xy = f (x) =
(0,0),
(1,1)
(4,2)
1—21—4
Figure 2.2
Figure 2.3
Figure 2.4
x
Calc2e_02_23-30.qxd 11/18/11 12:37 AM Page 24
SolutionThe function g is increasing up to the point at x � 2,
where it then decreases down to x � 8, and then
increases thereafter. Using interval notation, we say
that g increases on (�∞,2) and on (8,∞), and that g
decreases on (2,8).
The concavity of g is trickier to estimate. Clearly
g is concave down in the vicinity of x � 2 and concave
up starting around x � 7. The exact point where the
concavity changes is called a point of inflection. On this
graph, it seems to be at the point (5,4), though some
people might imagine it to actually be a bit on either
side. Thus, we say that g is concave down on (�∞,5)
and concave up on (5,∞).
Honestly, any information obtained by simply
eyeballing a graph is going to be a rough estimate. Is
the local maximum at (2,6), or is it at (2.0003,5.9998)?
There is no way to tell the difference without an actual
formula for f (x).
The point at (2,6) where g stops increasing and
begins to decrease is the highest point in its immedi-
ate vicinity and is called a local maximum. The point at
(8,3) is similarly a local minimum, the lowest point in
its neighborhood. These points tend to be of particu-
lar interest, especially in applications.
ExampleUse the graph of the function h(x) in Figure 2.6 to
identify the domain, where it is increasing and decreas-
ing, where it has local maxima and minima, where it is
concave up and down, and where it has points of
inflection.
25
1
2
3
4
5
6
–1–2 1 2 3 4 5 6–1
y
x
7
8
9 107 8
y = g (x)
Figure 2.5
MATHEMATICAL NOTATION NOTE
Out of context, an expression like (2,8) is ambiguous. Is this a single point with coordinates x � 2 and y � 8?Is this an interval consisting of all the real numbers between 2 and 8? Only the context can make clear whichis meant. If we read “at (2,8),” then this is a single point. If we read “on (2,8),” then it refers to an interval.
1
23
4
5
6
–1–2–3 1 2 3 4 5 6–1
–2
–3
–4
–5
–6
y
x–4
y = h(x)
Figure 2.6
Calc2e_02_23-30.qxd 11/18/11 12:37 AM Page 25
SolutionThe first thing to notice is that h has three breaks, or
discontinuities. If we wanted to trace the graph of h
with a continuous motion of a pencil, then we would
have to lift up the pencil at x � �2, x � 2, and at
x � 5. The little unshaded circle at (5,3) indicates a hole
in the graph where a single point has been taken out.
This means that x � 5 is not in the domain, just as x �
�2 has no point above or below it. The situation at x
� 2 is more interesting because x � 2 is in the domain,
with the point (the shaded-in circle) at (2,�2) repre-
senting h(2) � �2. All of the points immediately
before x � 2 have y-values close to y � 3, but then
there is an abrupt jump down to x � 2. Such jumps
occur often when describing real-life situations using
functions like the way the cost of postage leaps up as
soon as a letter weighs more than one ounce.
Because of the discontinuities, we must name
each interval separately, as in: h increases on (�∞,�2),
(�2,2), (2,5), and on (5,∞). As well, h is concave up on
(�∞,�2), (2,5), and on (5,∞), and concave down on
(�2,2).
There is a local minimum at (2,�2), because this
point there is the lowest in its immediate vicinity, say
for all 1 � x � 3. There is no local maximum in that
range because the y-values get really close to y � 3;
there is no highest point in the range because of the
unshaded circle.
Similarly, a point of inflection can be seen at x �
2 but not at x � �2 because there can’t be a point of
inflection where there is no point!
The line x = –2 is called a vertical asymptote
because the graph of f (x) begins to look more like this
line the closer the inputs get to –2. Because the graph
appears to flatten out like the straight horizontal line
y � 0 (the x-axis) as the graph goes off to the left, we
say that the graph of appears to have a hor-
izontal asymptote at y � 0. We will examine this more
closely in Lesson 13.
Practice
Use the graph of each function to determine the dis-
continuities, where the function is increasing and
decreasing, the local maximum and minimum points,
where the function is concave up and down, the points
of inflection, and the asymptotes.
y � h1x 2
–GRAPHS–
26
Calc2e_02_23-30.qxd 11/18/11 12:37 AM Page 26
13.
14.
15.
16.
–GRAPHS–
27
1
2
3
4
5
6
–1–2–3–4–5–6 1 2 3 4 5 6–1
–2
–3
–4
–5
–6
y
x
y = f (x)
1
2
3
–1
–2–3 1 2 3–1
y
x
4
y = g(x)
1
2
3
–1
–2–3 1 2 3–1
y
x
4
–2
y = h(x)
1
2
3
4
5
6
–1–2–3–4–5–6 1 2 3 4 5 6–1
–2
–3
–4
–5
–6
y
x
y = k(x)
Calc2e_02_23-30.qxd 11/18/11 12:37 AM Page 27
17.
18.
19.
20.
Note
We can obtain all sorts of useful information from a
graph, such as its maximal points, where it is increas-
ing and decreasing, and so on. Calculus will enable us
to get this information directly from the function. We
will then be able to draw graphs intelligently, without
having to calculate and plot thousands of points (the
method graphing calculators use).
–GRAPHS–
28
1
2
3
–1
–2–3 1 2 3–1
y
x
–2
–3
y = f(x)
(2,3)
1
2
–1
–2–3 1 2 3–1
y
x
–2
y = g(x)
1
2
3
4
5
6
–11 2 3 4 5 6–1
–2
–3
–4
y
x
y = j(x)
(2,5)
1
2
3
4
5
6
–1–2–3–4 1 2 3 4 5 6–1 7 8 9
y
x
y = h(x)
7
Calc2e_02_23-30.qxd 11/18/11 12:37 AM Page 28
Straight Lines
The most familiar and arguably most widely used of all
graphs are straight lines. Human beings tend to build
and move linearly. Given any two points, we can
immediately get a feel for the steepness of a line, as
seen in Figure 2.7.
“How much a line is increasing or decreasing” is
called the slope and is calculated by:
slope �
y-change�
x-change
�
ExampleWhat is the slope of the line passing through points
(2,7) and (�1,5)?
Solution
slope �
Practice
Find the slope between the following points.
21. (1,5) and (2,8)
22. (7,3) and (�2,3)
23. (�2,�4) and (�6,5)
24. (2,7) and (5,w)
Equation of a Line
No matter what two points you choose on a line, the
slope will always be the same. Thus, if a straight line
has slope m and goes through the point , then
using any other point (x,y) on the line, we get the same
slope, namely:
By cross-multiplying, we get the point-slope formula
for the equation of a straight line:
or equivalently
Going one step further, we get
y = mx +
This is called the slope-intercept formula for the line
because the point (0,b) is the y-intercept of the line
(that is, where it crosses the y-axis).
( )− +mx y
b
1 1
Call this 1 24 34
y � m1x � x1 2 � y1
y � y1 � m1x � x1 2
y � y1
x � x1� m
1x1, y1 2
5 � 7�1 � 2
��2�3
�23
y2 � y1
x2 � x1
riserun
–GRAPHS–
29
y
x
(x , y )1
(x , y )22
1
“rise”
y – y12
“run”x – x
12
Figure 2.7
Make sure to subtract the y’s in the top and the x’sin the bottom IN THE SAME ORDER. For instance,don’t use y2 – y1 for the rise and x1 – x2 for the run.
Calc2e_02_23-30.qxd 11/18/11 12:37 AM Page 29
ExampleFind the equation of the line with slope �2 through
point (�1,8). Graph the line.
Solution
(see Figure 2.8)
The slope of �2 � means the y-value goes
down 2 with every decrease of 1 unit in the x-value.
ExampleFind the equation of the straight line through (2,6)
and (5,7). Graph the line.
SolutionThe slope is , so the equation is
(see Figure 2.9).
The slope of means the y-value goes up 1 for
every increase of 3 units in the x-value.
Practice
Find the equation of the straight line with the given
information and then graph the line.
25. slope 2 through point (1,�2)
26. slope through point (6,1)
27. through points (5,3) and (�1,�3)
28. through points (2,5) and (6,5)
�23
13
y �13
1x � 2 2 � 6 �13
x �163
7 � 65 � 2
�13
�21
y � �2x � 6
y � �21x � 1�1 2 2 � 8
–GRAPHS–
30
1
2
3
4
5
6
–11 2 3 4–1
y
x
(0,6)
y = –2x + 6
(–1,8)
Figure 2.8
1
2
3
4
5
6
–1–2–3–4–5–6 1 2 3 4 5 6–1
y
x
0, y = x +
(2,6) (5,7)
16—3
16—31—3
Figure 2.9
Calc2e_02_23-30.qxd 11/18/11 12:37 AM Page 30
Exponents
Exponents frequently arise in calculations throughout calculus. If a is a positive real number and n is a pos-
itive integer (that is, n = 1, 2, 3, …), then an means “multiply the base a by itself n times.” Symbolically,
ExamplesReview the following examples.
25 � 2 # 2 # 2 # 2 # 2 � 32
34 � 3 # 3 # 3 # 3 � 81
n times
an � a # a # a p a
LE
SS
ON
EXPONENTS AND LOGARITHMS3
31
�
Do not multiply the base a times the exponent n.Symbolically, an ≠ a ⋅ n.
Calc2e_03_31-36.qxd 11/18/11 12:38 AM Page 31
106 = 1,000,000
When two numbers with the same base are multiplied,
their exponents are added.
ExamplesReview the following examples.
The rule about adding exponents has an inter-
esting consequence. We know that
because this is what “square root” means. Alternately,
. Because and act
exactly the same, they are equal: � . This works
for square roots, cube roots, and so on:
ExamplesReview the following examples.
When two numbers with the same base are divided,
their exponents are subtracted.
ExamplesWork through the following simplifications.
The rule about subtracting exponents has two inter-
esting consequences. First, � 1 because any
nonzero number divided by itself is one. Also,
. Thus, � 1. In general:
� 1
Simplify the following.
� 1
� 1
The second consequence follows from:
while
also . Thus, . In general:
ExamplesReview the following examples.
When an is itself raised to a power m, the exponents
are multiplied.
51
5
1
2−
= = 1
5 1
2
4�1 �141 �
14
3�2 �132 �
19
a�n �1an
2�4 �124
23
27 � 23�7 � 2�4
23
27 �2 # 2 # 2
2 # 2 # 2 # 2 # 2 # 2 # 2�
12 # 2 # 2 # 2
�124
2000
30
a0
5054
54 � 54�4 � 50
54
54
11
1111 11
15
615 6 9= =−
35
32 �3 # 3 # 3 # 3 # 3
3 # 3�
3 # 3 # 3 # 3 # 3
3 # 3� 3 # 3 # 3 � 33
an
am � an�m
64 64 413 3= =
9 9 31
2 = =
a a a a a a1
213
143 4= = = , , , K
5125
51255
12 # 5
12 � 5
12 �1
2 � 51 � 5
5 5 5 ⋅ =
72 # 74 # 73 � 79
53 # 5 � 53 # 51 � 54
2
3
2
3
2
3
2
3
2
3
2
3
16
81
2 2
⋅
= ⋅ ⋅ ⋅ =
410 # 47 � 417
m timesn times
an # am � 1a # a # a p a 2 # 1a # a # a p a 2 � an�m
1
2
1
2
1
2
1
2
1
8
3
+ ⋅ ⋅ =
51 � 5
–EXPONENTS AND LOGARITHMS–
32
� �
Again, don’t multiply a times the exponent 0 toconclude that a0 = 0.
Calc2e_03_31-36.qxd 11/18/11 12:38 AM Page 32
ExamplesReview the following examples.
Practice
Simplify the following.
1.
2.
3.
4.
5.
6.
7.
8. 5–1 ⋅ 5
9.
10.
11.
12.
13.
14.
Exponential Functions
We can form an exponential function by leaving the
base fixed and varying the exponent.
ExampleThe function has the graph shown in Fig-
ure 3.1. Note that is quite different from . For
example, when , the value of is
, while the
value of is .
ExampleThe function has the graph shown in Fig-
ure 3.2. Note that g (x) grows faster than f (x) = 2x as x
gets larger. For reasons that will become clear later, a
very nice base to use is the number e � 2.71828 . . . ,
which, just like p� 3.14159 . . . , can never be written
out completely.
g1x 2 � 3x
102 � 10 # 10 � 100x2
210 � 2 # 2 # 2 # 2 # 2 # 2 # 2 # 2 # 2 # 2 � 1,024
2xx � 10
x22x
f 1x 2 � 2x
9
81
12
2
2
−
−
4 4 44
2 0 5
1
−
−⋅ ⋅
8
8
4
2
12−( )
5 23−( )
2723
2�3
91
38 # 3 # 3�5
60
63
65
107
103
4 # 42
23 # 22
5 5 5 5 5
4 4 4 4 4 414
22
2 2 2 2 4
23
2 2 2 2 2 2 66
( ) = ⋅ = =
( ) = ⋅ ⋅ = = =
+
− − − − − + − + − −( ) ( ) ( )
a a a a a anm
n n n
m
n n n m
m
( ) = ⋅ ⋅ ⋅ = =+ + ⋅K1 244 344K
674 84
times
times
–EXPONENTS AND LOGARITHMS–
33
1
2
3
4
5
6
–1–2–3 1 2 3 4–1
y
x
(3,8)
(2,4)
(1,2)
(0,1)
xy = f(x) = 2
–1, 1—2–2, 1—4–3, 1—8
Figure 3.1
Calc2e_03_31-36.qxd 11/18/11 12:38 AM Page 33
Because 2 � e � 3, the graph of y � fits
between y � and y � (see Figure 3.3).
Another useful function is the inverse of ,
known as the natural logarithm ln(x). Just as subtract-
ing undoes adding, dividing undoes multiplying, and
taking a square root undoes squaring, the natural log-
arithm undoes .
If y � , then = x.
The graph of y � ln(x) comes from flipping the
graph of y � across the line y � x, as depicted in
Figure 3.4. In particular, since e0 = 1, it follows that
ln(1) = 0.
ex
ln1y 2 � ln1ex 2ex
ex
ex
3x2x
ex
34
1
2
3
4
5
6
–1–2 1 2 3–1
y
x
7
8
9 (2,3)
(1,3)
(0,1)
xy = g(x) = 3
–1, 1—3–2, 1—9
Figure 3.2
1
2
3
4
5
6
–1–2 1 2 3–1
y
x
7
8
9
(0,1)
x
x
y = 3
y = 2
y = e x
(1, e)
(1, e 2)
–2, 1—e2–1, 1—e
Figure 3.3
EXPONENTS AND LOGARITHMS
The exponential function f(x) = ex and the natural logarithm g(x) = ln(x) “undo” each other when composed.That is,
f (g(x)) = eln(x) = x and g(f(x)) = ln(ex) = x
We say that f and g are inverses.
ln(0) ≠ 1! In fact, ln is not even defined at x = 0.
Calc2e_03_31-36.qxd 11/18/11 12:38 AM Page 34
The laws of natural logarithms might appear unusual,
but they are natural consequences of the exponent rules.
The last of the three preceding laws is useful for
turning an exponent into a matter of multiplication.
ExampleSolve for x when .
SolutionTake the natural logarithm of both sides.
Use .
Divide both sides by ln(10).
A calculator can be used to find a decimal approxima-
tion: � 0.84509, if desired.
ExampleSimplify ln(25) � ln(4) � ln(2).
SolutionUse .
� ln(2)
Use .
� ln(50)
Practice
Simplify the following.
15.
16.
17.
18.
19. eln152
ln1e2 2
e0
e12
e5
e3 # e8
ln a25 # 4
2b
ln1a 2 � ln1b 2 � ln aabb
ln125 # 4 2
ln1a 2 � ln1b 2 � ln1a # b 2
ln17 2
ln110 2
x �ln17 2
ln110 2
x # ln110 2 � ln17 2
ln1an 2 � n # ln1a 2
ln110x 2 � ln17 2
10x � 7
ln1an 2 � n # ln1a 2
ln1a 2 � ln1b 2 � ln aabb
ln1a 2 � ln1b 2 � ln1a # b 2
–EXPONENTS AND LOGARITHMS–
35
1
2
3
–1
–2–3 1 2 3–1
–2
–3
y
x
(1,e)
(0,1) (e, 1)
(1,0)
y = e x
y = ln(x)
y = x
, –3 1—e3
, –2 1—e2
, –1 1—e
–3, 1—e3–2, 1—e2
–1, 1—e
Figure 3.4
These are the only three properties. Take particu-lar note of the following, which are often mistak-enly used in their place.ln(a + b) ≠ ln(a) + ln(b)ln(a – b) ≠ ln(a) – ln(b)(ln(a))b ≠ b ln(a)
Calc2e_03_31-36.qxd 11/18/11 12:38 AM Page 35
20.
21. ln(7) 1 ln(2)
22. ln(24) 2 ln(6)
23.
24. Solve for x when = 10.
25. Solve for x when = 100.
26. If loga x = 2 and loga y = –3, then what is
?loga
xy 3
3x # 35
2x
ln( ) ln( ) ln5 225
− +
ln 250( )
–EXPONENTS AND LOGARITHMS–
36
Calc2e_03_31-36.qxd 11/18/11 12:38 AM Page 36
LE
SS
ON
TRIGONOMETRY
Some very interesting and important functions are formed by dividing the length of one side of a right
triangle by the length of another side. These functions are called trigonometric because they come from
the geometry of a right triangle. Let H represent the length of the hypotenuse, A represent the length
of the side adjacent to the angle x, and O represent the length of the side opposite (away) from the angle x.
Such a triangle is depicted in Figure 4.1.
4
37
x
A
OH
Figure 4.1
Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 37
The six trigonometric functions, sine (abbrevi-
ated sin), cosine (cos), tangent (tan), secant (sec), cose-
cant (csc), and cotangent (cot), are defined at an angle
x by dividing the following sides:
The first thing to notice is that all of the func-
tions can be obtained from just sin(x) and cos(x) using
the following trigonometric identities.
Thus, all of the trigonometric functions can be evalu-
ated for an angle x if the sin(x) and cos(x) are known.
The next thing to notice is that the Pythagorean
theorem, which, stated in terms of the sides O, A, and
H, is . And, if we divide through by ,
we get the following:
Thus, . To save on paren-
theses, we often write this as .
Because no particular value of x was used in the cal-
culations, this is true for every value of x.
Drawing triangles and measuring their sides is an
impractical and inaccurate method to calculate the
values of trigonometric functions. It is better to use a
calculator. However, when using a calculator, it is very
important to make sure that it is set to the same format
for measuring angles that you are already using: that is,
degrees or radians.
There are 360 degrees in a circle, possibly because
ancient peoples thought that there were 360 days in a
sin21x 2 � cos21x 2 � 1
1sin1x 2 22 � 1cos1x 2 22 � 1
aOHb
2
� aAHb
2
� 1
O2
H2 �A2
H2 �H2
H2
H2O2 � A2 � H2
cot1x 2 �AO
�A
H
OH
�cos1x 2
sin1x 2
csc1x 2 �HO
� 1 OH
�1
sin1x 2
sec1x 2 �HA
� 1 AH
�1
cos1x 2
tan1x 2 �OA
�O
H
AH
�sin1x 2
cos1x 2
cot1x 2 �AO
csc1x 2 �HO
sec1x 2 �HA
tan1x 2 �OA
cos1x 2 �AH
sin1x 2 �OH
38
MNEMONIC HINT
Some people remember the first three trigonometric functions by SOA CAH TOA to remember
sin(x) � , cos(x) � , and tan(x) � .OA
AH
OA
Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 38
year. As the earth went around the sun, the position of
the sun against the background stars moved one degree
every day. The 2π radians in a circle correspond to the
distance around a circle of radius 1.
� To convert from degrees to radians, multiply by
.
� To convert from radians to degrees, multiply by
.
ExampleConvert 45° to radians.
Solution
radians � radians
Example
Convert radians to degrees.
Solution
radians �
Practice
Convert the following to radians.
1. 30°
2. 180°
3. 270°
4. 300°
5. 135°
Convert the following to degrees.
6.
7.
8. 2p
9.
10.11π
6
π10
π2
π3
2
3 120
ππ
⋅ ° = °1802π3
2π3
π4
45 45180
° = °°
⋅ π
3602
180° = °π π
2
360
π π°
=°180
39
CONVERSION HINT
To convert from degrees to radians, multiply by .
To convert from radians to degrees, multiply by .3602
180° = °π π
2360 180
π π°
=°
Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 39
Trigonometric Values of Nice Angles
There are a few nice angles for which the trigonomet-
ric functions can be easily calculated. If x = = 45°,
then the two legs of the triangle are equal. If the
hypotenuse is H � 1, then we have the triangle in Fig-
ure 4.2.
By the Pythagorean theorem, , so
and . This means that O � A �
. If we rationalize the denominator, we
get . Thus:
Another nice angle is x = 60° = , because it is
found in equilateral triangles such as the one seen in
Figure 4.3. This triangle can be cut in half by inserting
a segment from the top vertex down to the base, result-
ing in the triangle shown in Figure 4.4.
π3
cosπ4 1
2
2
22
= = = A
H
sinπ4 1
2
2
22
= = = O
H
1
2
1
2
2
2
2
2 = =⋅
1
2
1
2 =
A2 �12
2A2 � 1
A2 � A2 � 1
π4
–TRIGONOMETRY–
40
O = A
A
π–4
H = 1
Figure 4.2
1 1
1
–π3
Figure 4.3
1
–π3
1–2
O
H =
A =
–π6
Figure 4.4
Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 40
By the Pythagorean theorem, ,
so . Thus, . This
means that:
We can flip that last triangle around to calculate
the trigonometric functions for the other angle x = 30°
= (see Figure 4.5).
ExampleCompute sec .
SolutionUse the trigonometric identity for sec.
Use x = .
Use .
Simplify.
Practice
Use the trigonometric identities to evaluate the
following.
11. tan
12. tan
13. csc
14. sec
15. cot
16. cot
17. sec
18. cscπ4
π6
π6
π3
π3
π6
π3
π4
secπ4
2
2
2
2
2
2
2 22
2
= = ⋅ = =
secπ4 2
2
= 1
cosπ4 2
= 2
seccos( )
ππ4
1
4
=
π4
sec1x 2 �1
cos1x 2
π4
cosπ6 1
3
2
32
= = = A
H
sinπ6 1
1
2
12
= = = O
H
π6
cosπ3 1
1
2
12
= = = A
H
sinπ3 1
3
2
32
= = = O
H
O 3
4 = = 3
2O2 � 1 �
14
�34
a12b
2
� O2 � 12
–TRIGONOMETRY–
41
–π
1–2
3
2
6
A =
O =
H = 1
Figure 4.5
Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 41
Trigonometric Values for Angles Greater Than 90° =
Example
For example, for , we have the picture shown
in Figure 4.6.
The circle of radius 1 around the origin is called
the unit circle. As such, the hypotenuse has length 1,
and the sine is the y-value of the point where a ray of
the given angle intersects with the circle of radius 1.
Similarly, the cosine is the x-value. Note in Figure 4.7
that the angle of measure 0 runs straight to the right
along the positive x-axis, and every other positive angle
is measured counterclockwise from there.
This can be used to find the trigonometric values
of nice angles greater than 90° = . The trick is to
use either a 30°, 60°, 90° triangle (a , , trian-
gle) or else a 45°, 45°, 90° triangle (a , ,
triangle) to find the y-value(sine) and x-value (cosine)
of the appropriate point on the unit circle. As before,
calculating the trigonometric values for non-nice
angles is best done with a calculator.
ExampleFind the sine and cosine of 120° = .
Solution
For this angle, we use a , , triangle, as
shown in Figure 4.8 to find the x- and y-values. The
y-value of the point where the ray of angle hits2
3
π
π2
π3
π6
2
3
π
π2
π4
π4
π2
π3
π6
π2
cosπ6
= 3
2
sinπ6
= 1
2
π6
30= °
π2
–TRIGONOMETRY–
42
1––2
32
,
x
y
1
–1
1
–1
–π6
–π630 =
–32
11–2
Figure 4.6
–
π
00° = 360° =
30° =
45° =
60° = 90° =
120° =
135° =
150° =
180° =
210° =
225° =
240° = 270° =
300° =
315° =
330° =
6
4
32
ππ
π
–
–
π
– 2π
3π
5π
7π
5π
4π3π
5π7π
11π
3
4
6
6
4
32
3
4
6
Figure 4.7
Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 42
the unit circle is y � . Thus, .
The x-value is negative, x � , so .
ExampleFind the sine and cosine of = 225°.
SolutionBecause is a multiple of , we use a
triangle to find the x- and y-values. As
seen in Figure 4.9, both coordinates are negative, so
and .
ExampleFind all of the trigonometric values for 90° = .
Solution
Even though there isn’t a triangle here, there is still a
point on the unit circle. See Figure 4.10. We conclude
that cos = 0 and sin = 1 from the x- and
y-values of the point. Using the trigonometric identi-
ties, we can calculate that = 1 and
= 1. The tangent and secantcotcos( )
sin( )π π
π2
0
12
2
= =
cscsin( )
ππ2
1
2
=
π2
π2
π2
cos5
4
2
2
π
= − sin5
4
2
2
π
= −
45�, 45�, 90�
45�225�
5
4
π
cos2
3
1
2
π
= − �12
sin2
3
π
= 3
2
3
2
–TRIGONOMETRY–
43
120 =
–3π
1
2π–3
2π–3
32
1–232
,(– )
1–2–
Figure 4.8
225° =5π–4
5π–4
1
( )
4π–
2_2
2
2– 2–,
2_
2_ 2_
Figure 4.9
π90° =2–
(0,1)
Figure 4.10
Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 43
functions, however, involve division by 0 and thus are
undefined. The angle x = is not in the domain of
tan and sec.
Notice that all of the trigonometric functions are
the same at 0° = 0 and 360° = 2π. This is because turn-
ing 360° leaves you facing in your original direction.
Thus, everything repeats at this point.
Using the table along with the fact that every-
thing repeats, we can sketch the graphs of and
. See Figures 4.11 and 4.12.
The functions sine and cosine are classic exam-
ples of periodic, or oscillating, functions.
cos1x 2
sin1x 2π2
–TRIGONOMETRY–
44
1
– π6 4 3 2
ππ π–– – 2π 3π 5π π π π π π π π7 5 4 3 5 73 4 6 6 4 3 2 3 4
π π π π π π116
232
–– – –– –4
––6
1–2
1–2–
22
22
32
32
13 96 4
73
52
y = sin(x)
ππ ππ
––
–1
Figure 4.11
1
– π6 4 3 2
ππ π–– – 2π 3π 5π π π π π π π π7 5 4 3 5 73 4 6 6 4 3 2 3 4
π π π π π π116
232
–– – –– –4
––6
1–2
1–2–
22
22
32
32
13 96 4
73
52
y = cos(x)
ππ ππ
–
–1–
Figure 4.12
Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 44
Practice
Use the unit circle and the trigonometric identities to complete the following table. Find the answers to questions
19 through 38.
–TRIGONOMETRY–
45
0° = 0 0 1 0 1 undef. undef.
30° = *19* 2
45° = 1 1
60° = 2
90° = 1 0 undef. undef. 1 0
120° = � �2 *20* �
135° = *21* *22* *23* *24* *25* *26*
150° = � � � 2
180° = π *27* �1 0 �1 *28* undef.
210° = � � � �2 *29*
225° = � � 1 1
240° = *30* *31* *32* *33* *34* *35*
270° = �1 0 undef. undef. �1 0
300° = � 2 � �
315° = � �1 *36* �1
*37* *38* �2
360° = 2π 0 1 0 1 undef. undef.
Note: The numbers appearing in bold with asterisks are questions 19 through 38.
− 32 3
3− 3
3330
11
6o = π
− 22
2
2
27
4
π
3
3
2 3
3− 31
2
3
25
3
π
3
2
π
4
3
π
− 2− 22
2
2
2
5
4
π
2 3
3
3
3
3
2
1
27
6
π
− 32 3
3
3
3
3
2
1
2
5
6
π
3
4
π
3
3− 3
1
23
2
2
3
π
π2
3
32 3
331
23
2
π3
222
2
2
2
π4
32 3
3
3
2
1
2
π6
cot1x 2csc1x 2sec1x 2tan1x 2cos1x 2sin1x 2
Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 45
19. Find the value that goes in the position in the
table where you see *19*.
20. Find the value that goes in the position in the
table where you see *20*.
21. Find the value that goes in the position in the
table where you see *21*.
22. Find the value that goes in the position in the
table where you see *22*.
23. Find the value that goes in the position in the
table where you see *23*.
24. Find the value that goes in the position in the
table where you see *24*.
25. Find the value that goes in the position in the
table where you see *25*.
26. Find the value that goes in the position in the
table where you see *26*.
27. Find the value that goes in the position in the
table where you see *27*.
28. Find the value that goes in the position in the
table where you see *28*.
29. Find the value that goes in the position in the
table where you see *29*.
30. Find the value that goes in the position in the
table where you see *30*.
31. Find the value that goes in the position in the
table where you see *31*.
32. Find the value that goes in the position in the
table where you see *32*.
33. Find the value that goes in the position in the
table where you see *33*.
34. Find the value that goes in the position in the
table where you see *34*.
35. Find the value that goes in the position in the
table where you see *35*.
36. Find the value that goes in the position in the
table where you see *36*.
37. Find the value that goes in the position in the
table where you see *37*.
38. Find the value that goes in the position in the
table where you see *38*.
Solving Simply Trigonometric Equations
The chart can be used to solve some simple equations.
ExampleFind all values of x between 0 and 2π such that
.
SolutionNote that multiples of have cosines equal to
or . Of these, the values that solve the equation
are and .
Practice
For questions 39 and 40, find the value(s) of x between
0 and 2π that satisfy the given equation.
39. .
40. cos(x) = –1.
sin( )x = − 32
x = 54πx = 3
4π
− 22
22
π4
cos( )x = − 22
–TRIGONOMETRY–
46
Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 46
LE
SS
ON
LIMITS AND CONTINUITY
The notion of a limit is the single most important underlying concept upon which calculus is built. We
can use the notion of a limit to describe the behavior of a function near a particular input, even when
the function is not defined there.
Limits can be illustrated using graphs and tables of values. For example, consider the function whose
graph is shown in Figure 5.1. We can’t talk about f(x) at x = 2 because of the unshaded circle on its graph.
But, we can talk about what happens close to 2. The values of the function at x-values close to 2 are listed in
the table.
5
47
x
)
Figure 5.1
x f(x)
1.9 5.39
1.99 5.0399
1.999 5.003999
1.9999 5.000399999
2 ???
↑ ↑2.0001 4.99959999
2.001 4.995999
2.01 4.9599
2.1 4.59
↑↑
9
–2–3 1 2 3–1
y
x
?
Calc2e_05_47-54.qxd 11/18/11 12:41 AM Page 47
The domain of f is . We can’t plug x � 2
into f. However, the hole appears to be at (2,5). How do
we know the hole has a y-value of 5? Well, the points
on the curve with x-values near x � 2 have
y-values close to y � 5. The closer we get to x � 2, the
closer the y-values of the points come to y � 5.
The mathematical shorthand for this is
= 5, which is read as “the limit as x approaches 2 of
is 5.”
The utility of limits lies in the fact that f need not
be defined at the value a in order to have a limit as x
approaches a.
We can also approach points from either the left
or from the right. For example, consider the graph of
in Figure 5.2.
Here, and .
The little minus in means that we approach
x � 1 using numbers less than (to the left) of x � 1. As
we approach x � 1 from the left-hand side, we slide up
the graph through y-values that approach 4. Similarly,
the plus in means “approach from the right.”
From the right, the height of the graph slides down to
y � 2 as x approaches 1.
In this example, does not exist because
there is no single y-value to which all of the points
near x � 1 get close. Some are close to 4, and others
are close to 2. Because there is no agreement, there is
no limit.
As another example, consider the graph of
in Figure 5.3. Here, because
sliding up to x � 3 from the left has us pass through
points with y-values near 2. Similarly, .
Because there is agreement from the left and right, we
have the general limit, . Notice that what
happens exactly at x � 3 is irrelevant. Here ,
but the resulting point at (3,5) has no bearing on the
limit as x approaches 3.
Vertical asymptotes correspond with infinite
limits. For example, consider the graph of in
Figure 5.4.
y � k1x 2
h13 2 � 5
limxS3
h1x 2 � 2
lim ( )x
h x→ +
=3
2
limxS3�
h1x 2 � 2y � h1x 2
limxS1
g1x 2
limxS1�
limx → −1
limxS1�
g1x 2 � 2limxS1�
g1x 2 � 4
y � g1x 2
f 1x 2
lim ( )x
f x→2
x � 2
–LIMITS AND CONTINUITY–
48
1
2
3
–1
–2–3 1 2 3–1
y
x
4
y = g(x)
Figure 5.2
1
2
3
4
5
6
–11 2 3–1
y
x4 5
y = h(x)
(4,1)
Figure 5.3
Calc2e_05_47-54.qxd 11/18/11 12:41 AM Page 48
Here, we write and k(x) =
. These statements simply suggest what the graph
is doing on either side of 2. The limits technically do
not exist since they are infinite.
Look back at the graphs in this lesson. Would you
agree with the following?
The difference from the limits discussed previ-
ously is that now, in each case, the function is contin-
uous at the point where we are computing the limit,
meaning that there are no holes, jumps, or asymptotes
there. Symbolically, we say f is continuous at x = a if
.
Practice
Use the graphs in Figure 5.5 to evaluate the following.
1.
2.
3.
4.
5. Is f continuous at x � �1?
6. limxS3�
f 1x 2
f 1�1 2
lim
xS �1 f 1x 2
limxS �1�
f 1x 2
limxS �1�
f 1x 2
lim ( ) lim ( ) ( )x a x a
f x f x f a→ →− +
= =
lim ( ) , lim ( ) , lim ( )x x x
f x g x h x→ →− →
= = =0 3 4
9 0 1
−∞
limx → +2
lim ( )x
k x→ −
= ∞2
–LIMITS AND CONTINUITY–
49
1
2
3
–1
–2–3 1 2 3–1
y
x
4
y = k(x)
Figure 5.4
1
2
3
4
–1–2 1 2 3 4 5–1
y
x
2
3
4
5
6
–11 2 3 4 5 6–1
–2
y
x
y = f (x)
y = g(x)
1
Figure 5.5
Calc2e_05_47-54.qxd 11/18/11 12:41 AM Page 49
7.
8.
9.
10. Is f continuous at x � 1?
11.
12.
13.
14.
15.
16.
Evaluating Limits Algebraically
It is not necessary to have the graph of a function to
evaluate its limits. For instance, if f is continuous at a,
then its limit as x approaches a is simply f (a).
Technically, this works only with functions that
are polynomials like , roots like ,
rational functions (formed by dividing two poly-
nomials) like , trigonometric functions,
and transcendental functions like , and .
Because this works for any combination of these func-
tions added, subtracted, multiplied, divided, or com-
posed, it works also for every function considered in
this book.
ExampleEvaluate and .
SolutionBecause 4 can be plugged into without
there being a division by zero, the limit
� . Similarly, �
.
Practice
Evaluate the following limits.
17.
18.
19.
20.
21.
22.
Dividing by a tiny number is equivalent to mul-
tiplying by an enormous number. For example:
It is for this reason that if the denominator of a frac-
tion approaches zero while the numerator goes to a
nonzero number, the result is an infinite limit.
A classic example is (graphed in Fig-
ure 5.6).
f 1x 2 �1x
5 �1
10,000 � 5 # 10,000
1 � 50,000
limx
xe→−
−
2
3 2
lim( )a
x a→
+ +0
2 1
limxSp
6
sin1x 2
x
limx
x
xx
→
−+
+
2
324
10 3
limxS3
x � 3x2 � x
lim( )x
x x x→
+ − +1
3 210 4 5 7
31�2 22 � 1�2 2 � 7 � 3
lim ( )x
x x→−
+ −2
23 74 � 516 � 40
�9
56
limxS4
x � 5
x2 � 10x
x � 5x2 � 10x
lim ( )x
x x→−
+ −2
23 7limxS4
x � 5
x2 � 10x
exln1x 2
3x � 52x3 � x2 � 1
x4x5 � 10x3 � 7
limxS5�
g1x 2
limxS5�
g1x 2
limxS3
g1x 2
limxS3�
g1x 2
g11 2
limxS1
g1x 2
f 13 2
lim
xS3 f 1x 2
limxS3�
f 1x 2
–LIMITS AND CONTINUITY–
50
Calc2e_05_47-54.qxd 11/18/11 12:41 AM Page 50
Because the denominator goes to zero while the
numerator stays one in all of these cases, there is a ver-
tical asymptote at x � 0. The function therefore
approaches either positive or negative infinity from
either side. When x is less than zero, as it always is when
, the function is also negative. Thus,
. Similarly, as , is always
positive, so . Finally, because the limit
from the two sides are different, the undirected limit
does not exist.
Example
Evaluate .
SolutionThe numerator approaches 5 while the denominator
approaches 0 as x approaches 2 from the right. There-
fore, this limit from the right is either ∞ or �∞. What
we need to determine is whether the function is posi-
tive or negative at x-values just slightly larger than 2.
We do this by looking at each factor individually.
As , the values we are plugging into each
factor are slightly larger than 2. So, (x + 3) and (x – 2)
are both positive, while (x – 4) is negative. Because the
function is made of two positive parts
and one negative part, the entire fraction will be neg-
ative. Thus, = –∞.
Because this is negative, the limit is �∞. Another
method will be covered in Lesson 13.
Example
Evaluate .
SolutionHere, the numerator approaches �10, which isn’t zero,
while the denominator approaches zero, so the limit is
either q or �q. As , the values we are plug-
ging into each factor are slightly smaller than –3. So,
(x + 1) and (x + 3) are both negative, while (2 – x) and
(x + 5) are both positive.
The combination of two negative factors and two
positive factors is positive, thus:
lim( )( )
( )( )x
x x
x x→− −
+ −+ +
= ∞3
1 2
3 5
x → − −3
limxS �3�
1x � 1 2 12 � x 2
1x � 3 2 1x � 5 2
lim( )( )x
xx x→ +
+− −2
32 4
xx x
+− −
32 4( )( )
x → +2
lim( )( )x
xx x→ +
+− −2
32 4
limxS0
1x
limx x→ +
= ∞0
1
1x
x S 0�limx x→ −
= − ∞0
1
1x
x S 0�
–LIMITS AND CONTINUITY–
51
1
2
3
4
5
6
–1–2–3–4–5–6 1 2 3 4 5 6–1
–2
–3–4
–5
–6
y
x
y = 1__x
Figure 5.6
Calc2e_05_47-54.qxd 11/18/11 12:41 AM Page 51
Practice
Evaluate the following limits.
23.
24.
25.
26.
27.
28.
When both the numerator and the denominator
go to zero, there are some common tricks for simpli-
fying the limit. The first is to factor and cancel. The
second is to rationalize. These are illustrated next.
Example
Evaluate .
SolutionHere, both the numerator and denominator go to zero,
so we aren’t guaranteed an infinite limit. First, factor
the numerator and denominator.
Because as , we can cancel .
Now we can plug in without dividing by zero.
The following example utilizes the trick of
rationalizing.
Example
Evaluate .
SolutionBecause both numerator and denominator go to zero,
a trick is necessary. First, multiply the top and bottom
by the part with the square root, but with the opposite
sign between them.
Simplify.
= lim( )x
x
x x→
−− +( )9
9
9 3
lim lim( )( )x x
x
x
x x x
x x→ →
−−
= + − −− +9 9
3
9
3 3 9
9 3
lim limx x
x
x
x
x
x
x→ →
−−
= −−
++
⋅
9 9
3
9
3
9
3
3
limx
x
x→
−−9
3
9
limxS4
1x � 2 2
1x � 5 2�
69
�23
limxS4
x2 � 2x � 8x2 � x � 20
�
lim
xS4 1x � 4 2 1x � 2 2
1x � 4 2 1x � 5 2� lim
xS4 1x � 2 2
1x � 5 2
limxS4
x2 � 2x � 8x2 � x � 20
�
x � 4x � 4
� 1x S 4x � 4
limxS4
x2 � 2x � 8x2 � x � 20
� limxS4
1x � 4 2 1x � 2 2
1x � 4 2 1x � 5 2
limxS4
x2 � 2x � 8x2 � x � 20
limxS �5
x � 21x � 5 22
limxS2
1x � 5 2 1x � 5 2
1x � 3 2 1x � 4 2
limxS3�
1x � 2 2 1x � 5 2
1x � 6 2 1x � 3 2
limx → −3
limxS4�
x � 5x � 4
limx x→ − −1
1
1
–LIMITS AND CONTINUITY–
52
When computing a limit as x goes to a, if plug-ging in a results in 0
0, do NOT automatically con-clude the limit is 1. This means you must simplifythe expression somehow before plugging in a.
Calc2e_05_47-54.qxd 11/18/11 12:41 AM Page 52
Eliminate .
Plug in.
Factoring tricks can even be useful when dealing with
transcendental functions.
ExampleEvaluate
SolutionTo compute this limit, use the fact that and
factor the denominator as a difference of squares.
Then, cancel factors that are common to both the
numerator and denominator, and substitute x = 0 into
the simplified expression, as follows:
Practice
Evaluate the following limits.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40. limln( )z
z z
z
e e
e→
( ) − ( ) +
−2
26 8
2
limx
x x xx→
+ −−3
3 2
2
2 159
limh
x h x
h→
+( ) −0
3 32 2
limcos( )cos( )x
xx→
+−π
11
lima
a
x a x→ + −0
limaS0
1x � a 22 � x2
a
lim(x
x
x x→
−−25
5
2
5)( + 1)
limxS3�
x � 5x � 3
limxS3
x2 � 4x � 3
x2 � 2x � 15
limxS4
x2 � 9x � 3
limxS2
x � 2x2 � 4
1x � 6 2 1x � 2 2
1x � 2 2 1x � 1 2lim
xS �2�
limx
xe e→ + +=
+=
+=
00
11
11
11 1
12
lim lim limx
x
xx
x
x x
ee
e
e→ → →+ + +−−
= −− ( )
=0
20
20
11
1
1
e ex x22
= ( )
limx
x
x
ee→ +
−−0 2
11
lim lim( )x x
x
x x→ →
−−
=+
=9 9
3
9
1
3
1
6
lim lim( )
( )( )x x
x
x
x
x x→ →
−−
= −− +9 9
3
9
9
9 3
x � 9x � 9
� 1
–LIMITS AND CONTINUITY–
53
1
1 1
−−( ) +( )
e
e e
x
x x
Calc2e_05_47-54.qxd 11/18/11 12:41 AM Page 53
Calc2e_05_47-54.qxd 11/18/11 12:41 AM Page 54
LE
SS
ON
DERIVATIVES
Straight lines are convenient to deal with, but most functions have curved graphs. This does not stop
us from projecting straight lines on them! For example, at the point marked x on the graph in Figure
6.1, the function is clearly increasing. However, exactly how fast is the function increasing at that point?
Since “how fast” refers to a slope, we draw in the tangent line, the line straight through the point that heads
in the same direction as the curve (see Figure 6.2). The slope of the tangent line tells us how fast the func-
tion is increasing at the given point.
6
55
x
y = f(x)
Figure 6.1
Calc2e_06_55-60.qxd 11/18/11 12:43 AM Page 55
We can figure out the y-value of this point by
plugging x into f and getting . However, we
can’t get the slope of the tangent line when we have just
one point. To get a second point, we go ahead a little
further along the graph (see Figure 6.3). If we go ahead
by distance a, the second point will have an x-value of
and a y-value of .
Because this second point is on the curve and not
on the tangent line, we get a line that is not quite the
tangent line. Still, its slope will be close to the one we
want, so we calculate as follows:
slope �
To make things more accurate, we pick a second
point that is closer to the original point (x,f (x)) by
using a smaller a. This is depicted in Figure 6.4.
In fact, if we take the limit as a goes to zero from
the right, we will get the slope of the tangent line
exactly. The situation is completely similar if we take a
< 0 so that x + a is to the left of x. This process gives us
the derivative of and is written .
� slope of the tangent line at point
ExampleWhat is the derivative of ?
SolutionStart with the definition of the derivative.
Use .
f ¿ 1x 2 � limaS0
1x � a 22 � x2
a
f 1x 2 � x2
f ¿ 1x 2 � limaS0
f 1x � a 2 � f 1x 2
a
f 1x 2 � x2
1x,f 1x 2 2
f˛˛¿ 1x 2 � limaS0
f 1x � a 2 � f 1x 2
a
f ¿ 1x 2f 1x 2
f 1x � a 2 � f 1x 2
1x � a 2 � x�
f 1x � a 2 � f 1x 2
a
f 1x � a 2x � a
1x,f ˛1x 2 2
–DERIVATIVES–
56
x
y = f(x)tangentline
Figure 6.2
x
y = f(x)tangentline
x + a
(x, f(x))
(x + a, f(x + a))
not quite thetangent line
Figure 6.3
x
y = f(x)tangentline
x + a
(x, f(x))
(x + a, f(x + a))
closer to thetangent line
Figure 6.4
Remember, f (x + a) ≠ f (x) + f (a).
Calc2e_06_55-60.qxd 11/18/11 12:43 AM Page 56
Multiply out and simplify.
Factor and simplify.
Evaluate the limit.
The derivative is . This means that
the slope at any point on the curve is exactly
twice the x-coordinate. The situation at x � �2,
x � 0, and x � 1 is shown in Figure 6.5.
ExampleWhat is the slope of the line tangent to g(x) =
at x � 9?
SolutionStart with the definition of the derivative.
Use g(x) = .
Rationalize the numerator.
g′(x) =
Multiply and simplify.
g′(x) =
Simplify.
Plug in to evaluate the limit.
The derivative of g(x) = is thus
g′(x) . This means that at x � 9, the slope of
the tangent line is g′(9) . This is illus-
trated in Figure 6.6.
ExampleFind the equation of the tangent line to
at x � 3.h1x 2 � 2x2 � 5x � 1
= = 1
2 9
1
6
= 1
2 x
x
= 1
2 x
=+ +
1
0x x
1
x a x + +g¿ 1x 2 � lim
aS0
′ = // + +→
g xa
a x a xa( ) lim
( )
0
lim( )a
x a x x a x x a x
a x a x→
+ + ⋅ + − ⋅ + −+ +0
lima
x a x
a
x a x
x a x→
+ −
+ ++ +
0
′ = + −→
g xx a x
aa( ) lim
0
x
g¿ 1x 2 � limaS0
g1x � a 2 � g1x 2
a
x
y � x2
f ¿ 1x 2 � 2x
f ¿ 1x 2 � limaS0
2x � a � 2x
f ¿ 1x 2 � limaS0
12x � a 2a
a
f ¿ 1x 2 � limaS0
x2 � 2ax � a2 � x2
a
–DERIVATIVES–
57
1
2
3
–1
–2–3 1 2 3–1
y
x
4at (–2,4)
slope = –4
slope = 0at (0,0)
slope = 2at (1,1)
y = x2
Figure 6.5
Calc2e_06_55-60.qxd 11/18/11 12:43 AM Page 57
SolutionTo find the equation of the tangent line, we need a
point and a slope. The y-value at x � 3 is
, so the point is (3,4).
And to get the slope, we need the derivative. Start with
the definition of the derivative.
Thus, the derivative of is
. The slope at x � 3 is
. The equation of the tangent
line is therefore . This
is shown in Figure 6.7.
y � 71x � 3 2 � 4 � 7x � 17
h¿ 13 2 � 413 2 � 5 � 7
h¿ 1x 2 � 4x � 5
h1x 2 � 2x2 � 5x � 1
h¿ 1x 2 � limaS0
h1x � a 2 � h1x 2
a
h13 2 � 213 22 � 513 2 � 1 � 4
–DERIVATIVES–
58
1
2
3
1 2 3 4 5 6
y
x7 8 9 10
xy = g(x) =
slope = 1-6
at (9,3)
Figure 6.6
Use .
Multiply out and simplify.
Factor out and simplify.
Evaluate the limit.
′ = + − = −→
h x x a xa
( ) lim( )0
4 2 5 4 5
14x � 2a � 5 2a
ah¿ 1x 2 � lim
aS0
2x2 � 4ax � 2a2 � 5x � 5a � 1 � 2x2 � 5x � 1
ah¿ 1x 2 � lim
aS0
21x � a 22 � 51x � a 2 � 1 � 12x2 � 5x � 1 2
ah¿ 1x 2 � lim
aS0
h1x 2 � 2x2 � 5x � 1
Calc2e_06_55-60.qxd 11/18/11 12:43 AM Page 58
Practice
1. Find the derivative of .
2. If , then what is ?
3. Find the derivative of .
4. Find the derivative of f(x) = 3 .
5. If , then what is
6. Find the slope of at x � 2.
7. Find the slope of g(x) = at x � 16.
8. Where does the graph of
have a slope of 0?
9. Find the equation of the tangent line to
at (2,�3).
10. Find the equation of the tangent line to
when x � 1.
Derivatives Don’t Always Exist!
Because the definition of involves a full-blown
limit in order for it to exist, the following left- and
right-hand limits must be equal:
This is not always the case, though.
ExampleLet us try to compute the derivative of A(x) = |x| at
x = 0 (shown in Figure 6.8).
lim( ) ( )
lim( ) ( )
a a
f x a f xa
f x a f xa→ →− +
+ − = + −0 0
′f x( )
k1x 2 � 5x2 � 2x
h1x 2 � 1 � x2
g1x 2 � x2 � 4x � 1
3 x
f 1x 2 � 3x2 � x
k¿ 1x 2?k1x 2 � x3
x
g1x 2 � 10
h¿ 1x 2h1x 2 � x2 � 5
f 1x 2 � 8x � 2
59
–2 1 2–1
y
x
1
2
3
4
5
6
7
8
9
10
3 4–1
–2
(3,4)
h(x) = 2x2 – 5x + 1
y = 7x – 17
Figure 6.7
The absolute value of x, denoted x , tells you how far x is from zero. For instance, 5 = 5 (since 5 is five unitsfrom 0 on the right-hand side) and |–4| = 4 (since –4 is four units from 0, just on the left-hand side). Symboli-cally,
x =x, whenever x ≥ 0
–x, whenever x < 0
ABSOLUTE VALUE
Calc2e_06_55-60.qxd 11/18/11 12:43 AM Page 59
For a > 0, we have
So, .
Now, let us take a < 0. Doing so, we get:
So, .
But, the left- and right-hand limits aren’t equal. So,
the full-blown limit does not
exist. Thus, m′(0) doesn’t exist. Geometrically, the
sharp corner in the graph of y = m(x) at x = 0 is the rea-
son the derivative doesn’t exist there.
lim( ) ( )
a
m a ma→
+ −0
0 0
lim( ) ( )
lim ( )a a
m a ma→ →− −
+ − = − = −0 0
0 01 1
m a ma
aa
aa
( ) ( )0 0 01
+ − =−
= − = −
lim( ) ( )
lima a
m a ma→ →+ +
+ − = =0 0
0 01 1
m a ma
aa
aa
( ) ( )0 0 01
+ − =−
= =
–a ac
y
x
slope–1
m(x) = x
slope1
–DERIVATIVES–
60
Calc2e_06_55-60.qxd 11/18/11 12:43 AM Page 60
LE
SS
ON
BASIC RULES OF DIFFERENTIATION
Using the limit definition to find derivatives can be very tedious. Luckily, there are many shortcuts
available. For example, if function f is a constant, like or , then . This
can be proven for all constants c at the same time in the following manner.
If:
then:
All of the general rules in this chapter can be proven in such a manner, using the limit definition of the deriv-
ative, though we shall not actually do so. The first rule is the Constant Rule, which says that if where
c is a constant, then .
Before we go any further, a word needs to be said about notation. The concept of the derivative was dis-
covered by both Isaac Newton and Gottfried Leibniz. Newton would put a dot over a quantity to represent
its derivative, much like we have used the prime notation to represent the derivative of . Leibniz
would write the derivative of y (where x is the variable) as . Newton’s notation is certainly moredy
dx
f 1x 2f ¿ 1x 2
f ¿ 1x 2 � 0
f 1x 2 � c
f ¿ 1x 2 � lim
aS0 f 1x � a 2 � f 1x 2
a� lim
aS0 c � c
a� lim
aS0 0a
� 0
f 1x 2 � c
f ¿ 1x 2 � 0f 1x 2 � 18f 1x 2 � 5
7
61
Calc2e_07_61-66.qxd 11/18/11 12:44 AM Page 61
convenient, but Leibniz’s enables us to represent
“take the derivative of something” as (something).
Thus, if , then .
Using Leibniz’s notation, the Constant Rule where c is
a constant is expressed as .
The next rule is the Power Rule, which is stated:
. This rule says “multiply the expo-
nent in front and then subtract one from it.”
ExampleDifferentiate .
Solution
ExampleDifferentiate .
Solution
ExampleDifferentiate .
SolutionTo use the Power Rule, we need expressed as x
raised to a power, or:
Notice how much easier it is to use the Power
Rule to compute the derivative than it was using the
limit definition of the derivative in Lesson 6.
′ = = = =− − ⋅g x x x
x x( )
1
2
1
2
1
2
1
2
12
12
1 1
g1x 2 � x12
g1x 2
g x x( ) =
dy
dxx x = =−8 88 1 7
y � x8
f ¿ 1x 2 � 2x2�1 � 2x1 � 2x
f 1x 2 � x2
ddx1xn 2 � n # xn�1
ddx1c 2 � 0
ddx1f 1x 2 2 � f ¿ 1x 2
dy
dx �y � f 1x 2
ddx
62
CONSTANT RULE
If f (x) = c where c is a constant, then f ′(x) = 0.
And, using Leibniz’s notation, if c is a constant, then .ddx
c( ) = 0
POWER RULE
ddx
x n xn n( ) = ⋅ −1
Calc2e_07_61-66.qxd 11/18/11 12:44 AM Page 62
Example
Differentiate .
SolutionFirst, rewrite y as so that it becomes x raised to a
power. Then,
Notice that means “take the derivative with
respect to variable t.” While x is often used as the vari-
able, so the derivative of is ,
sometimes it is convenient to use other variables. If
, then is the derivative of f with
respect to u, for example.
ExampleDifferentiate .
Solution
ExampleDifferentiate .
SolutionRewrite using the exponent rules.
Now, use the Power Rule to differentiate.
Practice
Differentiate each of the following.
1.
2. y =
3.
4.
5. y = t12
6.
7.
8.
9.
10.
11.
12.
13. f xx
( ) 1=
y �1x
y u =
k1x 2 � 24 x
g1x 2 � x�45
f 1t 2 � �11
f 1x 2 � x100
y � x 75
h1x 2 � 8
g1u 2 � u�5
x21
f 1x 2 � x5
dydt
t= − −52
72
yt
ttt
t t t= = = ⋅ =− −3 3
3
12 1
252
yt
t=
3
ddt123 t 2 �
ddt1t
13 2 �
13
t 13 �1 �
13
t�23 �
1
3t 23
y t= 3
dy
du� f ¿ 1u 2y � f 1u 2
dy
dx� f ¿ 1x 2y � f 1x 2
ddt
�ddx1x�2 2 � �2x�2�1 � �2x�3 �
�2x3
dy
dx�
ddxa
1x2b
x�2
y �1x2
63
THE CONSTANT COEFFICIENT RULE
If a function has a constant multiplied in front, leave it while you take the derivative of the rest.Using Leibniz’s notation, , where c is a constant.d
dxcf x c
ddx
f x( ( )) ( ( )) =
Calc2e_07_61-66.qxd 11/18/11 12:44 AM Page 63
–BASIC RULES OF DIFFERENTIATION–
14.
15.
16.
The Constant Coefficient Rule
The Constant Coefficient Rule is stated as follows: If a
function has a constant multiplied in front, leave it
while you take the derivative of the rest. This means
that because , the derivative of
would be 5 • (8x7) = 40x7. Just imagine that the constant
steps aside and waits while you differentiate the rest.
ExamplesDifferentiate the following.
Solutions
f ′(x) = 11 ⋅ (4x3) = 44x3
(1) = 12
In that last example problem, don’t forget that p
is a constant, and thus should be treated just as
you would or .
The Additive Rule
Next, we will examine the Additive Rule, which says
that if parts of a function are added together, dif-
ferentiate the parts separately and add the results. We
know that and . The
Additive Rule then says that if , then
.
ExampleDifferentiate .
Solution′ = + = +f x
ddx
xddx
x x x( ) ( ) ( )4 30 20 605 2 4
f 1x 2 � 4x5 � 30x2
20x � 12
ddx110x2 2 �
ddx112x 2 �
dy
dx�
ddx110x2 � 12x 2 �
y � 10x2 � 12x
ddx112x 2 � 12
ddx110x2 2 � 20x
712r20r
2p r
′ = ⋅ =A r r r( ) ( )π π2 2
′ = ⋅
= =− −k u u u
u( )
154
13
54
5
4
23
23
23
dy
dx� 12
′ = ⋅ −( ) = − = −− −h t t tt
( ) 4 6 24247 7
7
′ = ⋅
= =− −g x x x
x( ) 3
12
32
3
2
12
12
dy
dxx x= ⋅ =1 20 (2 0)
A1r 2 � p r2
k1u 2 �1523 u
4�
154
u13
y � 12x
h tt
t( ) = = −44
66
g x x x( ) = =3 312
y x= 10 2
f x x( ) = 11 4
5x8ddx1x8 2 � 8x7
yt
t=
3
4
h ttt
( ) = −
5
2
g xx x
( ) = 1
64
Don’t make the mistake that
.ddx
c f xddx
cddx
f x( ( )) ( ) ( ( ))
⋅ = ⋅ =
= 0
0123
Calc2e_07_61-66.qxd 11/18/11 12:44 AM Page 64
ExampleDifferentiate .
SolutionThis can be rewritten as a sum:
g(x) = x3 + (–4)x2
thus:
The previous example shows that the Additive
Rule applies to cases of subtraction as well.
ExamplesDifferentiate the following.
Solutions
Practice
Differentiate the following.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29. y � 3 �2x
�1x2
h1u 2 � u5 � 4u4 � 7u3 � 2u2 � 8u � 2
g1x 2 � 3x15 � 5x3
F1x 2 � 6x100 � 10x50 � 4x25 � 2x10 � 9
s t t et t( ) ln( )= + + +π 3 2 3 3
y x x x= − −− −4 32
f 1x 2 � 8x3 � 3x2
4 3 703 2t t− +
k1x 2 � 1 � x2
g1t 2 �12t4
5
V1r 2 � 43p r3
f 1x 2 ��3x10
y x= 6 7
′ = − + = −− −k t t t
t t( )
125
2 012
5
215 2
5 2
h¿ 1x 2 � 40x4 � 40x3 � 9x2 � 14x � 5
dy
dx x x
1
2 +
1
2= =0
k t tt
e t t e( ) = + + = + +−32
3 245
45 1
h1x 2 � 8x5 � 10x4 � 3x3 � 7x2 � 5x � 4
y x x + + = =4 412
′ = ( ) + −( ) = + − ⋅
= −
g xddx
xddx
x x x
x x
( ) ( )
.
3 2 2
2
4 3 4 2
3 8
g1x 2 � x3 � 4x2
65
THE ADDITIVE RULE
If parts of a function are added together, differentiate the parts separately. Then, add the results. Using Leibniz’s notation, .
ddx
f x g xddx
f xddx
g x( ( ) ( )) ( ( )) ( ( )) + = +
Remember, e is a number (approximately equal to2.71828…).
Calc2e_07_61-66.qxd 11/18/11 12:44 AM Page 65
30.
31.
32.
The derivative of the derivative is called the sec-
ond derivative. The derivative of that is the third deriv-
ative, and so on. Using notation, if y = f(x), then the
derivative is , the second derivative
is , the third derivative is ,
and the tenth derivative, for example, is .
We put the 10 in parentheses because counting the ten
primes in is ridiculous.
ExampleFind the first three derivatives of y = .
Solution
When working on multiple derivatives like this,
it makes sense to leave the exponents negative and
fractional until all derivatives have been computed.
ExampleFind all the derivatives of .
Solution
All of the subsequent derivatives will also be zero, so
we can write
for .
Practice
33. Find the first four derivatives of .
34. Find the second derivative of s(t) = – 16t 2 +
at + b, where a and b are constants.
35. Find the third derivative of
.
36. Find the first three derivatives of .y � 623 t
y x x x= − +− − −3 23 2 1
f 1x 2 �1x
n � 4f ˛
1n2 1x 2 � 0
f 1x 2 � 0
f ‡ 1x 2 � 6
f – 1x 2 � 6x � 8
f ¿ 1x 2 � 3x2 � 8x � 5
f 1x 2 � x3 � 4x2 � 5x � 7
f 1x 2 � x3 � 4x2 � 5x � 7
d 3y
dx3 �38
x�52
d 2y
dx2 � �14
x�32
dy
dx�
12
x�12
y � x 12
x
f –1x 2
d10y
dx10 � f 1102 1x 2
d3y
dx3 � f ˛‡ 1x 2d2y
dx2 � f – 1x 2
dy
dx� f ¿ 1x 2
f xx x x
( ) = −−3
25
4
y x x 4 + = =9 93
y � u2 � u�2
–BASIC RULES OF DIFFERENTIATION–
66
Calc2e_07_61-66.qxd 11/18/11 12:44 AM Page 66
LE
SS
ON
RATES OF CHANGE
It is useful to contemplate slopes in practical situations. For example, suppose the following graph in Fig-
ure 8.1 is for , a function that gives the price y for various amounts x of cheese. Because the
straight line goes through the points (1 lb.,$2) and (2 lbs.,$4), the slope � = = $2 per
pound.
$21 lb.
$ $4 22 1
−− lbs. lb.
y � f 1x 2
8
67
1
2
3
4
5
1 2 3
y
y = f(x)
Amount (in pounds)
(in dollars)
Costs of Cheese
x
6
Price
Figure 8.1
Calc2e_08_67-74.qxd 11/18/11 12:45 AM Page 67
The slope is therefore the rate at which the
price of cheese changes per pound. Because slope �
, a slope will always be a rate measured in
y-units per x-unit.
For example, suppose a passenger on a bus writes
down the exact time she passes each highway mile
marker. She then sketches the graph shown in Figure
8.2 of the bus’s position on the highway over time. The
slope at any point on this graph will be measured in y-
units per t-unit, or miles per hour. The steepness of the
slope represents the speed of the bus.
Practice
For each of the following four graphs, describe the rate
that a slope of the curve represents.
1.
2.
–RATES OF CHANGE–
68
y
y = s(t)
50
100
150
200
250
300
Posi
tion
Giv
en b
y M
arke
r (i
n m
iles)
Time on Bus (in hours)
Mile Markers
t
1 2 3
Figure 8.2
y
Time Worked (in hours)
t
1 2
Money Earned
Pay
(in
dol
lars
)
10
20
30
40
50
60
3 4 5 6 7 8 9 10
y
1 2
Gasoline Use
10
20
30
40
50
3 4
60
x
5
Dis
tan
ce D
rive
n (
in m
iles)
Gasoline Used (in gallons)
y-change
x-change
Calc2e_08_67-74.qxd 11/18/11 12:45 AM Page 68
3.
4.
Because the derivative of a function gives the
slope of its tangent lines and the tangent line indicates
how fast the function is increasing or decreasing, these
practice problems show that the derivative of a func-
tion gives its rate of change.
An excellent example comes from position func-
tions. A position function states the position of an
object along a straight line at any given time. The
derivative states the rate at which that object’s
position is changing—that is, the velocity of the func-
tion. Thus, s ′(t) = v(t). The second derivative
tells how fast the velocity is changing, or
the acceleration. Thus, s ′′(t) = v ′(t) = a(t) where s(t) is
the position function, v(t) is the velocity function, and
a(t) is the acceleration function.
ExampleSuppose an object rolls along beside a tape measure so
that after t seconds, it is next to the inch marked
. Where is the object after 1 sec-
ond? After 3 seconds? What is the velocity function?
How fast is the object moving after 2 seconds? What is
the acceleration function?
SolutionThe position function tells us
where the object is. After 1 second, the object is next to
the s(1) = 17-inch mark on the tape measure. After 3
seconds, the object is at the -inch mark.
The velocity function is .
Thus, after 2 seconds, the object is moving at the rate
of inches per second. Realize that this
velocity of 24 inches per second is an instantaneous
velocity, the speed just at a single moment. If a car’s
speedometer reads 60 miles per hour, this does not
mean that it will drive for 60 miles or even for a full
hour. The car might speed up, slow down, or stop.
However, at that instant, the car is traveling at a rate
that, if unchanged, will take it 60 miles in one hour. A
derivative is always an instantaneous rate, telling you
v12 2 � 24
v1t 2 � s¿ 1t 2 � 8t � 8
s13 2 � 65
s1t 2 � 4t2 � 8t � 5
s1t 2 � 4t2 � 8t � 5
s– 1t 2 � v¿ 1t 2
s¿ 1t 2
s1t 2
–RATES OF CHANGE–
69
yGrowth of a Baby
10
20
30
x
Age (in months)
40
Wei
ght
(in
pou
nds
)
6 12 18 24 30 36
Time (in hours)
Size of Snowball(on a 40°F day)
5
y
t
Dia
met
er(i
n in
ches
)
Calc2e_08_67-74.qxd 11/18/11 12:45 AM Page 69
the slope at a particular point, but not making any
promises about what will happen next.
The acceleration function is
. Because this is a constant, it tells us that
the object increases in speed by 8 inches per second
every second.
The most popular example of constant accelera-
tion is gravity, which accelerates objects downward
by every second. Because of this, an object
dropped with an initial velocity of b feet per second
from a height of h feet above the ground will have (after
t seconds) a height of feet.
The starting time is , at which point the
object is feet off the ground, the correct
initial height. The velocity function is
. At the starting time t � 0, the velocity is
, the desired initial velocity. The function
means that 32 feet per second are
subtracted from the initial velocity b every second. The
acceleration function is .
This is the desired constant acceleration.
Note: A negative velocity means the object is moving
backward (or in the direction of decreasing y-value).
The speed is the absolute value of the velocity.
ExampleSuppose a brick is thrown straight upward with an ini-
tial speed of from a 150-foot rooftop. What are
its position, velocity, and acceleration functions?
Solution
Because the initial velocity is and the ini-
tial height is h � 150 feet, the position function is s(t)
= –16t 2 + 10t + 150. The velocity function is b(t) =
s′(t) = –32t + 10. The acceleration is a(t) = –32, a con-
stant 32 feet per second downward each second. The
negative sign indicates that gravity is acting to decrease
the height of the brick, pulling it downward.
ExampleSuppose a rock is dropped from a 144-foot tall bridge.
When will the rock hit the water? How fast will it be
going then?
SolutionBecause the rock is dropped, the initial velocity is
b � 0. The initial height is h � 144. Thus,
gives the height function. The
rock will hit the water (have a height of zero) when:
And because �3 seconds doesn’t make any sense, the
rock will hit the water after 3 seconds.
The velocity function is ;
therefore, the rock will have a velocity of
after 3 seconds. This means that it will be traveling at
a rate of 96 feet per second downward when it hits
the water.
Example
If gives the value, in
thousands of dollars, of a start-up company after t
days, then how fast is its value changing after 30 days?
After 500 days?
p1t 2 �t2
10� 80t � 50,000
v13 2 � �96
v1t 2 � s¿ 1t 2 � �32t
t � ;3
144 � 16t2
�16t2 � 144 � 0
s1t 2 � �16t2 � 144
b � 10ft
sec
10ft
sec
s– 1t 2 � �32a1t 2 � v ¿ 1t 2 �
v1t 2 � �32t � b
v10 2 � b
� �32t � b
� s¿ 1t 2v1t 2
s10 2 � h
t � 0
s1t 2 � �16t2 � bt � h
32ft
sec
s– 1t 2 � 8
v¿ 1t 2 �a1t 2 �
–RATES OF CHANGE–
70
Calc2e_08_67-74.qxd 11/18/11 12:45 AM Page 70
Solution
The derivative gives the rate of
change in value, measured in thousands of dollars per
day. After 30 days, , so the company
will be losing value at a rate of $74,000 per day. After
500 days, , so the company will be gain-
ing value at the instantaneous rate of $20,000 a day.
Practice
5. The height of a tree after t years is h(t) = 30 –
feet when . How fast is the tree growing
after 5 years?
6. The level of a river t days after a heavy
rainstorm is feet. How
fast is the river’s level changing after 7 days?
7. When a company makes and sells x cars, its
profit is dollars.
How fast is its profit changing when the com-
pany makes 50 cars?
8. When a container is made x inches wide, it costs
dollars to make. How is the
cost changing when x � 3 inches?
9. An electron in a particle accelerator is
meters from the start
after t seconds. Where is it after 3 seconds? How
fast is it moving then? How fast is it accelerating
then?
10. A brick is dropped from 64 feet above the
ground. What is its position function? What is its
velocity function? What is its acceleration? When
will it hit the ground? How fast will it be travel-
ing then?
11. A bullet is fired directly upward at 800 feet per
second from the ground. How high is it when it
stops rising and starts to fall?
12. A rock is thrown 10 feet per second down a
1,000-foot cliff. How far has it gone down in the
first 4 seconds? How fast is it traveling then?
Derivatives of Sine and Cosine
Examining rates and slopes at various points can help
us determine the derivative of . Look at the
slopes at various points on its graph in Figure 8.2. It
appears that the derivative function of must
oscillate between �1 and 1, and must go through the
following points (see Figure 8.3). The function
is exactly such an oscillating function (see Figure 8.4).
This suggests that .
A similar study of the slopes of would
show that . The slopes of the
cosine function are not the values of the sine function,
but rather their exact negatives. These two results
could be obtained using the limit definition of the
derivative, but involve the use of trigonometric
identities and certain limit formulas not covered in
this book.
ExamplesDifferentiate the following
y � 2 � cos1t 2
f 1x 2 � 5sin1x 2 � 4x2
ddx1cos1x 2 2 � �sin1x 2
cos1x 2
ddx1sin1x 2 2 � cos1x 2
cos1x 2
sin1x 2
sin1x 2
s1t 2 � t3 � 2t2 � 10t
C1x 2 � 0.8x2 �24x
P xx
x x( ) , = − +3
2
1060 9 000
L1t 2 � �t2 � 8t � 26
t � 125t
p¿ 1500 2 � 20
p¿ 130 2 � �74
p¿ 1t 2 �t5
� 80
–RATES OF CHANGE–
71
, not cos(c), when c is a
constant. Similarly, .ddx
ccos( ) = 0
ddx
csin( ) = 0
Calc2e_08_67-74.qxd 11/18/11 12:45 AM Page 71
–RATES OF CHANGE–
72
1
–1
5π–
at x =slope = 0 5π–2
at x =slope = 1
2π
slope = 03π–2
at x =
2π3π–2slope = –1
at x = π
π π2–slope = 1
at x = 0–π
2–
slope = 0at x = – π
2–
y = sin(x) slope = 0at x =π2–
2
Figure 8.2
,0)(3π–2
�1
π2
π– 3π–22π–π
2–5π–2
(–π2–,0) ( ,0)
( ,�1)
,0)(π2–
π
2π( ,1)
5π–2
y =ddx
(sin(x)) = slopes of sin(x)1 (0,1)
Figure 8.3
π2–
�1
1
π2
π– 3π–
22π 5π–
2
y = cos(x)
Figure 8.4
Calc2e_08_67-74.qxd 11/18/11 12:45 AM Page 72
Solutions
Practice
For questions 13 through 17, compute the derivative.
13.
14.
15. g(x) =
16.
17.
18. Find the equation of the tangent line to
at x = .
Derivatives of the Exponential and Natural
Logarithm Functions
The reason why the nicest exponential function is
where e � 2.71828 . . . is because this makes for the
following very nice derivative:
It is only with this exact base that the derivative of the
exponential function is itself (see Figure 8.5).
The derivative of its inverse function is as
follows:
A proof of this formula is given in Lesson 11.
ddx1 ln1x 2 2 �
1x
ln1x 2
ddx1ex 2 � ex
ex
π2
f 1x 2 � sin1x 2 � cos1x 2
h1x 2 � cos1x 2 � cos15 2
r( ) sin( ) cos( )θ θ θ = +1
2
1
2
352
3x
xx− + − πcos( )
f 1t 2 � 3sin1t 2 �2t
y � 4x5 � 10cos1x 2 � 3
g¿ 1x 2 � cos1x 2 � sin1x 2
dy
dt� �sin1t 2
f ¿ 1x 2 � 5cos1x 2 � 8x
g x x x( ) sin( ) cos( ) sin= − +
π6
–RATES OF CHANGE–
73
1
2
3
4
5
6
–1
–2 1 2 3–1
y
x
7
8
9
y = ex
(1,e)
(2,e2)at
slope = e2
atslope = e
atslope = 1
at
slope =
(0,1)–1, 1—e1—e
Figure 8.5
Calc2e_08_67-74.qxd 11/18/11 12:45 AM Page 73
ExamplesDifferentiate the following.
Solutions
Practice
For questions 19 through 23, compute the derivative.
19.
20.
21.
22.
23.
24. Find the second derivative of
.
25. Find the 100th derivative of .
26. What is the slope of the tangent line to
at x � 10?f 1x 2 � ln1x 2
g1x 2 � 3ex
f 1x 2 � ex � ln1x 2
k x x e x( ) ln( )= + +3 552 π
h x x x e( ) ln( )= − + −8 3
y � cos1x 2 � 10ex � 8x
g1t 2 � 12ln1t 2 � t2 � 4
f 1x 2 � 1 � x � x2 � x3 � ex
dydx
= 0
dydu u
e eu= − +8
g ¿ 1t 2 � 3et �2t
dy
dx� 10ex
f ¿ 1x 2 � 4ex
y u e eu
y e
u= − +
= +
8
53
ln( )
ln( )
g1t 2 � 3et � 2ln1t 2
y � 10ex � 10
f 1x 2 � 4ex
–RATES OF CHANGE–
74
ex (e raised to the power x) and ex (the numbere times x) are very different functions. Note that
, whereas .ddx
ex e( ) =ddx
e ex x( ) =
and whenever c is
a positive constant.
ddx
c(ln( )) = 0ddx
ec( ) = 0
Calc2e_08_67-74.qxd 11/18/11 12:45 AM Page 74
The Product Rule
When a function consists of parts that are added together, it is easy to take its derivative: Simply take the deriv-
ative of each part and add them together. We are inclined to try the same trick when the parts are multiplied
together, but it does not work.
For example, we know that and . The derivative of their product is
. This shows that the derivative of a product is not the product of the derivatives:
Instead, we take the derivative of each part, multiply by the other part left alone, and add these results together:
This time, we did get the correct answer.
ddx1x2 # x3 2 �
ddx1x2 2 # x3 �
ddx1x3 2 # x2 � 12x 2 # x3 � 13x2 2 # x2 � 5x4
5x4 �ddx1x2 # x3 2 �
ddx1x2 2 # d
dx1x3 2 � 12x 2 # 13x2 2 � 6x3
ddx1x2 # x3 2 �
ddx1x5 2 � 5x4
ddx1x3 2 � 3x2d
dx1x2 2 � 2x
LE
SS
ON
THE PRODUCT AND QUOTIENT RULES9
75
Calc2e_09_75-80.qxd 11/18/11 12:47 AM Page 75
ExampleDifferentiate .
SolutionHere, the “first part” is and the “second part” is
. Thus, by using the Product Rule,
�
. This could be simplified as
.
ExampleDifferentiate .
Solution
Thus, the derivative is:
Example
Differentiate .
Solution
Using the product rule with can be a little bit
confusing because there is no difference between the
derivative of and “left alone.” Still, if you write
everything out, the correct answer should fall into
place, even if it looks weird.
ExampleDifferentiate y = .
Solution
ExampleDifferentiate .
SolutionWe’ll use the Product Rule with as the first part and
as the second part. However, in takingsin1x 2cos1x 2
x5
y � x5sin1x 2cos1x 2
dydt
ddt
t tddt
t t
t tt
t
tt
t
t
t
=
⋅ + ( ) ⋅
= ⋅ +
⋅
= ⋅ +
= +
−
13
13
23
13
23
23
23
13
1
1
3
1
13
1
ln( ) ln( )
ln( )
ln( )
ln( )
t t3 ⋅ ln( )
exex
ex
� 35x6 # ex � ex # 5x7 � 5x6ex17 � x 2
ddx15x7 2 # ex �
ddx1ex 2 # 5x7g¿ 1x 2 �
g1x 2 � 5x7 # ex
′ = −f xx
xx x( )
cos( )sin( )ln( )
�1x
# cos1x 2 � sin1x 2 # ln1x 2
ddx1 ln1x 2 2 # cos1x 2 �
ddx1cos1x 2 2 # ln1x 2f ¿ 1x 2 �
f 1x 2 � ln1x 2 # cos1x 2
dy
dx� x213sin1x 2 � xcos1x 2 2
3x2sin1x 2 � cos1x 2 # x3
ddx1x3sin1x 2 2 �
ddx1x3 2 # sin1x 2 �
ddx1sin1x 2 2 # x3
sin1x 2
x3
y � x3sin1x 2
76
THE PRODUCT RULE
The Product Rule can be stated “the derivative of the first times the second, plus the derivative of the secondtimes the first.” It can be proven directly from the limit definition of the derivative, but only with a few tricks anda lot of algebra. Using Leibniz’s notation, the Product Rule is stated as follows:
ddx
f x g x f x g x g x f x( ( ) ( ) ( ) ( ) ( ) ( )) =⋅ ′ ⋅ + ′ ⋅
Calc2e_09_75-80.qxd 11/18/11 12:47 AM Page 76
the derivative of , we’ll have to use the
Product Rule a second time. This can get messy, but it
will be fine if everything is written down carefully.
Practice
For questions 1 through 12, compute the derivative.
1.
2.
3. y =
4.
5.
6.
7.
8. h(t) =
9.
10.
11. . (Hint: .)
12.
13. What is the slope of the tangent line to
at (0,2)?
14. Find the equation of the tangent line to
at x = π.
The Quotient Rule
The Quotient Rule for functions where the parts are
divided is slightly more complicated than the Product
Rule. The Quotient Rule can be stated:
Just as with the Product Rule, each part is differ-
entiated and multiplied by the other part. Here, how-
ever, they are subtracted, so it matters which one is
differentiated first. It is important to start with the
derivative of the top.
ddxa
f 1x 2
g1x 2b �
f˛ ¿ 1x 2g1x 2 � g¿ 1x 2f 1x 2
1g1x 2 22
y � xsin1x 2
f 1x 2 � x2ex � x � 2
g1x 2 � 3x4ln1x 2cos1x 2
y x e xx= ⋅( )sin( )y � xexsin1x 2
f 1x 2 � sin21x 2 � sin1x 2 # sin1x 2
y � 5x3 � xln1x 2
t t t3 4+( ) −( )sin( ) cos( )
y � 8ln1x 2sin1x 2 � cos1x 2
k x x xx
x( ) cos( )
sin( )= −4
h u u u eu( ) = +( )2 3
g1x 2 � 3x2ln1x 2 � 5x4 � 10
π sin( )cos( )x x
y � 8t3et
f 1x 2 � x2cos1x 2
= + − ⋅4 5 2 2 5x x x x x xsin( )cos( ) (cos ( ) sin ( ))
Bcos1x 2 # cos1x 2 � sin1x 2 # sin1x 2R # x5
= 4 5x x xsin( )cos( )
ddx1cos1x 2 2 # sin1x 2 R # x5
B ddx1sin1x 2 2 # cos1x 2 �= 4 5x x xsin( )cos( )
dy
dx�
ddx1x5 2 # sin1x 2cos1x 2 �
ddx1sin1x 2cos1x 2 2 # x5
sin1x 2cos1x 2
77
THE QUOTIENT RULE
ddx
f xg x
f x g x g x f x
g
( )( )
( ) ( ) ( ) ( )
= ′ − ′
( (x))2
Calc2e_09_75-80.qxd 11/18/11 12:47 AM Page 77
Example
Differentiate .
Solution
Here, the top part is and the bottom
part is . Therefore, by the Quotient Rule:
=
=
Example
Differentiate .
Solution
Example
Differentiate .
SolutionHere, the Product Rule is necessary to differentiate
the top.
Example
Differentiate .
Solution
= −12
ln( )t
t
=⋅ − ⋅1
1
2t
t t
t
ln( )
dy
dt�
ddt1 ln1t 2 2 # t � d
dt 1t 2 # ln1t 2
t2
y �ln1t 2
t
=⋅ + ⋅[ ]⋅ −2 2
2
x x x x x x x
x
sin( ) cos( ) ln( ) sin( )
(ln( ))
dy
dx�
ddx 1x
2sin1x 2 2 # ln1x 2 � ddx 1 ln1x 2 2 # x2sin1x 2
1 ln1x 2 22
y �x2sin1x 2
ln1x 2
= − −−
= −−
30 3 2010 1
10 310 1
4 2 4
2 2
4 2
2 2
x x xx
x xx( ) ( )
= ⋅ − ⋅−
( ) ( )( )
( )
3 10 1 20
10 1
2 2 3
2
x x x x
x
f ¿ 1x 2 �ddx 1x
3 2 # 110x2 � 1 2 � ddx 110x2 � 1 2 # x3
110x2 � 1 22
f 1x 2 �x3
10x2 � 1
15x4 � 6x 2 # cos1x 2 � sin1x 2 # 1x5 � 3x2 � 1 2
cos21x 2
15x4 � 6x 2 # cos1x 2 � 1�sin1x 2 2 # 1x5 � 3x2 � 1 2
cos21x 2
dydx
x x x x x x
x
ddx
ddx
=
− + ⋅ − ⋅ − +( ) cos( ) (cos( )) ( )
(cos( ))
5 2 5 2
2
3 1 3 1
cos1x 2
x5 � 3x2 � 1
y �x5 � 3x2 � 1
cos1x 2
–THE PRODUCT AND QUOTIENT RULES–
78
Just as with the Product Rule,
.ddx
f xg x
f x
g x
ddxddx
( )( )
( ( ))
( ( ))
≠
=⋅ + ⋅
⋅ − ⋅d
dxd
dx xx x x x x x x
x
( ) sin( ) (sin( )) ln( ) sin( )
(ln( ))
2 2 1 2
2
Calc2e_09_75-80.qxd 11/18/11 12:47 AM Page 78
Practice
For questions 15 through 26, compute the derivative.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27. Find the second derivative of y = xex + ex.
28. What is the slope of the tangent line to
f(x) = x2 ln(x) at x = e?
Derivatives of Trigonometric Functions
We can find the derivatives of the rest of the trigono-
metric functions using the Quotient Rule.
ExampleDifferentiate .
Solution
Use .
Differentiate using the Quotient Rule.
Simplify.
Use .
Use .
Thus:ddx1tan1x 2 2 � sec21x 2
dy
dx� sec21x 2
sec1x 2 �1
cos1x 2
dy
dx�
1cos21x 2
sin21x 2 � cos21x 2 � 1
dy
dx�
cos21x 2 � sin21x 2
cos21x 2
dy
dx�
cos1x 2 # cos1x 2 � 1�sin1x 2 2 # sin1x 2
cos21x 2
dy
dx�
ddx1tan1x 2 2 �
ddxa
sin1x 2
cos1x 2b
tan1x 2 �sin1x 2
cos1x 2
y � tan1x 2
f 1x 2 �x2ex
cos1x 2
y �xln1x 2
ex
h tt t
t( )
ln( )
sin ( )= +
2
yx x
=+( ) −( )
1sin( ) cos( )π π
g uu
u u( )
sin( )=−3 3
yx x
x x= −
+1
1
g t tt
( )sin( )
=3
π
y �4et � t
t3 � 2t � 1
y �x5
ln1x 2
f 1x 2 �x � ln1x 2
ex � 1
f 1x 2 �x2 � 1x2 � 1
h1x 2 �x3 � 10x � 73x2 � 5x � 2
–THE PRODUCT AND QUOTIENT RULES–
79
Calc2e_09_75-80.qxd 11/18/11 12:47 AM Page 79
ExampleDifferentiate .
Solution
Use .
Differentiate using the Quotient Rule.
Simplify.
Use and .
Thus:
Practice
Differentiate the following.
29.
30.
31.
32. g(x) = ex sec(x)
33. h(t) = et ln(t)tan(t)
34. j xx x
x( )
sec( )= +3 4
f 1x 2 � xtan1x 2
y � cot1x 2
y � csc1x 2
ddx1sec1x 2 2 � sec1x 2tan1x 2
dy
dx� sec1x 2tan1x 2
tan1x 2 �sin1x 2
cos1x 2sec1x 2 �
1cos1x 2
dy
dx�
sin1x 2
cos21x 2�
1cos1x 2
# sin1x 2
cos1x 2
dy
dx�
0 # cos1x 2 � 1�sin1x 2 2 # 1
cos21x 2
dy
dx�
ddx1sec1x 2 2 �
ddxa
1cos1x 2
b
sec1x 2 �1
cos1x 2
y � sec1x 2
–THE PRODUCT AND QUOTIENT RULES–
80
et ⋅ ln(t) and eln(t) are NOT the same function. Thefirst one is a product, whereas the second is acomposition.
Calc2e_09_75-80.qxd 11/18/11 12:47 AM Page 80
LE
SS
ON
CHAIN RULE
We have learned how to compute derivatives of functions that are added, subtracted, multiplied,
and divided. Next, we will learn how to compute the derivative of a composition of functions
For example, it would be difficult to multiply out just to take the
derivative. Instead, notice that looks like put inside . Therefore, in
terms of composition, f(x) = (h ° g)(x) = h(g(x)).
The trick to differentiating composed functions is to take the derivative of the outermost layer first, while
leaving the inner part alone, and then multiplying that by the derivative of the inside.
Using Leibniz’s notation, the Chain Rule can be stated as follows:
If this is confusing, think of the Chain Rule in the following way:
ddx
h1something 2 � h¿ 1something 2 # ddx1something 2
ddx1h1g1x 2 2 2 � h¿ 1g1x 2 2 # g¿ 1x 2
h1x 2 � x5g1x 2 � x3 � 10x � 4f 1x 2
f 1x 2 � 1x3 � 10x � 4 25
10
81
Calc2e_10_81-84.qxd 11/18/11 12:48 AM Page 81
The usual key to figuring out what is inside and
what is outside is to watch the parentheses. Imagine
that the parentheses form the layers of an onion, and
that you must peel (differentiate) the outermost layers
one at a time before reaching the inside.
Example
Differentiate .
Solution
Here, where the something �
. Because , the Chain
Rule gives:
= 5(x3 + 10x + 4)4 ⋅ (x3 + 10x + 4)
= 5(x3 + 10x + 4)4 ⋅ (3x2 + 10)
ExampleDifferentiate .
SolutionHere, the function is essentially sin(something) where
the “something” � . The deriva-
tive of sine is cosine, so the Chain Rule gives:
ExampleDifferentiate .
SolutionThis is tricky because of the lack of parentheses. It
might look like the “outside” function is cos(some-
thing), but it is actually .
Thus, this function is really . So, the
Chain Rule gives:
= 3(cos)(x))2 ⋅ (cos(x))
= 3(cos)(x))2 ⋅ (–sin(x))
= –3(cos)2 ⋅ (x)sin(x)
ExampleDifferentiate .y � cos1x3 2
d
dx
dy
dx� 31something 22 # d
dx1something 2
1something 23y � cos31x 2 � 1cos1x 2 23
y � cos31x 2
132x3 � 6x � 2 2
g¿ 1x 2 � cos18x4 � 3x2 � 2x � 1 2 #
ddx18x4 � 3x2 � 2x � 1 2
g¿ 1x 2 � cos18x4 � 3x2 � 2x � 1 2 #
g¿ 1x 2 � cos1something 2 # ddx1something 2
8x4 � 3x2 � 2x � 1
g1x 2 � sin18x4 � 3x2 � 2x � 1 2
d
dx
f ¿ 1x 2 � 51something 24 # ddx1something 2
ddx1x5 2 � 5x4x3 � 10x � 4
f 1x 2 � 1something 25
f 1x 2 � 1x3 � 10x � 4 25
82
THE CHAIN RULE
or ddx
h hddx
( (something)) (something) something= ′ ⋅ ( )ddx
h g x h g x g x( ( ( ))) ( ( )) = ′ ⋅ ′( )
Make certain to not mix the derivatives of the lay-ers to get and mistakenly say f ′(x) = 5(3x2 + 10)4.
Calc2e_10_81-84.qxd 11/18/11 12:48 AM Page 82
SolutionIn this example, our function is cos(something).
Because , the Chain Rule gives:
= –sin(x3) ⋅ (x3)
= –sin(x3) ⋅ 3x2
ExampleDifferentiate .
Solution
so:
= e5x ⋅ (5x) = e5x ⋅ 5 = 5e5x
Practice
Differentiate the following.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10. h(x) = sin(πx)
11.
12.
13.
14.
15.
16.
17. y =
18. y = ln(x sin(x))
e xx e
22( ) +
f 1x 2 � sec110x2 � ex 2
y � xe2x
f 1u 2 �sin12u 2
u
g1x 2 �ex � e�x
2
f 1x 2 � ex � e2x � e3x
y � 1 ln1x 2 25
y � ln13t � 5 2
g x x( ) tan( )=
f x x( ) tan= ( )y � 23 ex � 1
g x x x( ) + + = 2 9 1
y � 1u5 � 3u4 � 7 272
h1t 2 � 1t8 � 9t3 � 3t � 2 210
y � 1x2 � 8x � 9 23
f 1x 2 � 18x3 � 7 24
d
dx
h¿ 1x 2 � e1something2 # ddx1something 2
h1x 2 � e1something2
h1x 2 � e5x
d
dx
dy
dx� �sin1something 2 # d
dx1something 2
ddx1cos1x 2 2 � �sin1x 2
83
“SOMETHING” HINT
It is important that the “something” in the parentheses appear somewhere in the derivative, just as it does inthe original function. If it doesn’t appear, then a mistake has been made.
tan(cos(x)) ≠ tan(x) ⋅ cos(x).
You CANNOT cancel the θ’s to conclude that sin(2θ)
θ = sin(2).
Calc2e_10_81-84.qxd 11/18/11 12:48 AM Page 83
19. s(u) =(sin(u) + cos(u))3
20. y = tan(cos(x))
This rule is called the Chain Rule because it works in
long succession when there are many layers to the
function. It helps to write out the function using lots
of parentheses, and then work patiently to take the
derivative of each outermost layer.
Example
Differentiate .
SolutionWith all of its parentheses, this function is
. The outermost layer is “some-
thing to the seventh power,” the second layer is “the
sine of something,” the third layer is “e raised to the
something,” and the last layer is 5x. Thus:
= 7(sin(e(5x)))6 ⋅ cos(e(5x)) ⋅ e(5x) ⋅ 5
= 35e(5x)sin6(e(5x))cos(e(5x))
ExampleDifferentiate .
Notice once again that every part except the out-
ermost layer (the natural logarithm) appears some-
where in the derivative.
Practice
Differentiate the following.
21.
22.
23.
24.
25.
26. h xe
e
x
x( ) cos= −
−
2
4 2
11
k1u 2 � sec1 ln18u3 2 2
y � sin1sin1sin1x 2 2 2
g1t 2 � ln1tan1et � 1 2 2
y � 1e9x2�2x�1 24
f 1x 2 � cos318x 2
y � ln1x3 � tan13x2 � x 2 2
= ⋅ ⋅ ⋅7 55 6 5 5(sin( )) cos( ) ) ( )( ) ( ) ( )e e e xx x x d
dx
= ⋅ ⋅7 5 6 5 5(sin( )) cos( ) ( )( ) ( ) ( )e eddx
ex x x
f ¿ 1x 2 � 71sin1e15x2 2 26 # ddx1sin1e15x2 2 2
f 1x 2 � 1sin1e15x2 2 27
f 1x 2 � sin71e5x 2
–CHAIN RULE–
84
Solution
= + + ++ +
⋅3 3 6 1
3
2 2 2
3 2
x x x x
x x x
sec ( ) ( )
tan( )
13x2 � sec213x2 � x 2 # 16x � 1 2 2=+ +
⋅133 2x x xtan( )
3 3 32 2 2 2x x xd
dxx x+ + +
⋅ sec ( ) ( )=+ +
⋅133 2x x xtan( )
dy
dx x x x
d
dxx x x=
+ ++ +⋅1
33
3 23 2
tan( )( tan( ))
Calc2e_10_81-84.qxd 11/18/11 12:48 AM Page 84
LE
SS
ON
IMPLICIT DIFFERENTIATION
Acommon complaint about the Chain Rule is “I don’t know where to stop!” For example, why do we
use the Chain Rule for to get , but not for ,
which has ? The honest answer is that we could use the Chain Rule as follows:
When we get down to , we know we are done. The advantage to this way of thinking is that it
explains what really means. This isn’t merely a symbol that says “we took the derivative.” This is the result
of differentiating both sides of an equation.
dy
dx
ddx1x 2 � 1
� cos1x3 2 # 3x2 # 1 � cos1x3 2 # 3x2f ¿ 1x 2 � cos1x3 2 # ddx1x3 2 � cos1x3 2 # 3x2 # d
dx1x 2
g¿ 1x 2 � cos1x 2 # ddx1x 2 � cos1x 2 # 1 � cos1x 2
g¿ 1x 2 � cos1x 2
g1x 2 � sin1x 2f ¿ 1x 2 � cos1x3 2 # 3x2f 1x 2 � sin1x3 2
11
85
Calc2e_11_85-90.qxd 11/18/11 12:49 AM Page 85
ExampleDifferentiate .
SolutionStart with the equation.
Differentiate both sides of the equation.
Use .
Simplify.
Now if , then there is a relationship
between y and x. This relationship is given explicitly
because we know exactly what y is in terms of x. How-
ever, if the variables x and y are all mixed up on both
sides of the equals sign, then the relationship is said to
be implicit. The relationship is implied, but it is up to
us to figure out what the relationship between x and y
is explicitly. For example, the equation of the unit cir-
cle is:
There is a relationship between the values of x
and y, because what y can be depends on the value of
x. If x � 0, for instance, then y could be either 1 or �1.
We could take the implicit description of y in
and make it explicit by solving for y:
Solving for y is not always possible, though.
In fact, it rarely is. If our equation were ln(y) + cos(y)
= 3ex – x3, then we would not be able to solve for y.
Fortunately, we can still find the slope of the tan-
gent line, , without having to solve the original
equation for y. The trick is to use implicit differentia-
tion by taking the derivative of both sides and making
sure to include wherever the Chain Rule
dictates.
ExampleFind the slope of the tangent line to .
SolutionStart with the equation.
Differentiate both sides.
Use the Chain Rule everywhere.
Use and .
Solve for .
It might make you uneasy to have given in
terms of both x and y, but this is necessary. If we were
dy
dx
dy
dx�
�2x2y
� �xy
dy
dx
2x # 1 � 2y # dy
dx� 0
ddx1y 2 �
dy
dxddx1x 2 � 1
2x # ddx1x 2 � 2y # d
dx1y 2 � 0
ddx1x2 � y2 2 �
ddx11 2
x2 � y2 � 1
x2 � y2 � 1
ddx1y 2 �
dy
dx
dy
dx
y x 1 = ± − 2
y2 � 1 � x2
x2 � y2 � 1
x2 � y2 � 1
y � 4x5 � ex
dy
dx� 20x4 # 1 � ex # 1 � 20x4 � ex
dy
dx� 20x4 # d
dx1x 2 � ex # d
dx1x 2
ddx1y 2 �
dy
dx
ddx1y 2 �
ddx14x5 � ex 2
y � 4x5 � ex
y � 4x5 � ex
–IMPLICIT DIFFERENTIATION–
86
Calc2e_11_85-90.qxd 11/18/11 12:49 AM Page 86
asked, “What is the slope of the tangent line to
at ?” We would have to reply,
“Which one?” There are two tangent lines when
! See Figure 11.1. If we want the slope of
the tangent line at , then
.
Example
Find when .
SolutionStart with the equation.
Differentiate both sides of the equation.
Use the Chain Rule everywhere.
Use and .
Factor out a .
Solve for .
To get rid of the fraction-in-a-fraction, we can
multiply the top and bottom by the denominator y
that we want to eliminate:
�3yex � 3x2y
1 � ysin1y 2
� a3ex � 3x2
1y � sin1y 2
b # a y
yb
dy
dx�
3ex � 3x2
1y � sin1y 2
dy
dx�
3ex � 3x2
1y � sin1y 2
dy
dx
a1y
� sin1y 2bdy
dx� 3ex � 3x2
dy
dx
1y
# dy
dx� sin1y 2 # dy
dx� 3ex � 3x2
ddx1y 2 �
dy
dxddx1x 2 � 1
3ex # ddx1x 2 � 3x2 # d
dx1x 2
1y
# ddx1y 2 � sin1y 2 # d
dx1y 2 �
ddx1 ln1y 2 � cos1y 2 2 �
ddx13ex � x3 2
ln1y 2 � cos1y 2 � 3ex � x3
ln1y 2 � cos1y 2 � 3ex � x3dy
dx
dy
dx
x
y = =
−= =
12
32
1
3
3
3
1
2
3
2, −
x �12
x �12
x2 � y2 � 1
–IMPLICIT DIFFERENTIATION–
87
1
1–1
y
x__2
x2 y2= 1+
)32–( 1– –
2,
1
–1
)32–( 1– –
2,
Figure 11.1
Don’t forget to apply the Chain Rule when com-
puting dx d (cos(y)) when y depends on x.
Calc2e_11_85-90.qxd 11/18/11 12:49 AM Page 87
ExampleFind the slope of the tangent line to
at (1,�5).
SolutionStart with the equation.
Differentiate both sides of the equation.
Use the product rule on .
Use and .
Plug in x � 1 and y � �5.
Use .
Thus, the slope of the tangent line at (1,�5) is 25.
Example
Find when cos(x ⋅ sin(y)).
SolutionStart with the equation.
Differentiate both sides of the equation.
Compute the derivatives on both sides.
Use (x) = 1 and .d
dxy
dy
dx( ) =
d
dx
sec ( ) ( )
sin sin( ) sin( ) cos( ) ( )
2 yddx
y
x y y yddx
y x
⋅
= − ⋅( )⋅ + ⋅
⋅
ddx
yddx
x ytan( ) cos sin( )( ) = ⋅( )( )
tan( ) cos sin( )y x y= ⋅( )
dy
dx
25 �dy
dx
ln11 2 � 0
2 5 11
15
0
2( ) ln( ) ( )− ⋅ ⋅ + ⋅ − ==
dy
dx
dy
dx{
2y # dy
dx# ln1x 2 �
1x
# y2 �dy
dx
ddx1y 2 �
dy
dxddx1x 2 � 1
2y # ddx1y 2 # ln1x 2 �
1x
# ddx1x 2 # y2 �
ddx1y 2 � 0
y2ln1x 2
ddx1y2ln1x 2 2 �
ddx1y � 5 2
y2ln1x 2 � y � 5
y2ln1x 2 � y � 5
–IMPLICIT DIFFERENTIATION–
88
Simplify the right-hand side.
sec ( )
sin sin( ) sin( ) sin sin( ) cos( )
2 ydydx
x y y x y y xdydx
⋅
= − ⋅( )⋅ − ⋅( ) ⋅ ⋅ ⋅
sec ( ) sin sin( ) sin( ) cos( )2 ydydx
x y y ydydx
x⋅ = − ⋅( )⋅ + ⋅
⋅
Bring both instances of to the same side.
Factor out a .
sin sin( ) sin( )x y y= − ⋅( )⋅
sec ( ) sin sin( ) cos( )2 y x y y xdydx
+ ⋅( ) ⋅ ⋅[ ] =
dy
dx
sec ( ) sin sin( ) cos( )
sin sin( ) sin( )
2 ydydx
x y y xdydx
x y y
⋅ + ⋅( ) ⋅ ⋅ ⋅
= − ⋅( )⋅
dy
dx
Calc2e_11_85-90.qxd 11/18/11 12:49 AM Page 88
Solve for .
ExampleUse implicit differentiation and the fact that
to prove that .
SolutionIf , then the derivative of ln(x) is .
Raise both sides as powers of e.
Since and are inverses, .
Differentiate both sides.
Use the Chain Rule.
Solve for .
Use .
So, .
Practice
For questions 1 through 14, compute .
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15. Find the tangent line slope of
at (�3,1).
16. Find the tangent line slope of
at (1,�2).x3 � y3 � 3y � x
y3 � x2 � y2 � 5y � 14
sec1y 2 � 9y � x3cos1y 2
xy
� xy � x � y
x2y � y4x4
1y � x2 24 � 10x
y x y y y− = − − −ln( ) 3 2 1
sin( ) sin( ) sin( )x y x y− = −
y x y = +
tan1y 2 � cos1x 2
e e e ex x y y+ = +2 2
y2 � x � 3x4 � 8y
y y x − = ln( )
sin1y 2 � 4x � 7
y3 � y � sin1x 2
1y � 1 23 � x4 � 8x
dy
dx
ddx
xx
ln( )( ) = 1
dy
dx�
1x
ey � eln1x2 � x
dy
dx�
1ey
dy
dx
ey # dy
dx� 1
ddx1ey 2 �
ddx1x 2
ey � x
eln1x2 � xexln1x 2
ey � eln1x2
y � ln1x 2
dy
dxy � ln1x 2
ddx1 ln1x 2 2 �
1x
ddx1ex 2 � ex
dydx
x y y
y x y y x=
− ⋅( )⋅+ ⋅( ) ⋅ ⋅
sin sin( ) sin( )
sec ( ) sin sin( ) cos( )2
dy
dx
–IMPLICIT DIFFERENTIATION–
89
. To see this, let a = 16 and b
= 9. The left-hand side is , but the
right-hand side is .16 4 9 + 3 = 7+ =
25 5 =
a b a b+ ≠ +
Calc2e_11_85-90.qxd 11/18/11 12:49 AM Page 89
17. Find the slope of the tangent line to
at (4,2).
18. Find the slope of the tangent line at
on the graph of .
19. Find the equation of the tangent line to
at the point .
20. Find the equation of the tangent line to
at (0,0).ln ln( )e e yx y+( ) = −2
1
2 6,π
sin1y 2 � x
cos sin( )π
2 2y x
=
2
2 4,π
ln13y � 5 2 � x � y2
–IMPLICIT DIFFERENTIATION–
90
. To see this, let A
= π and B = . The left-hand side is sin
= 1, but the right-hand side is sin(π) – sin= 0 – 1 = –1.
π2
π2
π2
s A B s A s Bin in( ) in( )( )− ≠ −
Calc2e_11_85-90.qxd 11/18/11 12:49 AM Page 90
LE
SS
ON
RELATED RATES
Sometimes, both variables x and y depend on a third variable t. An equation relating x and y is often
able to be determined geometrically. Once you have gotten the hang of implicit differentiation, it
should not be difficult to take the derivative of both sides with respect to the variable t. This enables
us to see how x and y vary with respect to time t. The only difference is that , , and
so on. Only can be simplified.
ExampleAssume x and y depend on some variable t. Differentiate with respect to t.
SolutionStart with the equation.
Us the Chain Rule to differentiate both sides with respect to t.
ddt1y2 � cos1x 2 2 �
ddt14x2y 2
y2 � cos1x 2 � 4x2y
y2 � cos1x 2 � 4x2y
ddt1t 2 � 1
ddt1y 2 �
dy
dtddt1x 2 �
dxdt
12
91
Calc2e_12_91-96.qxd 11/18/11 12:51 AM Page 91
Use and .
Example
Assume x and y depend on some variable t. Differen-
tiate ex + y = y 3 + with respect to t.
SolutionStart with the equation.
ex + y = y 3 +
Differentiate both sides with respect to t.
Use the Chain Rule everywhere.
Use and .
The variables need not be x and y, and the variable
upon which they depend need not be t.
ExampleAssume A, B, and C depend on some variable r. Dif-
ferentiate .
Solution
Practice
Assume all variables depend on t. Differentiate with
respect to t.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11. S = 6e2
12. D x y= +2 2
A �12
bh
C � 2p r
A � 4p r2
V �43p r3
A2 � B2 � C2
z �25
x2 �25
y2 �3
5x
ln1y 2 � ex � x2y2
2x � 2y � 10x3 � 7x
y4 � 3x2 � cos1y 2
y � 1x3 � x � 1 25
ddr
A Bddr
AC
dAdr
BdBdr
dAdr
CdCdr
A
C
3 4
3 8
2
2
+( ) = +
⋅ + ⋅ =⋅ − ⋅
π
3 4 2A BAC
+ = + π
edx
dt
dy
dty
dy
dt x
dx
dtx ⋅ ⋅ ⋅= + + 3
1
2
2
ddt1y 2 �
dy
dtddt1x 2 �
dxdt
ed
dtx
d
dty y
d
dty
x
d
dtxx ⋅ ⋅ ⋅=( ) ( ) ( ) ( )+ +3
1
2
2
d
dte y
d
dty xx( ) ( )+ = + 3
x
x
2y # dy
dt� sin1x 2 # dx
dt� 8xy # dx
dt�
dy
dt# 4x2
ddt1y 2 �
dy
dtddt1x 2 �
dxdt
8x # ddt1x 2 # y �
ddt1y 2 # 4x2
2y # ddt1y 2 � sin1x 2 # d
dt1x 2 �
–RELATED RATES–
92
= 0 because π is a constant.ddr
( )π
Calc2e_12_91-96.qxd 11/18/11 12:51 AM Page 92
Just as is a rate, so are , ,
, and so on. Because t typically represents time,
represents how fast y is changing
over time. Thus, if A is a variable that represents an
area, represents how fast that area is increasing or
decreasing over time.
Differentiating an equation with respect to t
results in a new equation, which shows how the rates
of change of the variables are related. For example, the
area A and radius r of a circle are related by:
Differentiating both sides with respect to t gives:
If a circle is growing in size, this equation details how
the rate at which the radius is changing, ,
relates to the rate at which the area is growing, .
ExampleA rock thrown into a pond makes a circular ripple that
travels at 4 feet per second. How fast is the area of the
circle increasing when the circle has a radius of 12 feet?
SolutionWe know that for circles A = πr2, so that,
. And we know that the radius is
increasing at the rate of feet per second, so
when the radius is r � 12 feet, the area is increasing at:
square feet per second
ExampleA spherical balloon is inflated with 40 cubic inches of
air every second. When the radius is 12 inches, how fast
is the radius of the balloon increasing? (Hint: The vol-
ume of a sphere with radius r is .)
Solution
We know that the volume of the balloon is increasing
at the rate of . We want to know the
value of when r � 12 inches. Differentiating
with respect to t gives:
When we plug in and r � 12 in, we get:
The radius of the balloon is increasing at the very
slow rate of inches per second.5
72p� 0.022
drdt
=⋅
=40 5 in.
4 144 sec in.
72 secπ π
40in3
sec� 4p 112 in 22 # dr
dt
dVdt
� 40in3
sec
dVdt
� 4p r2 # drdt
V �43p r3
drdt
dVdt
� 40in3
sec
V � 43p r3
� 96p � 301.6
� 96pft2
sec
dAdt
� 2p 112 feet 2 # 4feet
second
drdt
� 4
dAdt
� 2p r # drdt
dAdt
drdt
dAdt
� 2p r # drdt
A � p r2
dAdt
dy
dt�
y-change
t-change
dAdt
dy
dtdxdt
dy
dx�
y-change
x-change
–RELATED RATES–
93
Calc2e_12_91-96.qxd 11/18/11 12:51 AM Page 93
ExampleSuppose the base of a triangle is increasing at a rate of
8 feet per minute while the height is decreasing by 1
foot every minute. How fast is the triangle’s area
changing when the height is 5 feet and the base is 20
feet?
SolutionIf we represent the length of the base by b, the height
of the triangle as h, and the area of the triangle as A,
then the formula that relates them all is . The
base is increasing at and the height is
changing at . The �1 implies that 1
foot is subtracted from the height every minute, that is,
the height is decreasing. We are trying to find ,
which is the rate of change in area. When we differen-
tiate the formula with respect to t, we get:
When we plug in all of our information, includ-
ing the h � 5 feet and b � 20 feet, we get:
Thus, at the exact instant when the height is 5 feet
and the base is 20 feet, the area of the triangle is
increasing at a rate of 10 square feet every minute.
ExampleA 20-foot ladder slides down a wall at the rate of 2 feet
per minute (see Figure 12.1). How fast is it sliding
along the ground when the ladder is 16 feet up the
wall?
Solution
Here, because the ladder is sliding
down the wall at 2 feet per minute. We want to know
, the rate at which the bottom of the ladder is mov-
ing away from the wall. The equation to use is the
Pythagorean theorem.
If we plug in y � 16 ft and ft/min, we get:
We still need to know what x is at the particular instant
that y � 16, and for this, we go back to the
Pythagorean theorem.
, so x � ;12x2 � 144
x2 � 116 22 � 120 22
2 2 16 2 0xdxdt
⋅ + ⋅( )⋅ −
=ftft
min
dy
dt� �2
2x # dxdt
� 2y # dy
dt� 0
ddt1x2 � y2 2 �
ddt1202 2
x2 � y2 � 202
dxdt
dy
dt� �2
ftmin
dAdt
= ⋅
⋅( ) + −
⋅ ⋅ ( )
= − =
12
8 5 112
20
20 10 102 2 2
ftft
ftft
ft ft ft
min min
min min min
dAdt
�12
# dbdt
# h �dhdt
# 12
b
A �12
bh
dAdt
dhdt
� �1ft
min
dbdt
� 8ft
min
A �12
bh
–RELATED RATES–
94
ladder
20 feet
wall
y
x ground
Figure 12.1
Calc2e_12_91-96.qxd 11/18/11 12:51 AM Page 94
Using x � 12 (a negative length here makes no sense),
we get:
At the moment that y � 16 ft, the ladder is sliding
along the ground at feet per minute.
In the previous example, it was okay to say that the
hypotenuse was 20 because the length of the ladder
didn’t change. However, if we replace y with 16 in the
equation before differentiating, we would have implied
that the height was fixed at 16 feet. Because the height
does change, it needs to be written as a variable, y. In
general, any quantity that varies needs to be represented
with a variable. Only after all derivatives have been com-
puted can the information for the given instant, like
, be substituted.
Practice
13. Suppose and .
What is when x � �1 and y � 2?
14. Suppose . What is when
, x � 3, and y � �2?
15. Let . If and ,
what is when L � 0 and I � 3?
16. Suppose , , and
. What is when A � 2, B � 2,
and C � 1?
17. Suppose . If I increases by 4 feet
per minute and R increases by 2 square feet every
minute, how fast is A changing when I � 20?
18. Suppose . Every hour, K
decreases by 2. How fast is R changing when K
� 3 and ?
19. The height of a triangle decreases by 2 feet every
minute while its base shrinks by 6 feet every
minute. How fast is the area changing when the
height is 15 feet and the base is 20 feet?
20. The surface area of a sphere with radius r is
. If the radius is decreasing by
2 inches every hour, how fast is the surface area
shrinking when the radius is 20 inches?
21. A circle increases in area by 20 square feet
every hour. How fast is the radius increasing
when the radius is 4 feet?
22. The volume of a cube grows by 1,200 cubic
inches every minute. How fast is each side
growing when each side is 10 inches?
23. The surface area of a cube is decreasing at a
rate of 2 square inches per second. How fast is
an edge shrinking at the instant when each side
is 40 inches? (Hint: The surface area of a cube
with edge e is S = 6e2.)
A � 4p r2
R �14
K3 �1R2 � 11
A � I2 � 6R
dBdt
dCdt
� �2
dAdt
� 8A3 � B2 � 4C2
dKdt
dIdt
� 4dLdt
� 5K � eL � L � I2
dxdt
� 8
dy
dtxy2 � x2 � 3
dxdt
dy
dt� 5y2 � 3y � 6 � 4x3
y � 16
83
2 12 2 16 2 0
8
3
⋅ ( ) ⋅ + ⋅( ) ⋅ −
=
=
ft ftft
ft
dxdt
dxdt
min
min
95
CHANGING VALUES HINT
It is important to use variables for all of the values that are changing. Only after all derivatives have been com-puted can they be replaced by numbers.
Calc2e_12_91-96.qxd 11/18/11 12:51 AM Page 95
24. The height of a triangle grows by 5 inches each
hour. The area is increasing by 100 square
inches each hour. How fast is the base of the
triangle increasing when the height is 20 inches
and the base is 12 inches?
25. One end of a 10-foot long board is lifted
straight off the ground at 1 foot per second
(see Figure 12.2). How fast will the other end
drag along the ground after 6 seconds?
26. A kite is 100 feet off the ground and moving
horizontally at 13 feet per second (see Figure
12.3). How quickly is the string being let out
when the string is 260 feet long?
–RELATED RATES–
96
1 ft__sec
?
board
10 ft
Figure 12.2
ftsec
100 ftstring
?
12
Figure 12.3
Calc2e_12_91-96.qxd 11/18/11 12:51 AM Page 96
LE
SS
ON
LIMITS AT INFINITY
This lesson will serve as a preparation for the graphing in the next lesson. Here, we will work on ways
to identify asymptotes from the formula of a rational function, a quotient of two polynomials.
We’ve encountered vertical asymptotes informally in Lesson 5. They are easy to recognize for
rational functions because they occur at precisely those x-values at which the denominator equals zero and
the numerator does NOT equal zero. If both top and bottom are zero when evaluated at an x-value, you get
a small unshaded circle on its graph at that point. For example, has vertical asymp-totes at x � �3 and x � 4.
Horizontal asymptotes take a bit more work to identify. The graph will flatten out like a horizontal line
if large values of x all have essentially the same y-value.
In the graph of , in Figure 13.1 for example, if x is bigger than 5, then y will be very close to y
� 1. Similarly, if x is a large negative number, the corresponding y-value will be close to zero. Horizontal
asymptotes are related to the limits as x gets really big. For given in the graph:
and
In such case, we say that y = 1 and y = 0 are horizontal asymptotes of f.
lim ( )x
f x→−∞
= 0lim ( )x
f x→∞
= 1
f˛˛1x 2
y � f˛ 1x 2
f˛ 1x 2 �13x � 2 2 1x � 1 2
1x � 3 2 1x � 4 2
13
97
Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 97
These limits at infinity (and negative infinity) identify
what the ends of the graph do. For example, if
, then the graph of will typi-
cally look something like that in Figure 13.2. If
, then the graph of will look
like that in Figure 13.3.
Notice that the infinite limits say only what hap-
pens way off to the left and to the right. Other calcu-
lations must be done to know what happens in the
middle of the graph.
The general trick to evaluating an infinite limit is
to focus on the most powerful part of the function.
Take for example.
There are several terms being added in this function.
However, the most powerful part is the term .
When x gets big enough, like when x � 1,000,000, then
This clearly rounds to 2,000,000,000,000,000,000,
which is the value of 2x3 at x = 1,000,000. It is in this
sense that is called the most powerful part of the
function. As x gets big, is the only part that counts.2x3
2x3
� 1,999,899,999,989,995,000
100,000,000,000,000 � 10,000,000 � 5,000
� 2,000,000,000,000,000,000 �
2x3 � 100x2 � 10x � 5,000
2x3
lim ( , )x
x x x→∞
− − −2 100 10 5 0003 2
y � h1x 2lim ( )x
h x→−∞
= ∞
y � g1x 2lim ( )x
g x→∞
= 3
98
ASYMPTOTE HINT
Notice that the graph of y = f (x) crosses both horizontal asymptotes. Vertical asymptotes cannot be crossedbecause they are, by definition, not in the domain. Horizontal asymptotes can be crossed, as illustrated in thisexample. Think of “asymptote” as meaning “flattens out like a straight line” and not “a line not to be crossed.”
1
2
3
1 2 3 4 5 6
y
x7 8 9 10–1–2–3–4–5
–1
–2
–3
y = f(x)
Figure 13.1
Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 98
As x gets huge, is clearly even larger, and is
twice that. Thus, as x goes to infinity, so does . Basi-
cally, the higher the exponent of x, the more powerful
it is. With that in mind, the rules for infinite limits of
rational functions are fairly simple:
� If the numerator is more powerful, the limit goes
to or � .� If the denominator is more powerful, the limit
goes to 0.� If the numerator and denominator are evenly
matched, the limit is formed by the coefficients of
the most powerful parts.
∞∞
2x3
2x3x3
lim( , ) limx x
x x x x→∞ →∞
− − − = = ∞2 100 10 5 000 23 2 3
99
1
2
3
y
–1
y = g(x)
4
5
1
2
3
y
x
–1
y = g(x)4
5
OR
levels out like y = 3
levels out like y = 3
toward ∞ toward ∞
x
Figure 13.2
y
x
y = h(x)
does notlevel out
toward –∞
Figure 13.3
RULES FOR INFINITE LIMITS
The rules for Infinite Limits of Rational Functions are as follows:� If the numerator is more powerful, the limit goes to ∞ or �∞.� If the denominator is more powerful, the limit goes to 0.� If the numerator and denominator are evenly matched, the limit is formed by the coefficients of the
most powerful parts.
Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 99
Example
Evaluate .
SolutionThe most powerful part of the numerator is , and
in the denominator is . Thus:
This limit is zero because the numerator is overpow-
ered by the denominator. Also, as x gets really big,
gets really close to zero. For example,
when x = 1,000, then .
Example
Evaluate .
SolutionHere, the numerator and denominator are evenly
matched, with each having as its highest power
of x.
The limit is formed by the coefficients of the
most powerful parts: 3 in the numerator and �4 in the
denominator.
Example
Evaluate .
SolutionHere,
As x goes to infinity, also gets really large, but the
negative in the �5 reverses this and makes
approach negative infinity.
Practice
Evaluate the following infinite limits.
1.
2. limx
x x x
x x→−∞
+ ++ −
4 3
5 8 1
2
3
3 10
limx
x x
x→∞
+ −+
5 2
8 1
2
4
3 10
�5x8
x8
lim lim
lim
x x
x
x x
x
x
x
x
→∞ →∞
→∞
− +−
=−
= − = −∞
5 7
1
5
5
5
2
10
2
8
10 4
limx
x x
x→∞
− +−
5 7
1
5
2
10 4
lim( )( )
lim
lim
lim
x x
x
x
x xx x
x xx
xx
→ − → −
→ −
→ −
∞ ∞
∞
∞
+ −− +
= + −−
=−
=−
= −
3 2 51 2 1 2
3 2 51 4
34
34
34
2 2
2
2
2
x2
lim( )( )x
x xx x→−∞+ −
− +3 2 5
1 2 1 2
2
1 1
1 0000 001
x = =
,.
1x
lim lim limx x x
x
x x
x
x x→∞ →∞ →∞
−+ +
= − = − =1
3 2
10
2
3
2
3
x3
�x2
limx
x
x x→∞
−+ +1
3 2
2
3
100
GOING TO INFINITY
The whole concept of “going to infinity” might be a bit confusing. This really means “going toward infinity,”because infinity is not reachable. Just know that “going to infinity” means that we see what happens when weplug really large numbers into the function, and that “going to negative infinity” means that we see what hap-pens when we plug really large negative numbers into the function.
Make certain to fully expand the polynomials inthe top and bottom of a rational function beforeidentifying the dominating terms in each.
Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 100
3.
4.
5.
6.
7.
8.
9.
10.
The infinite limits of and can be seen
from their graphs in Figure 13.4.
In general, as x goes to infinity, is more pow-
erful than x raised to any number. The natural loga-
rithm, however, goes to infinity slower than any power
of x. It may look as though is beginning to
level out toward a horizontal asymptote, but actually,
it will eventually surpass any height as it slowly goes up
to infinity.
In more complicated situations, we use L’Hôpi-
tal’s rule. This states that if the numerator and
denominator both go to infinity (positive or negative),
then the limit remains the same after taking the deriv-
ative of the top and the bottom.
y � ln1x 2
ex
lim ln( )x
x→∞
= ∞ limx
e x
→−∞= 0lim
xe x
→∞= ∞
ln1x 2ex
limx
t
t→∞
+− 0, 000
6, 000, 000
22
limx
x
x→−∞
−+
2
2
1
1
limx
x x x
x x→−∞
+ −+
4 2
2
3 8
2
+ 4
+ 1
lim( )
x
x x
x→ ∞+
−5 2
1
2
2
lim( )( )t
t tt t→ ∞− +
− +8 3 11
1 3 1 3
4 3
2 2
lim( )( )t
tt t t→ −∞
++ −
14 1
limx
x x
x→∞
− −+
10 3 100
2
3
5
limx
x
x→∞
+−
5 2
2 1
–LIMITS AT INFINITY–
101
1
2
3
–2–3 1–1
y
x
y = ex
1
2
3
–1
1 2 3
–2
–3
y
x
y = ln(x)
Figure 13.4
Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 101
Example
Evaluate .
Solution
Since and , we can use
L’Hôpital’s Rule.
Note: The little H over the equals sign indicates that
L’Hôpital’s Rule has been used at that point of the
computation. Examples like this demonstrate how
goes to infinity even slower than x does.
Example
Evaluate .
SolutionHere, and ,
so we can use L’Hôpital’s Rule.
Here, we need to use L’Hôpital’s Rule several more
times:
Hx
xe= ∞→
= ∞ lim6
lim limx
H
x
e
x x
e
x
x x
→∞ →∞+ +=
+3 4 5 6 42
=+ +→∞
limx
e
x x
x
3 4 52
H
x
ddx
ddx
e
x x x
x
=+ + +
→∞lim
( )
( )3 22 5 2
limx
e
x x x
x
→∞ + + +3 22 5 2
lim( )x
x x x→∞
+ + + = ∞3 22 5 2limx
xe→ ∞
= ∞
limx
e
x x x
x
→∞ + + +3 22 5 2
ln1x 2
limln( )
lim(ln( ))
( )
lim lim
x
H
x
x x
x
x
d
dxx
d
dxx
xx
→∞ →∞
→∞ →∞
−=
−
=−
= − =
11
1
10
1
lim( )x
x→∞
− = −∞1lim ln( )x
x→∞
= ∞
limln( )
x
x
x→∞ −1
102
L’HOPITAL’S RULE
If the numerator and denominator both go to infinity (positive or negative), the limit remains the same after tak-ing the derivative of the top and bottom. Using notation,
if and lim ( )x
g x→±∞
= ± ∞ lim ( )x
f x→±∞
= ± ∞ lim( )( )
lim( )( )x x
f xg x
f xg x→± →±∞ ∞
= ′′
When applying L’Hôpital’s Rule, we differentiatetop and bottom separately and form the quotientof them. We do NOT apply the Quotient Rule.
Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 102
This example shows how is more powerful than
. If the denominator had an , we’d have to use
L’Hôpital’s Rule 100 times, but in the end, would
drive everything to infinity.
Example
Evaluate .
SolutionThe limit is not infinite. So we can’t use
L’Hôpital’s Rule. The function is only powerful
when x goes to positive infinity. Instead, we use the old
“plug in” method.
Example
Evaluate .
SolutionThis has the same problem as the previous example.
No matter what x may be, will always be
between �1 and 1. Thus, and so
Because and , the
function is squeezed between them to zero as
well: . This is called the Squeeze Theo-
rem or the Sandwich Theorem because of the way
is squished between two curves, both going
to zero.
Practice
Evaluate the following limits.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20. limx
x x
e xx→−∞
+ +−
4 5 2
7
3 2
3
limx
x x
e xx→∞
+ +−
4 5 2
7
3 2
3
limcos( )
x
x
x→∞
limx x
xe→ −∞
limln( )
x x
x x
e→∞+
2
limx
x
x
ee→∞
++
2
3
32
limx
x
x l x→∞
+−
3 22
n( )
limx
x x
x→−∞
+ −+
2 5 10
4 2
limx
x
x→∞
+−
5
1
limln( )
ln( )x
x
x→∞ +
3
5
sin1x 2
x2
limsin( )
x
x
x→∞=
20
sin1x 2
x2
limx x→∞
=10
2limx x→∞
− =10
2
�1x2 �
sin1x 2
x2 �1x2
�1 � sin1x 2 � 1
sin1x 2
limsin( )
x
x
x→∞ 2
limx
e
x x
x
→−∞ + −= =
5 7 10
0
something not zero
ex
limx
e x
→−∞= 0
limx
e
x x
x
→−∞ + −5 7 1
ex
x100x3
ex
–LIMITS AT INFINITY–
103
y
x
Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 103
Sign Diagrams
In order to calculate the limits at vertical asymptotes,
it is necessary to know where the function is positive
and negative. The key is this: A continuous function
cannot switch between positive and negative without
being zero or undefined. Functions are zero when the
top is zero and the bottom is NOT zero, and undefined
where the denominator is zero. Mark all of these points
on a number line. Between these points, the function
must be entirely positive or negative. This can be
found by testing any point in each interval.
For example, consider .
This function is zero at x � 4 and undefined at both
x � �2 and x � 1. We mark these on a number line
(see Figure 13.5).
In between x � �2 and x � 1, the function is
either always positive or always negative. To find out
which it is, we test a point between �2 and 1, such as
0. Because is negative, the func-
tion is always negative between �2 and 1. Similarly,
we check a point between 1 and 4, such as
; a point after 4, such as
; and a point before �2, such
as . The sign diagram for this
function is shown in Figure 13.6.
This makes calculating the limits at the
vertical asymptotes very easy. Not only does
have vertical asymptotes at
x � �2 and x � 1, but the limits are:
As before, we can calculate the limits at infinity:
Note: We do not use a sign diagram when determining
horizontal asymptotes.
Thus, has a horizontal asymptote of y � 0.
With all of this, we begin to get a picture of
, which can be seen in
Figure 13.7.
f˛1x 2 �x � 4
1x � 2 2 11 � x 2
f˛1x 2
lim( )( )x
x
x x→−∞−
+ −=
4
2 10
lim( )( )x
x
x x
x
x xx→∞−
+ −= −
− − +=
→∞
lim
4
2 1
4
20
2
lim( )( )x
x
x x→ +
−+ −
= ∞1
4
2 1
lim( )( )x
x
x x→ −
−+ −
= − ∞1
4
2 1
lim( )( )x
x
x x→− +
−+ −
= − ∞2
4
2 1
lim( )( )x
x
x x→− −
−+ −
= ∞2
4
2 1
f˛1x 2 �x � 4
1x � 2 2 11 � x 2
f˛1�3 2 ��7
�114 2�
74
f˛15 2 �1
71�4 2� �
128
f˛12 2 ��2
41�1 2�
12
f˛10 2 ��4211 2
� �2
f˛1x 2 �x � 4
1x � 2 2 11 � x 2
–LIMITS AT INFINITY–
104
–2 1 4
Figure 13.5
++ --
–2 1 4
f(x)
Figure 13.6
Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 104
Notice that the horizontal asymptote y � 0 is
approached from above as , because is
always positive when x � �2. At the other end, the
asymptote is approached from below as
because the function is negative when x � 4.
We shall deal with graphing more thoroughly in
the next lesson.
Practice
For questions 21 through 26, determine all asymptotes,
vertical and horizontal, of the following functions.
Also, make a sign diagram for each.
21.
22.
23.
24.
25.
26. m xx
x x x( )
( )( )( )=
+ + +
6
2 29 1 2
j xx x
( )( )( )
= −+ +
1
1 52
k1x 2 �2x � 1
x2 � 4x � 3
h1x 2 �x2 � 11x � 3 22
g1x 2 �x � 3x2 � 4
f˛1x 2 �x � 2x � 4
x → ∞
f˛1x 2x → −∞
–LIMITS AT INFINITY–
105
1
2
3
4
5
6
–1–2–3–4–5–6 1 2 3 4 5 6–1
–2
–3
–4
–5
–6
y
x
Figure 13.7
Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 105
Evaluate the following limits.
27.
28.
29.
30.
31.
32. lim( )( )( )x
x
x x x→− + + +2
6
2 29 1 2
lim( )( )x x x→ +−
−+ +5
1
1 52
limxS3�
x � 1
x2 � 4x � 3
limxS �3�
x2 � 11x � 3 22
limxS2�
x � 3x2 � 4
limxS4�
x � 2x � 4
–LIMITS AT INFINITY–
106
Remember, x2 + a2 ≠ (x – a)(x + a).
Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 106
LE
SS
ON
USING CALCULUS TO GRAPH
Here is where everything comes together! We know how to find the domain, how to identify asymp-
totes, and how to plot points. With the help of the sign diagrams from the previous lesson, we shall
be able to tell where a function is increasing and decreasing, and where it is concave up and down.
Quite simply, where the derivative is positive, the function is increasing. The derivative gives the slope
of the tangent line at a point, and when this is positive, the function is heading upward, viewed from left to
right. When the derivative is negative, the function slopes downward and decreases.
When the second derivative is positive, the function is concave up. This is because the second deriva-
tive says how the first derivative is changing. If the second derivative is positive, then the slopes are increas-
ing. If the slopes, from left to right, increase from –2, to –1, to 0, to 1, to 2, and so on, then the graph must
curve like the one in Figure 14.1. In other words, the curve must be concave up.
Similarly, if the second derivative is negative, the function curves downward like the one in Figure 14.2
and is concave down.
14
107
Calc2e_14_107-114.qxd 11/18/11 12:53 AM Page 107
The concavity governs the shape of the graph,
depending on whether the function is increasing
or decreasing. If is increasing and concave up
(thus, both and are positive), then the
graph has the shape shown in Figure 14.3.
If is increasing and concave down (thus,
is positive and is negative), then the
graph has the shape shown in Figure 14.4.
If is decreasing and concave down (thus,
both and are negative), then the graph
has the shape shown in Figure 14.5.
If is decreasing and concave up (thus, f ′(x)
is negative and f ″(x) is positive), the graph has the
shape of the one in Figure 14.6.
ExampleGraph .
Solution
This function is defined everywhere and thus has no
vertical asymptotes. Because
and , there
are no horizontal asymptotes.
The derivative
is zero at x �
�5 and x � 1. To form the sign diagram, we test:
, f ′(0) = –15, and . Note:
These points were chosen arbitrarily. Any point less
than �5 will give the same information as the value
x � �6, for instance, and any point between �5
and �1 will give the same information as the value at
x � 0. Thus, the sign diagram for is shown in
Figure 14.7.
f ¿ 1x 2
f ¿ 12 2 � 21f ¿ 1�6 2 � 21
31x2 � 4x � 5 2 � 31x � 5 2 1x � 1 2
f ¿ 1x 2 � 3x2 � 12x � 15 �
lim ( )x
x x x→−∞
+ − + = −∞3 26 15 1010) = ∞
lim(x
x x x→∞
+ − +3 26 15
f˛1x 2 � x3 � 6x2 � 15x � 10
f˛1x 2
f – 1x 2f ¿ 1x 2f˛1x 2
f – 1x 2f ¿ 1x 2f˛1x 2
f – 1x 2f ¿ 1x 2f˛1x 2
f˛1x 2
–USING CALCULUS TO GRAPH–
108
slope = –2
slope = –1
slope = 0
slope = 1
slope = 2
Figure 14.1
Figure 14.2
increasing concave up
+ =
Figure 14.3
concave downincreasing
+ =
Figure 14.4
decreasing concave down
+ =
Figure 14.5
concave updecreasing+ =
Figure 14.6
Calc2e_14_107-114.qxd 11/18/11 12:53 AM Page 108
Because the function increases up to x � �5 and
then decreases immediately afterward, there is a local
maximum at x � �5. The corresponding y-value is
. Thus, (�5,110) is a local maxi-
mum. Similarly, because the graph goes down to x �
1 and then goes up afterward, there is a local minimum
at x � 1. The corresponding y-value is , so
(1,2) is a local minimum.
A guideline for identifying local minimum and
maximum points is shown in Figure 14.8.
The second derivative is 6x + 12 =
6(x + 2), which is zero at x � �2. If we test the sign at
x � �3 and x � 0, we get and
. Thus, the sign diagram for is as
shown in Figure 14.9.
f – 1x 2f – 10 2 � 12
f – 1�3 2 � �6
f – 1x 2 �
f˛11 2 � 2
y � f˛1�5 2 � 110
109
–5 1
increasing decreasing increasing
+–
f(x)
+f '(x)
Figure 14.7
INCREASING OR DECREASING
Remember, the sign of f ′(x) determines whether f (x) is increasing or decreasing.
Note: We use f ′(x) to see if the graph is increasing or decreasing, but f (x) to find the y-value at a point.
increasing decreasing
local maximum increasingdecreasing
local minimum
Figure 14.8
+–f (x) –2
concave down concave up
f "(x)
Figure 14.9
Calc2e_14_107-114.qxd 11/18/11 12:53 AM Page 109
Clearly x � �2 is a point of inflection, because
this is where the concavity switches from concave
down to concave up. The y-value of this point is
.
Before we draw the axes for the Cartesian plane,
we should consider the three interesting points we
have found: the local maximum at (�5,110), the local
minimum at (1,2), and the point of inflection at
(�2,56). If our x-axis runs from x � �10 to x � 10,
and our y-axis runs from 0 to 120, then all of these
points can be plotted on our graph (see Figure 14.10).
Example
Graph .
SolutionThe domain is . There is a vertical asymptote at x
� 2. The sign diagram for is shown in Figure 14.11.
Thus, and .
Because and ,limx
x
x→−∞+−
=
3
21lim
x
x
x→∞+−
=
3
21
limx
x
x→ +
+−
= ∞2
3
2
lim
x
x
x→ −
+−
= − ∞2
3
2
g1x 2
x � 2
g1x 2 �x � 3x � 2
f˛1�2 2 � 56
–USING CALCULUS TO GRAPH–
110
–2–3–4–5–6 1 2–1
y
x
10
20
30
40
50
60
70
80
90
100
110
120(–5,110)
(–2,56)
(1,2)
f(x) = x3 + 6x2 – 15x + 10
–7–8–9–10 3 4 5 6 7 8 9 10
–5 1
–2
increasing/decreasing
concavity
Figure 14.10
+-+g(x)
–3 2 above x-axisbelow x-axisabove x-axis
Figure 14.11
Calc2e_14_107-114.qxd 11/18/11 12:53 AM Page 110
there is a horizontal asymptote at y � 1, both to the
left and to the right. The derivative
has the sign
diagram shown in Figure 14.12.
The second derivative has the
sign diagram shown in Figure 14.13.
Because we have no points plotted at all, it makes
sense to pick one or two to the left and right of the ver-
tical asymptote at x � 2. At x � 1, , so
(1,�4) is a point. At x � 3, , so (3,6) is
another point. At x � �3, , so (�3,0) is
another nice point to know. Judging by these, it will
be useful to have both the x- and y-axes run from
�10 to 10.
To graph , it helps to start with the points
and the asymptotes as shown in Figure 14.14.
g1x 2
g1�3 2 � 0
g13 2 � 6
g11 2 � �4
g– 1x 2 �10
1x � 2 23
1 # 1x � 2 2 � 1 # 1x � 3 2
1x � 2 22�
�51x � 2 22
g¿ 1x 2 �
–USING CALCULUS TO GRAPH–
111
– –g'(x)
2
decreasing decreasing
Figure 14.12
g "(x)
2
+–
concave down concave up
Figure 14.13
–2–3–4–5–6 1 2–1
y
x
1
2
3
4
5
6
7
8
9
10
–7–8–9–10 3 4 5 6 7 8 9 10–1
–2
–3
–4
–5
–6
–7
–8
–9
–10
(1,–4)
(–3,0)y = 1
x = 2
(3,6)
Figure 14.14
Don’t automatically assume that the signs in a signdiagram will alternate. In fact, they don’t preciselywhen the number on the line comes from a factorraised to an even power, like (x – 3)2.
Calc2e_14_107-114.qxd 11/18/11 12:53 AM Page 111
Then we establish the shapes of the lines through
these points using the concavity and the intervals of
decrease (see Figure 14.15).
Example
Graph .
Solution
To start, . Thus,
has vertical asymptotes at x � 1 and x � �1. The
sign diagram for is shown in Figure 14.16.h1x 2
h1x 2
h1x 2 �x2 � 1x2 � 1
�x2 � 1
1x � 1 2 1x � 1 2
h1x 2 �x2 � 1x2 � 1
–USING CALCULUS TO GRAPH–
112
–2–3–4–5–6 1 2
y
x
1
2
3
5
6
7
8
9
10
–7–8–9–10 3 4 5 6 7 8 9 10–1
–2
–3
–4
–5
–6
–7
–8
–9
–10
(3,6)
(1,–4)
(–3,0)
g(x) = _____x – 2
2
2
increasing/decreasing
concavity
–1
4
x + 3
Figure 14.15
Calc2e_14_107-114.qxd 11/18/11 12:53 AM Page 112
Note: can never be zero. The limits at the ver-
tical asymptotes are thus:
Because and ,
there is a horizontal asymptote at y � 1.
The derivative is as follows:
Its sign diagram is shown in Figure 14.17. This
indicates that there is a local maximum at x � 0. The
corresponding y-value is .
The second derivative is as follows:
.
The sign diagram is shown in Figure 14.18. It looks like
there ought to be points of inflection at x � �1 and x
� 1, but these are asymptotes not in the domain, so
there are no actual points where the concavity changes.
Before we graph the function, it will be useful to
have a few more points. When x � �2, then
and when x � 2, as
well. Thus, it will be useful to have the x- and y-axes
run from about �3 to 3. We start with just the points
and asymptotes (see Figure 14.19).
Then we add in the actual curves, guided by the
concavity and the intervals of increase and decrease
(see Figure 14.20).
y � h12 2 �53
y � h1�2 2 �53
12x2 � 41x � 1 231x � 1 23
�12x2 � 41x2 � 1 23
�
��41x2 � 1 2 � 2 # 2x1�4x 2
1x2 � 1 23
�41x2 � 1 22 � 21x2 � 1 2 # 2x1�4x 2
1x2 � 1 24h– 1x 2 �
y � h10 2 � �1
�4x1x � 1 221x � 1 22
2 1 2 1
1
2 2
2 2
x x x x
x
( ) ( )
( )
− − +−
=h¿ 1x 2 �
limx
x
x→−∞
+−
=2
2
1
11
lim
x
x
x→∞
+−
=2
2
1
11
limx
x
x→ +
+−
= ∞1
2
2
1
1
limx
x
x→ −+−
= − ∞1
2
2
1
1
limx
x
x→− +
+−
= − ∞1
2
2
1
1
limx
x
x→− −
+−
= ∞1
2
2
1
1
x2 � 1
–USING CALCULUS TO GRAPH–
113
+h(x)
–1 1 above x-axisbelow x-axisabove x-axis
+ –
Figure 14.16
–1
h'(x)
1
increasing decreasingincreasing
0
decreasing
h(x)
+ + – –
Figure 14.17
+ – +h"(x)
1concave down concave up
–1concave up
Figure 14.18
Calc2e_14_107-114.qxd 11/18/11 12:53 AM Page 113
Practice
Use the asymptotes, concavity, and intervals of increase
and decrease, and concavity to graph the following
functions.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10. f˛1x 2 �x
x2 � 1
j1x 2 �x2 � 1
x
k1x 2 �x
x2 � 1
h1x 2 �1
x2 � 9
g1x 2 �x
x � 2
f˛1x 2 � x4 � 8x3 � 5
k1x 2 � 3x � x3
h1x 2 � 2x3 � 3x2 � 36x � 5
g1x 2 � �4x � x2
f˛1x 2 � x2 � 30x � 10
–USING CALCULUS TO GRAPH–
114
1
2
3
–1
–2–3 1 2 3–1
y
x
–2
–3
h(x) = x2 + 1______
0 1–1
1–1
decreasing
concavity
x2 – 1
increasing
–2, 5—32, 5—3
Figure 14.20
2
3
–1
–2–3 1 2 3–1
y
x
–2
–3
y = 1
x = 1x = –1
1
(0,–1)
–2, 5—3 2, 5—3
Figure 14.19
Calc2e_14_107-114.qxd 11/18/11 12:53 AM Page 114
LE
SS
ON
OPTIMIZATION
Knowing the minimum and maximum points of a function is useful for graphing and even more for
solving real-life problems. Businesses want to maximize their profits, builders want to minimize their
costs, drivers want to minimize distances, and people want to get the most for their money. If we can
represent a situation with a function, then the derivative will help find optimal points.
If the derivative is zero or undefined at exactly one point, then this is very likely to be the optimal point.
The first derivative test states that if the function increases before that point and decreases afterward, it is max-
imal (see Figure 15.1). Similarly, if the function decreases before the point and increases afterward, then the
point is minimal.
The second derivative test states that if the second derivative is positive, then the function curves up, so
a point of slope zero must be a minimum (see Figure 15.2). Similarly, if the second derivative is negative, the
point of slope zero must be the highest point on the graph. Remember that we are assuming that only one
point has slope zero or an undefined derivative.
If there are several points of slope zero and the function has a closed interval for a domain, then plug
all the critical points (points of slope zero, points of undefined derivative, and the two endpoints of the inter-
val) into the original function. The point with the highest y-value will be the absolute maximum, and the one
with the smallest y-value will be the absolute minimum.
15
115
Calc2e_15_115-120.qxd 11/18/11 12:55 AM Page 115
ExampleA manager calculates that when x employees are work-
ing at the same time, the store makes a profit of
dollars each hour. If there
are ten employees and at least one must be working at
any given time, how many employees should be sched-
uled to maximize profit?
SolutionThis is an instance of a function defined on a closed
interval because limits the options for x.
The derivative of the profit function is
= = –3(x –
2)(x – 8). Thus, the derivative is zero at x � 2 and
at x � 8.
Because the function is defined on a closed inter-
val, we cannot use the first or second derivative tests.
Instead, we evaluate f(x) at each of our critical points.
These are the points of slope zero, x � 2 and x � 8, plus
the endpoints of the interval, namely x � 1 and x � 10.
These are evaluated as follows: ,
, , and . If the
manager wants to maximize the store profit, eight
employees should be scheduled at the same time,
because this will result in a maximal profit of $64
each hour.
ExampleA coffee shop owner calculates that if she sells cookies
at $p each, she will sell cookies each day. If it costs
her 20¢ to make each cookie, what price p will give her
the greatest profit?
SolutionProfit is computed as: Profit � Revenue � Costs. If she
charges $p per cookie, then she’ll make and sell
cookies each day. Thus, her revenue will be200p2
200p2
P110 2 � 20P18 2 � 64P12 2 � �44
P11 2 � �34
�31x2 � 10x � 16 2� 30x � 48 � 3x2
P¿ 1x 21 � x � 10
P1x 2 � 15x2 � 48x � x3
–OPTIMIZATION–
116
increasing decreasing
increasingdecreasing
slope = 0
slope = 0
MAX
MIN
Figure 15.1
slope = 0
slope = 0
MAX
MIN
concave down concave up
Figure 15.2
Calc2e_15_115-120.qxd 11/18/11 12:55 AM Page 116
and her costs will be
= 40p2 . Therefore, her profit function is
. We limit this to p � 0.20
because the only optimal situation would be when the
cookies were sold for more than it cost to make them.
The derivative is , which is
zero when and therefore , so
either p � 0 or 0.40. Because p � 0 is not
in the domain, the only place where the derivative is
zero is at p � 0.40.
Using the first derivative test, we see that
and . So the sign dia-
gram for P ′ is as shown in Figure 15.3. Thus, the
absolute maximal profit occurs when the cookies are
sold at 40¢ each.
ExampleAt $1 per cup of coffee, a vendor sells 500 cups a day.
When the price is increased to $1.10, the vendor sells
only 480 cups. If every 1¢ increase in price reduces the
sales by two cups, what price per cup of coffee will
maximize income?
SolutionHere, the income is Income � Price � Cups Sold. So
if x � the number of pennies by which the price is
increased, then . This
simplifies to . And, the deriva-
tive is . This is zero only when
. The second derivative is
, which is always negative, so x � 75 is
maximal by the second derivative test. Thus, the max-
imal income will occur when the price is raised by x �
75¢ to $1.75 per cup.
′′ = −I x( ) .0 04
x �3
0.04� 75
′ = −I x x( ) .3 0 04
I x x x( ) .= + −500 3 0 02 2
I x x x( ) ( . ) ( )= + ⋅ −1 0 01 500 2
′ = −P ( . )0 50 160′ =P ( . )0 30 740
80200
�p �
80p2 � 200p380p3 �
200p2
′ = − +P pp p
( )200 80
2 3
P pp p
( ) = −200 402
a200p2 b # 10.20 2
a200p2 b # p �
200p
–OPTIMIZATION–
117
0.40
P�(p)
increasing decreasing
+ –
Figure 15.3
Calc2e_15_115-120.qxd 11/18/11 12:55 AM Page 117
ExampleA farmer wants to build a rectangular pen with 80 feet
of fencing. The pen will be built against the side of a
barn, so one side won’t need a fence. What dimensions
will maximize the area of the pen? See Figure 15.4.
SolutionThe area of the pen is . We can’t take the
derivative yet because there are two variables. We need
to use the additional information regarding how much
fencing exists; there are 80 feet of fencing. Because no
fencing will be required against the barn wall, the total
lengths of the fence will be , thus
. We can plug this into the formula for
area in order to obtain .
Now we have a function of one variable
. The derivative is .
This is zero only when y � 20. Using the second deriv-
ative test, . So, the curve is concave down
and the point y � 20 is the absolute maximum. The
corresponding x-value is
. Therefore, the pen with the maximal area will be
x � 40 feet wide (along the barn) and y � 20 feet out
from the barn wall.
ExampleA manufacturer needs to design a crate with a square
bottom and no top. It must hold exactly 32 cubic feet
of shredded paper. What dimensions will minimize the
material needed to make the crate (the surface area)?
See Figure 15.5.
SolutionWe want to minimize the surface area of the crate. The
surface area of the box consists of four sides, each of area
, plus the bottom, with an area of . Thus,
the surface area is . Again, we need to
reduce this to a formula with only one variable in
order to differentiate. We know that the volume must
be 32 cubic feet, so . Thus,
. When we plug this into the surface area func-
tion, we get:
.
So, we have a function of one variable .A xx
x( ) = +128 2
Surface Area = 4 432 1282
22 2xy x x
xx
xx+ =
+ = +
y �32x2
Volume � x2y � 32
Area � 4xy � x2
x # x � x2x # y
�40
x � 80 � 2y � 80 � 2120 2
′′ = −A y( ) 4
′ = −A y y( ) 80 4A y y y( ) = −80 2 2
Area � x # y � 180 � 2y 2 # y
x � 80 � 2y
y � x � y � 80
Area � x # y
–OPTIMIZATION–
118
y y
x
barn wall
(overhead view)
pen
Figure 15.4
y
x
x
Figure 15.5
Calc2e_15_115-120.qxd 11/18/11 12:55 AM Page 118
The derivative is:
.
which is zero when
or , so x � 4.
The second derivative is:
,
which is positive when x � 4. So, the curve is concave
up and the sole point of slope zero is the absolute min-
imum. Thus, the surface area of the crate will be min-
imized if x � 4 feet and feet.
Practice
1. Suppose a company makes a profit of P(x) =
dollars when it makes
and sells x � 0 items. How many items should
it make to maximize profit?
2. When 30 orange trees are planted on an acre,
each will produce 500 oranges a year. For every
additional orange tree planted, each tree will
produce 10 fewer oranges. How many trees
should be planted to maximize the yield?
3. An artist can sell 20 copies of a painting at
$100 each, but for each additional copy she
makes, the value of each painting will go down
by a dollar. Thus, if 22 copies are made, each
will sell for $98. How many copies should she
make to maximize her sales?
4. A garden has 200 pounds of watermelons
growing in it. Every day, the total amount of
watermelon increases by 5 pounds. At the same
time, the price per pound of watermelon goes
down by 1¢. If the current price is 90¢ per
pound, how much longer should the
watermelons grow in order to fetch the highest
price possible?
5. A farmer has 400 feet of fencing to make three
rectangular pens. What dimensions x and y will
maximize the total area?
6. Four rectangular pens will be built along a
river by using 150 feet of fencing. What
dimensions will maximize the area of the pens?
x x
, , − +1 000 5 000
1002
y �32x2 �
3242 � 2
′′ = +A xx
( ) 256
23
x3 � 64�128x2 � 2x � 0
′ = − +A xx
x( ) 128
22
–OPTIMIZATION–
119
y
x
y
x
river (no fence needed)
Calc2e_15_115-120.qxd 11/18/11 12:55 AM Page 119
7. The surface area of a can is Area = 2πr2 + 2πrh,
where the height is h and the radius is r. The
volume is Volume = πr2h. What dimensions
minimize the surface area of a can with volume
16π cubic inches?
8. A painter has enough paint to cover 600 square
feet of area. What is the largest square-bottom
box that could be painted (including the top,
bottom, and all sides)?
9. A box with a square bottom will be built to
contain 40,000 cubic feet of grain. The sides of
the box cost 10¢ per square foot to build, the
roof costs $1 per square foot to build, and the
bottom will cost $7 per square foot to build.
What dimensions will minimize the building
costs?
10. A printed page will have a total area of 96
square inches. The top and bottom margins
will be 1 inch each, and the left and right
margins will be 32 inches each. What overall
dimensions for the page will maximize the area
of the space inside the margins?
h
r
–OPTIMIZATION–
120
y
x
printed
area
1
1
in.
in.
in.2__3
in.2__3
Calc2e_15_115-120.qxd 11/18/11 12:55 AM Page 120
LE
SS
ON
THE INTEGRAL AND AREAS UNDER CURVES
A round the same time that many great mathematicians focused on figuring out the slopes of tangent
lines, other mathematicians were working on an entirely different problem. They wanted to be able
to compute the area underneath any curve , such as the one shown in Figure 16.1.y � f˛1x 2
16
121
x
y = f(x)
y
a b
What is this area?
Figure 16.1
Calc2e_16_121-126.qxd 11/18/11 12:56 AM Page 121
The curvy nature of the upper curve y = f(x)
presents a problem when finding area. But we know
how to find areas of rectangles. Therefore, the approach
we shall take is to approximate the region using better
and better rectangular staircases, as follows.
–THE INTEGRAL AND AREAS UNDER CURVES–
122
Area of rectangular staircase
Number of Rectangles Diagram (Dashed Portion of the Region)
1 Area =
Bad approximation of the area of
the original region.
2 Area =
Better approximation than with
1 rectangle, but not great.
4 Area =
Even better approximation.
+ ⋅ + ⋅f f( ) ( )4 1 5 1height base height base{ { { {
f f( ) ( )2 1 3 1height base height base{ { { {⋅ + ⋅
1 2 3 4 5 x
y
y = f(x)
f f( ) ( )3 2 5 2height base height base{ { { {⋅ + ⋅
1 3 5 x
y
y = f(x)
f ( )5 4height base{ {⋅
5 x
y
y = f(x)
1
Figure 16.2
If we keep using more rectangles to dissect the original
region, the rectangles gobble up more of the region,
and so the approximation to the actual area is better.
The number to which these approximate areas get
close as we do so is called the integral of f on [a,b], and
is denoted .
Note: The “elongated S” symbol (called the integral
sign), , is used because the process involves a sum.
The integrand f (x)dx resembles the form of what we
are adding, namely height × base. The dx portion
doesn’t actually enter into the computation, though.
Example
Evaluate the integral .
Solution
This represents the area between the curve ,
the x-axis, the line x � 0, and the line x � 4 (see Fig-
ure 16.3). This region happens to be a triangle with a
height of 2 and a base of 4. The area of the triangle is
. Thus, .�4
0
12
x dx � 41212 2 14 2 � 4
y �12
x
�4
0
12
x dx
�
x
y
1 2 3
4
5
1
2
3
4
y = x_21
Figure 16.3
� f(x)dxb
a
Calc2e_16_121-126.qxd 11/18/11 12:56 AM Page 122
If the region is below the x-axis, the area is
counted negatively. Therefore, really repre-
sents “the area between the curve , the x-axis,
x � a, and x � b, where area below the x-axis is
counted negatively.”
Example
Evaluate the integrals f(x)dx, f(x)dx, f(x)dx,
and f(x)dx where the graph of is shown
in Figure 16.4.
Solution
First, f(x)dx = 2(2) = 4 because this area is a square
above the x-axis (see Figure 16.5).
Next, f(x)d = 2(2) + . (1) . 2 = 5 because this area
is a square plus a triangle (see Figure 16.6).
For f(x)dx, we must calculate how much area is
above the x-axis and how much is below (see Figure
16.7).
6
1�
1
2
4
1�
3
1�
y � f˛1x 23
1�
6
1�
4
1�
3
1�
y � f˛1x 2
–THE INTEGRAL AND AREAS UNDER CURVES–
123
1
2
3
1 2 3 4
y
x
y = f(x)
Figure 16.5
1
2
3
1 2 3 4
y
x
y = f(x)
Figure 16.6
1
2
3
1 2 3 4 5 6
y
x7–1–2
–1
–2
–3
y = f(x)
–4
Figure 16.4
� f(x)dxb
a
Calc2e_16_121-126.qxd 11/18/11 12:56 AM Page 123
There are 5 units of area above the x-axis and 4 units
below, so f(x)dx = 5 – 4 = 1.
Finally, f(x)dx represents a rectangle of area
4 that is entirely below the x-axis. Thus, f(x)dx =
(–4)(1) = –4 (see Figure 16.8).
Practice
Evaluate the following integrals.
Use the following graph to solve practice problems 1,
2, and 3.
1. f(x)dx 2. f(x)dx 3. f(x)dx
Use the following graph to solve practice problems 4,
5, and 6.
4. g(x)dx 5. g(x)dx 6. g(x)dx6
0�
6
4�
4
0�
2
0�
1
0�
2
1�6
1�
7
6�
6
1�
–THE INTEGRAL AND AREAS UNDER CURVES–
124
1
2
3
1 2 3 4 5
y
x
y = f(x)
5
1
2
3
1 2 3 4 6
y
x7–1–2
–1
–2
–3
y = f(x)
–4
5 square units of areaabove the x-axis
4 square units ofarea below the x-axis
Figure 16.7
1
2
3
1 2 3 4 5
6
y
x
7
–1–2–1
–2
–3
y = f(x)
–4
Figure 16.8
1
2
3
1 2 3 4 5
y
x
y = g(x)
6
Calc2e_16_121-126.qxd 11/18/11 12:56 AM Page 124
Use the following graph to solve practice problems 7,
8, and 9.
7. h(t)dt 8. h(t)dt 9. h(t)dt
Use the following graph to solve practice problems 10
through 12.
10. k(x)dx 11. k(x)dx 12. k(x)dx
For questions 13 through 18, compute the integral.
13. (x + 2)dx 14. 2dx 15. (t – 3)dt
16. xdx 17. 2xdx 18. (2x – 2)dx
You might have noticed that:
f(x)dx + f(x)dx = f(x)dx
The area between a and c is the area from a to b plus
the area from b to c, assuming, of course, that
a � b � c (see Figure 16.9).
Similarly, f(x)dx – f(x)dx = f(x)dx.
We can use these to perform calculations, even when
the exact functions are unknown.
Example
If f(x)dx = 7 and f(x)dx = 15, then what is
f(x)dx?
Solution
f(x)dx = f(x)dx + f(x)dx
= 7 + 15 = 22
10
3�
5
3�
10
3�
10
3�
10
3�
5
3�
b
a�
c
b�
c
a�
c
a�
c
b�
b
a�
8
0�
6
1�
5
–2�
5
1�
4
1�
4
0�
5
4�
6
4�
7
0�
6
4�
4
–1�
6
–1�
–THE INTEGRAL AND AREAS UNDER CURVES–
125
x
y = f(x)
y
a b c
Figure 16.9
1
1 2 3 4 5t
y = h(t)
6–1–1
–2
1
1 2 3 4 5x
y = k(x)
6–1–1
–2
2
7
Calc2e_16_121-126.qxd 11/18/11 12:56 AM Page 125
Example
If g(x)dx = 38 and g(x)dx = –12, then what is
g(x)dx ?
Solution
g(x)dx = g(x)dx – g(x)dx
= 38 – (–12) = 50
Practice
For questions 19 through 21, suppose f(x)dx ,
f(x)dx , and f(x)dx . Evaluate the following.
19. f(x)dx 20. f(x)dx 21. f(x)dx
For questions 22 through 24, suppose g(t)dt = –3,
g(t)dt = 8, and g(t)dt = –10. Evaluate the
following.
22. g(t)dt 23. g(t)dt 24. g(t)dt
For questions 25 through 27, suppose h(x)dx = 20,
h(x)dx = 12, and h(x)dx = –5. Evaluate the
following.
25. h(x)dx 26. h(x)dx 27. h(x)dx
For questions 28 through 30, suppose j(x)dx = 2,
j(x)dx = 3, j(x)dx = –4, and j(x)dx = 12.
Evaluate the following.
28. j(x)dx 29. j(x)dx 30. j(x)dx2
–1�
2
0�
3
2�
3
–1�
2
1�
1
0�
0
–1�
11
10�
10
1�
11
1�
10
–2�
1
–2�
11
–2�
10
5�
10
1�
14
5�
5
1�
14
10�
14
1�
11
0�
11
6�
7
0�
11
6�
7
6�
6
0�
10
8�
10
0�
8
0�
8
0�
10
8�
10
0�
–THE INTEGRAL AND AREAS UNDER CURVES–
126
Calc2e_16_121-126.qxd 11/18/11 12:56 AM Page 126
LE
SS
ON
THE FUNDAMENTALTHEOREM OF CALCULUS
Here comes the resounding climax of calculus. It would be best to read this lesson with some bombastic
orchestral music like that of Wagner or Orff! The initial question here is innocent enough: If we make
a function using that “area under a curve” stuff, what would its derivative be? To make this precise,
suppose that our curve is (see Figure 17.1). We use the variable t in order to save x for something
else later.
y � f1t 2
17
127
t
y = f(t)
y
Figure 17.1
Calc2e_17_127-132.qxd 11/18/11 12:58 AM Page 127
Now let our “area under the curve function” be
� the area under the curve between 0
and some point x. That is, . This
area is illustrated in Figure 17.2.
ExampleIf and , then what is g(3)?
Solution
� the area beneath the
curve from 0 to 3. The graph of f (t) � 2t
and the region are shown in Figure 17.3. This region
is a triangle with base 3 and height 6, so
.
Practice
For questions 1 through 6, suppose and
. Evaluate the following.
(Hint: The area of a trapezoid with bases b1 and b2 and
height h is .
1.
2.
3.
4.
5.
6. g15 2
g14 2
g13 2
g12 2
g11 2
g10 2
A h b b= +1
2 1 2( )
g1x 2 � �x
0
f˛1t 2 dt
f t t( ) = +3 1
g13 2 � �3
0
2t dt �1213 2 16 2 � 9
y � 2t
g13 2 � �3
0
f˛1t 2 dt � �3
0
2t dt
g1x 2 � �x
0
f˛1t 2 dtf˛1t 2 � 2t
g1x 2 � �x
0
f˛1t 2 dt
y � f˛1t 2g1x 2
–THE FUNDAMENTAL THEOREM OF CALCULUS–
128
t
y = f(t)
y
x
This area is g(x)
Figure 17.2
1
2
3
4
5
6
–11 2 3–1
y
t4 5
y = f(t) = 2t
(3,6)
Figure 17.3
Calc2e_17_127-132.qxd 11/18/11 12:58 AM Page 128
For questions 7 through 12, suppose and
. Evaluate the following.
7.
8.
9.
10.
11.
12.
Now that you are familiar with , we
can answer the next question: What is the derivative of
?
Begin with the definition of the derivative.
Use .
Use .
Now the integral represents the
skinny little area just to the right of point x (see Figure
17.4). This is almost a rectangle with a base of h and a
height of , so the integral is almost
. As h goes to zero, this approximation gets
better. Therefore,
� lim
˛
hS0 h # f˛1x 2
h� lim
˛
hS0 f˛1x 2 � f˛1x 2
g¿ 1x 2 � lim
hS0
�x�h
x
f˛1t 2 dt
h
h # f˛1x 2
�x�h
x
f˛1t 2 dtf˛1x 2
�x�h
x
f˛1t 2 dt
g¿ 1x 2 � limhS0
�x�h
x
f˛1t 2 dt
h
�c
a
f˛1t 2 dt � �b
a
f˛1t 2 dt � �c
b
f˛1t 2 dt
g¿ 1x 2 � limhS0
�x�h
0
f˛1t 2 dt � �x
0
f˛1t 2 dt
h
g1x 2 � �x
0
f˛1t 2 dt
g¿ 1x 2 � limhS0
g1x � h 2 � g1x 2
h
g1x 2
g1x 2 � �x
0
f˛1t 2 dt
g15 2
g14 2
g13 2
g12 2
g11 2
g10 2
g1x 2 � �x
0
f˛1t 2 dt
f˛1t 2 � 7
–THE FUNDAMENTAL THEOREM OF CALCULUS–
129
t
y = f(t)
y
x x + h
x
x + hf(t) dt
(x, f(x))
Figure 17.4
Calc2e_17_127-132.qxd 11/18/11 12:58 AM Page 129
What does this mean? It means that the deriva-
tive of the function , which represents “the area
under the curve from 0 to x,” is the very function
used to draw the curve. It came as an amazing surprise
to the world of mathematics that the process of find-
ing the slope of a tangent line and the process of find-
ing the area under a curve were such inverses. In order
to find the area under a curve , we need to
find a function whose derivative is .
We can use this to evaluate using the
Fundamental Theorem of Calculus.
For example, the derivative of is
. Thus, the area under between
x � 3 and x � 5 is
. This is exactly the area of the trapezoid
under the line between x � 3 and x � 5.
Example
The derivative of is . Use this
to evaluate .
SolutionBy the Fundamental Theorem of Calculus,
, where . Thus,
.
Even if we had drawn out the graph of ,
how would we have been able to guess that the area of
the two shaded curves add up to exactly three? This is
why the Fundamental Theorem of Calculus is so
powerful! (See Figure 17.5.)
y � x2
�1312 23 �
131�1 23 �
83
�13
� 3
�2
�1
x2 dx � g12 2 � g1�1 2
g¿ 1x 2 � f˛1x 2�b
a
f˛1x 2 dx � g1b 2 � g1a 2
�2
�1
x2 dx
g¿ 1x 2 � x2g1x 2 �13
x3
y � 2x
52 � 32 � 16
�5
3
2x dx � g15 2 � g13 2 �
f 1x 2 � 2xg¿ 1x 2 � 2x
g1x 2 � x2
�b
a
f˛1x 2 dx
f˛1x 2g1x 2
y � f˛1x 2
f˛1x 2
g1x 2
130
THE FUNDAMENTAL THEOREM OF CALCULUS
The Fundamental Theorem of Calculus can be written as follows:
f(x)dx = f(x)dx – f(x)dx = g(b) – g(a), where g′(x) = f (x)a
0�
b
0�
b
a�
1
2
3
–1
–2–3 1 2 3–1
y
x
4
y = x2
Figure 17.5
Calc2e_17_127-132.qxd 11/18/11 12:58 AM Page 130
Example
If , then . Use this to evaluate
4x3dx.
Solution
4x3dx = g(1) – g(–1)
The answer is zero because there is exactly as much
area above the x-axis (which counts positively) as there
is below the x-axis (which counts negatively).
Practice
For questions 13 through 16, use
to evaluate the following.
13. (2x + 1)dx
14. (2x + 1)dx
15. (2x + 1)dx
16. (2x + 1)dx
For questions 17 through 20, use to
evaluate the following.
17. dx
18. dx
19. dx
20. dx
For questions 21 through 24, use
to evaluate the following.
21. cos(x)dx
22. cos(x)dx
23. cos(x)dx
24. cos(x)dx2π
3π2
�
π
3π4
�
π2
π6
�
π4
0�
ddx
x x(sin( )) cos( )=
x100
0�
x9
4�
x4
0�
x1
0�
ddxa
23
x 32b � 2x
4
0�
6
2�
1
–3�
3
1�
ddx
x x x( )2 2 1+ = +
� 14 � 1�1 24 � 1 � 1 � 0
1
–1�
1
–1�
g¿ 1x 2 � 4x3g1x 2 � x4
131
–THE FUNDAMENTAL THEOREM OF CALCULUS–
Calc2e_17_127-132.qxd 11/18/11 12:58 AM Page 131
Calc2e_17_127-132.qxd 11/18/11 12:58 AM Page 132
LE
SS
ON
ANTIDIFFERENTIATION
The Fundamental Theorem of Calculus shows that the area under the curve, , can be calculated
using a function whose derivative is . Symbolically,
Because of this, the symbol , without the limits of integration, is used to represent the opposite of
taking the derivative. An integral like is called a definite integral because it represents a definite area.
An integral like is called an indefinite integral because it represents another function.
Thus, means the antiderivative of or “the function whose derivative is .” For exam-
ple, asks “whose derivative is ?” This could be because . However, it could also
be because . In fact, because the derivative of a constant is zero, could be
plus any constant. Therefore, we write where c is any constant.�2x dx � x2 � cx2
�2x dxddx1x2 � 5 2 � 2xx2 � 5
ddx1x2 2 � 2xx22x�2x dx
f˛1x 2f˛1x 2� f˛1x 2 dx
� f˛1x 2 dx
�b
a
f˛1x 2 dx
�
�b
a
f˛1x 2 dx � �g1x 2�ba � g1b 2 � g1a 2
g¿ 1x 2 � f˛1x 2g1x 2�
b
a
f˛1x 2 dx
18
133
Calc2e_18_133-138.qxd 11/18/11 12:59 AM Page 133
We usually simply write with-
out actually saying that c stands for “some constant.” In
many ways, the “plus c” is the trademark of the indef-
inite integral because every problem that begins
with ends with .
If we are dealing with a definite integral like
, then it does not matter what constant we
use. For example:
The “plus c” will always cancel out in the subtraction.
As such, we simply use c � 0 and write:
Example
Use to
evaluate and
.
Solution
Because , we
know that:
Also,
=
=
The general process for finding antiderivatives of
powers is fairly simple. To take the derivative of
, we first multiply by the exponent 5, and
then we subtract one from the exponent. Thus,
.
To antidifferentiate , we must do the
exact opposite of this process. First, we add one to the
exponent, and then we divide the result by the new
exponent. Thus, . In
general, we write:
if n � �1�xnˇdx �
xn�1
n � 1� c
�5x4ˇdx �
5x4�1
4 � 1� c � x5 � c
�5x4ˇdx
f ¿ 1x 2 � 5x4
f˛1x 2 � x5
8 � 40 � 6 � 11 � 10 � 3 2 � 54 � 14 � 40
1 11 23 � 10 # 11 2 � 3 # 11 2 21 12 23 � 10 # 12 22 � 3 # 12 2 2 �
�2
1
13x2 � 20x � 3 2 dx � �x3 � 10x2 � 3x�21
� x3 � 10x2 � 3x � c� 13x2 � 20x � 3 2 dx
ddx1x3 � 10x2 � 3x 2 � 3x2 � 20x � 3
�2
1
13x2 � 20x � 3 2 dx
� 13x2 � 20x � 3 2 dx
ddx1x3 � 10x2 � 3x 2 � 3x2 � 20x � 3
�5
3
2x dx � �x2�53 � 52 � 32 � 25 � 9 � 16
= + − − =25 9 16 c c/ /
� 152 � c 2 � 132 � c 2
�5
3
2x dx � �x2 � c�53
�5
3
2x dx
� c� 1 p . 2 dx
�2x dx � x2 � c
134
BRACKET NOTE
The brackets are just a way of keeping track of the limits of integration a and b before they are pluggedinto g(x) and subtracted.
K[ ]ab
Calc2e_18_133-138.qxd 11/18/11 12:59 AM Page 134
Example
Evaluate x7dx.
Solution
Example
Evaluate
Solution
Example
Evaluate x3dx.
Solution
Practice
Evaluate the following integrals.
1. x 4dx
2. x12dx
3. u 6du
4. x5dx
5. xdx
6. t –3dt
ADDITION AND CONSTANT
�
9
–1�
1
–1����
�14
# 16 �14
# 0 � 4 � 0 � 4
= ⋅ − ⋅14
214
04 4
x dx x3 4
0
21
4=
2
0�
2
0�
�x
12 �1
12 � 1
� c �x
32
32
� c �23
x 3 2 � c
x dx12�xdx =�
xdx�
x dxx
c x c77 1
8
7 1
1
8
=+
+ = ++
�
�
135
VERIFICATION HINT
You can verify your answer by taking its derivative. If the derivative of your answer is what you were trying tointegrate, then you are correct.
The derivative of is . This verifies that
.xdx x c = +23
32�
ddx
x c x x x23
23
32
032
12
12 +
= ⋅ + = =23
32x c +
Calc2e_18_133-138.qxd 11/18/11 12:59 AM Page 135
RULES
g x dx( )b
a�f x dx( ) +
b
a�f x g x dx( ( ) ( )) + =
b
a�f x dx( ) ,
b
a�cf x dx c( ) =
b
a�
136
7. t–2dt
8.
9.
10.
11.
12.
Just as with derivatives, constants can stand aside,
and the terms of sums can be dealt with separately.
Example
Evaluate .
Solution
Example
Evaluate .
Solution
Example
Evaluate .
SolutionIt always helps to first write everything in exponential
form.
= + = 261
1283 329128,
= +
− +
4 82
2564
21
( )
= ⋅
−
−
= +
−6
23
84
42
32
4
1
4
32
4
1
4
xx
xx
6 8 512x x dx − −( )
4
1�6
85
xx
dx −
=4
1�
68
5x
xdx −
4
1�
214
812
7
12
4 7
4 2
4 2
t t t c
t t t c
= ⋅ − ⋅ + +
= − + +
7dt �tdt + �83t dt − �( )2 8 7 23t t dt− + = �
( )2 8 73t t dt− + �
513
53
2 3 3x dx x c x c= +
= + �5 52x dx =�
5 2x dx�
8dx4
–1�
5dx�
udu3�x dx4�
x dx53�
–1
–4�
Calc2e_18_133-138.qxd 11/18/11 12:59 AM Page 136
Practice
Evaluate the following integrals.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
The antiderivatives of , , and
follow directly from their derivatives:
because
because
because
The integral of will have to wait until Les-
son 20, though we can use the fact that
right now. We are inclined to say that
, but this is not entirely correct. The
derivative of is
as well. It does not matter if the
x inside the natural logarithm is positive or negative, so
we can generalize with the absolute value |x|.
Incidentally, this nicely fills a hole in an earlier
formula:
if n � �1x dxx
ncn
n
=+
++ 1
1�
1x
dx x c = +ln�
=−
− =⋅ 1
11
x x( )
�1
�x# d
dx1�x 2
ddx1 ln1�x 2 2ln1�x 2
x cln( ) +
1x dx =�
ddx1 ln1x 2 2 �
1x
ln1x 2
ddx1�cos1x 2 2 � sin1x 2
( )10 4 14u u du − +2
0�
ddx1sin1x 2 2 � cos1x 2
cos( )x dx s x c in( ) = +�
ddx1ex 2 � exe dx e cx x = +�
cos1x 2sin1x 2ex
( )3 8103
47x x dx− �
12 xdx9
1�
( )10 4 14u u du − +2
0�
2 43 2t t t dt− −− −( )�
( )1 2 − t dt3
0�
( )3 911 2t t t dt+ + �
( )3 42x dx+ 2
1�
12 3x dx2
0�
( )6 10 52x x dx− + �x x dx3 −( )�
8 2u du�
9 4x dx�
–ANTIDIFFERENTIATION–
137
Calc2e_18_133-138.qxd 11/18/11 12:59 AM Page 137
and if then
Example
Evaluate .
Solution
Example
Evaluate .
Solution
Example
Evaluate .
Solution
Practice
Evaluate the following integrals.
25.
26.
27.
28.
29.
30.
31.
32. 8cos( )x dx5π2
π6
�
4e dxxln(3)
ln(2)�
( )x e dxx +1
0�
5 2sin( )x e dxx+( )�( sin( ))θ θ θ + 2 d�
2u du
e3
e�
( )3 2 3e x dxx + �( cos( ))x x dx2 5− �
= + + + − +
= + + + − +
−
13
12
13
12
1
3 2 1 1
3 2
x x x x x c
x x x xx
c
ln
ln
x x x x x dx2 1 1 2+ + + +( )− − 0�
x xx x
dx22
11 1+ + + +
= �
x xx x
dx22
11 1+ + + +
�
= − + = − 1 5 5 6 5e e
= − − − ( ) ( )1 5 0 53 1 3 0e e
( ) [ ]3 5 52 301t e dt t et t− = −
1
0�
( )3 52t e dtt−1
0�
�3cos1x 2 � 5sin1x 2 � c
( sin( ) cos( ))3 5x x dx + =�
( sin( ) cos( ))3 5x x dx +�
xdx x c= +1
ln�x dx− =1 �n � �1
–ANTIDIFFERENTIATION–
138
Calc2e_18_133-138.qxd 11/18/11 12:59 AM Page 138
LE
SS
ON
INTEGRATION BY SUBSTITUTION
The opposite of the Chain Rule is an integration technique called substitution. Using the Chain Rule,
for example, the derivative of is � �
. The corresponding antiderivative is thus � . It is easy
to recognize this after seeing the derivative worked out, but how should we know this otherwise?
The mantra of the Chain Rule is “multiply by the derivative of the inside.” So the first step to undoing
it is to identify what “the inside” must have been. We substitute a new variable u for this and then try to rewrite
the whole integral in terms of u.
For example, when confronted by , we first notice that if we multiplied out the
fourth power, then it would be a polynomial that we know how to evaluate, but doing so would be very tire-
some! Instead, we guess that “the inside” is the stuff inside the parentheses, and substitute .u � 3x2 � 7
�240x13x2 � 7 24 dx
813x2 � 7 25 � c�240x13x2 � 7 24 dx240 3 72 4x x( )+
8 # 513x2 � 7 24 # 6xddx1813x2 � 7 25 2813x2 � 7 25
19
139
Calc2e_19_139-144.qxd 11/18/11 1:00 AM Page 139
To convert the integral entirely in terms of u, we
must actually get a du into the integrand. Because u =
3x2 + 7, we know that = 6x, so du = 6xdx, or equi-
alently, dx = . So what? The steps below will illus-
trate why this is so useful.
Start with the original integral.
Substitute and .
Simplify.
Note, in particular, that substituting in dx =
resulted in the cancellation of all remaining x’s in the
integrand, so that the integral is not entirely in terms
of u.
Evaluate.
Replace .
Thus, � , as
we saw earlier.
In general, try using something inside parenthe-
ses with u. If every x doesn’t cancel out when replacing
dx with the expression involving du, then try using
something else as u. Sometimes, the entire denomina-
tor can be used as u. Sometimes, nothing works and a
different technique must be tried.
Example
Evaluate .
SolutionIf we use the stuff inside the only set of parentheses,
then , and thus and .
Start with the original integral.
Substitute and .
Simplify.
Every x is gone, so we can evaluate.
Replace .
Thus, � . This can be
verified by differentiating �
� .x2sin1x3 2�131�sin1x3 2 # 3x2 2 � 0
ddx
x c− +
13
3cos( )
�13
cos1x3 2 � c�x2sin1x3 2 dx
�13
cos1x3 2 � c
u � x3
�13
cos1u 2 � c
� 13
sin1u 2 du
�x2sin1u 2 du3x2
dx �du3x2u � x3
�x2sin1x3 2 dx
dx �du3x2du � 3x2
dxu � x3
�x2sin1x3 2 dx
813x2 � 7 25 � c�240x13x2 � 7 24 dx
813x2 � 7 25 � c
u � 3x2 � 7
8u5 � c
du
x6
�40u4 du
�240x1u 24 du6x
dx �du6x
u � 3x2 � 7
�240x13x2 � 7 24 dx
du
x6
du
dx
–INTEGRATION BY SUBSTITUTION–
140
Before actually computing the integral, all of thex’s must be gone! You cannot integrate with mixedvariables.
Calc2e_19_139-144.qxd 11/18/11 1:00 AM Page 140
If we had been faced with in the last
example, then substituting would have
resulted in . This cannot be evaluated
because it is not entirely in terms of u. In fact, this inte-
gral is very difficult to solve and requires the advanced
technique of replacing with an infinitely long
polynomial called a power series. Many such integrals
exist that are difficult to solve. This book will focus on
the ones that can be evaluated with basic techniques.
Example
Evaluate .
SolutionBecause there are no parentheses, try using the
denominator: . Here, , so
and .
Start with the original integral.
Substitute and .
Simplify.
Every x is gone, so we can evaluate.
Replace .
Thus, � .
Basically, the goal is to find a u whose derivative
is essentially the rest of the integral, except for possibly
a constant, so that between the u and the du, every x
goes away. This leads to some clever tricks, as will be
demonstrated in the following examples.
Example
Evaluate .
SolutionHere, we use . This is not because it is in
parentheses but because its derivative makes
up the rest of the integral. Here, , so
.
Start with the original integral.
Substitute and .
�ux
1x du 2
dx � x duu � ln1x 2
� ln1x 2
x dx
dx � x du
du �1x
dx
dudx
�1x
u � ln1x 2
� ln1x 2
x dx
32
ln|2x � 7| � c� 32x � 7
dx
32
ln|2x � 7| � c
u � 2x � 7
32
ln|u| � c
� 32
# 1u
du
� 3u
du2
dx �du2
u � 2x � 7
� 32x � 7
dx
dx �du2
du � 2 dx
dudx
� 2u � 2x � 7
� 32x � 7
dx
sin1x3 2
�sin1u 2 du3x2
u � x3
�sin1x3 2 dx
–INTEGRATION BY SUBSTITUTION–
141
Calc2e_19_139-144.qxd 11/18/11 1:00 AM Page 141
Simplify.
Every x is gone, so we can evaluate.
Replace .
Thus, � .
Example
Evaluate .
Solution
Here, the trick is to use so that
and .
Start with the original integral.
Substitute and .
Simplify.
Every x is gone, so we can evaluate.
Replace .
Thus, � .
To use substitution on a definite integral, evalu-
ate the indefinite integral first, and then compute at the
limits.
Example
Evaluate .
Solution
First, evaluate the following indefinite integral.
Use u = x3, du = 3x2dx, and so .
Simplify.
Every x is gone, so we can evaluate the indefinite integral.
Replace u = x3.
Now, because , it follows that
x e dx e e e ex x2 3 3
0
113
13
13
13
11 0=
= − = −( )1
0�
x e dx e cx x2 3 313
= +�
13
3e cx +
13
e cu +
13
e duu�
x edux
u223
�dx
dux
=3 2
x e dxx2 3�
x e dxx2 31
0�
�14
cos41x 2 � csin( )cos ( )x x dx3�
�14
cos41x 2 � c
u � cos1x 2
�14
u4 � c
−u du3�
sin( )sin( )
x udu
x ⋅ −
3�
dx � �du
sin1x 2u � cos1x 2
sin( )cos ( )x x dx3�
dx � �du
sin1x 2dudx
� �sin1x 2
u � cos1x 2
sin( )cos ( )x x dx3�
121 ln1x 2 22 � c
ln( )x
xdx�
121 ln1x 2 22 � c
u � ln1x 2
12
u2 � c
udu�
–INTEGRATION BY SUBSTITUTION–
142
Calc2e_19_139-144.qxd 11/18/11 1:00 AM Page 142
You will know that you chose the wrong u if
either some of the variables x still remain or the sim-
plified integral will still be hard to solve. If this happens,
go back to the beginning and try a different u. Don’t
forget that many integrals, like those of the previous
lesson, don’t require substitution at all. Like much of
mathematics, learning to integrate often requires
patience and a knack that is developed with practice.
Practice
Evaluate the following integrals.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24. ee
dxx
x
2
21+�
sin( )
cos( )
x
xdx�tan( )x dx =�
ex
dxx�
1
x xdx
ln( )�
(ln( ))x
xdx
3
�
e e dxx xsin( )�
sin( )7 2x dx −�
4cos( )x dx�
cos( )4x dx�
sin ( )cos( )2 x x dx�
sin( )cos( )x x dx�
−+2
4 10xdx
�
x
xdx
( )4 52 3+ �
( )( )8 5 4 5 12 3x x x dx + + −�
6 1
3 2 1
3
4
x
x xdx
−
− +
�
2 3 4x x dxcos( )�
9 5
3 5
2
3
x
xdx
−−
�
15 − x dx1
0�
2 1x dx+�
x x dx3 23 1⋅ − �
( )x x dx3 9 4− + �
x x dx2 3 41( )− 1
0�
( )4 34 10x dx+ �
x x dx4 5 71( )+ �
–INTEGRATION BY SUBSTITUTION–
143
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Calc2e_19_139-144.qxd 11/18/11 1:00 AM Page 144
LE
SS
ON
INTEGRATION BY PARTS
The integral of the product of two functions is unfortunately not the product of the integrals. For
example, the integral is not ( ) .( ) . We
know this because the derivative of is, by the Product Rule, �
, which is not equal to . It is unfortunate that this does not work because,
if it did, evaluating integrals would be simple and would not require so many different techniques.
The integration technique that undoes the Product Rule is called integration by parts. We derive the for-
mula as follows.
The product rule for differentiating f(x) ⋅ g(x) says
.
Integrating both sides of the formula gives .
This simplifies to .g x f x dx( ) ( )′ ⋅�f x g x dx( ) ( )′ ⋅ + �f x g x( ) ( )⋅ =
f x g x g x f x dx( ) ( ) ( ) ( )′ ⋅ + ′ ⋅[ ]�ddx
f x g x dx( ) ( )⋅( ) =�
ddx
f x g x f x g x g x f x( ) ( ) ( ) ( ) ( ) ( )⋅( ) = ′ ⋅ + ′ ⋅
x # cos1x 2x x x x⋅ + ⋅sin( ) cos( )12
2
ddxa
12
x2sin1x 2 � cb12
x2sin1x 2 � c
x x c=
⋅( ) +sin( )1
22x dxcos( )�x dx�x x dx ⋅ cos( )�
20
145
Calc2e_20_145-150.qxd 11/18/11 1:02 AM Page 145
–INTEGRATION BY PARTS–
146
Now, we introduce two variables, u and v, as follows to simplify the formula:
Plug these into the above formula to get preceding.
Simply put, uv = vdu + udv. Move the first integral on the right-hand side to the left and voilà!, we have
the integration by parts formula:
vdu.�udv uv= −�
��
f x g x dxu dv
( ) ( ){ 124 34⋅ ′�g x f x dx
v du
( ) ( ){ 124 34⋅ ′ +�f x g x
u v
( ) ( ){ {⋅ =
u f xdu
dxf x du f x dx
v g xdv
dxg x dv g x dx
= = ′ = ′
= = ′ = ′
( ) ( ), ( )
( ) ( ), ( )
so that which is equivalent to
so that which is equivalent to
Work through the following examples to see how this
is applied to help us compute more complicated inte-
grals.
This can often be used to transform a difficult
integral into one that is solvable. For example, take
. This looks just like udv if and
. In order to use the formula, we
will need to get by differentiating u. Because
u = x, we know that , so . We will
also need to get v from dv by integrating. And because
dv = cos(x)dx, it must be that . Thus:
�
This is the correct answer, as can be verified by taking
the derivative � 1 ⋅ sin(x) +
cos(x) ⋅ x – sin(x) + 0 = x ⋅ cos(x).
Example
Evaluate .
SolutionThis cannot be solved by basic integration or by sub-
stitution. Since it is a product, there is a good chance
that integration by parts will work. First, try u = x. The
dv must then be everything else after the integral sign,
so . After differentiating u and integrating
dv, we get:
And:
Note: Although there is technically a “+ c ” here, we will
wait until the end to add the “+ c” once all integrals
have been computed.
v � ex
dv � ex dx
du � dx
u � x
dv � ex dx
xe dxx�
ddx1xsin1x 2 � cos1x 2 � c 2
xsin1x 2 � cos1x 2 � c
x dxsin( )�= − sin( ) x x
vdu�= − uv
udv �x x dx ⋅ =cos( )�v � sin1x 2
du � dxdudx
� 1
du
dv � cos1x 2 dx
u � x�x x dx ⋅ cos( )�
The dx must be part of what you decide to letequal dv.
Calc2e_20_145-150.qxd 11/18/11 1:02 AM Page 146
Thus, using the integration by parts formula
, we evaluate as follows:
�
�
�
�
Example
Evaluate .
SolutionHere, since we don’t know how to integrate ln(x)
(yet!), we can’t let it equal dv. So, we set and
. Thus:
And:
And then we evaluate.
�
�
�
�
�
Example
Evaluate .
SolutionBecause there seems to be only one part to this inte-
gral, one wouldn’t think to try integration by parts
first. However, because nothing else will work, we can
try . The only thing left for the dv is dx, so
we use dv � dx, which leads to v � x.
And:
dv � dx
v � x
And now evaluate as follows.
�
�
�
�
�
Sometimes, integration by parts needs to be done
more than once to compute an integral.
xln1x 2 � x � c
xln1x 2 � �1 dx
ln1x 2 # x � �x # 1x
dx
uv � �v du
�u dv� ln1x 2 dx
du �1x
dx
u � ln1x 2
u � ln1x 2
� ln1x 2 dx
14
x4ln1x 2 �1
16x4 � c
14
x4ln1x 2 � � 14
x3 dx
ln1x 2 # 14
x4 � � 14
x4 # 1x
dx
uv � �v du
�u dv�x3ln1x 2 dx
v �14
x4
dv � x3 dx
du �1x
dx
u � ln1x 2
dv � x3 dx
u � ln1x 2
�x3ln1x 2 dx
xex � ex � c
xex � �ex dx
uv � �v du
�u dv�xex dx
�u dv � uv � �v du
–INTEGRATION BY PARTS–
147
Don’t forget that there is a minus in the formula!
Calc2e_20_145-150.qxd 11/18/11 1:02 AM Page 147
Example
Evaluate .
Solution
Here, , so , and , so
.
�
�
�
�
In order to compute , we have to
use integration by parts a second time, but this time,
with and .
And:
Now we evaluate as follows.
�
�
�
�
�
Thus,
�
The final example utilizes a clever trick.
Example
Evaluate .
SolutionThe terms ex and sin(x) are equally good candidates for
u. Let us use
u = ex and dv = sin(x)dx.
And:
v � �cos1x 2
dv � sin1x 2 dx
du � exˇdx
u � ex
�exsin1x 2 dx
x2sin1x 2 � 2xcos1x 2 � 2sin1x 2 � c
�x2cos1x 2 dx
x2sin1x 2 � 2xcos1x 2 � ��2cos1x 2 dx
x2sin1x 2 � 12x # 1�cos1x 2 2 � � 1�cos1x 2 2 # 2 dx 2
x2sin1x 2 � 1uv � �v du 2
x2sin1x 2 � �u dv
x2sin1x 2 � �2xsin1x 2 dx�x2cos1x 2 dx
v � �cos1x 2
dv � sin1x 2 dx
du � 2 dx
u � 2x
dv � sin1x 2 dxu � 2x
�2xsin1x 2 dx
x2sin1x 2 � �2xsin1x 2 dx
x2sin1x 2 � �sin1x 2 # 2x dx
uv � �v du
�u dv�x2cos1x 2 dx
v � sin1x 2
dv � cos1x 2 dxdu � 2x dxu � x2
�x2cos1x 2 dx
–INTEGRATION BY PARTS–
148
Calc2e_20_145-150.qxd 11/18/11 1:02 AM Page 148
Now, evaluate as follows.
�
�
�
�
To evaluate , we use integration by parts
again, but with u = ex and dv = cos(x)dx.
And:
And then the evaluation:
�
�
�
�
Thus, we have:
�
Here is the moment of despair: To evaluate
, we need to be able to evaluate
! And yet, the trick here is to bring both
integrals to one side of the equation:
� �
�
�
= − +( ) +1
2e x x cx cos( ) sin( )
121�excos1x 2 � exsin1x 2 2 � c�exsin1x 2 dx
�excos1x 2 � exsin1x 22�exsin1x 2 dx
�excos1x 2 � exsin1x 2
�exsin1x 2 dx�exsin1x 2 dx
�exsin1x 2 dx
�exsin1x 2 dx
�excos1x 2 � exsin1x 2 � �exsin1x 2 dx
�exsin1x 2 dx
�excos1x 2 � exsin1x 2 � �sin1x 2 # ex dx
�excos1x 2 � uv � �v du
�excos1x 2 � �u dv
�excos1x 2 � �excos1x 2 dx�exsin1x 2 dx
v � sin1x 2
dv � cos1x 2 dx
du � exˇdx
u � ex
�excos1x 2 dx
�excos1x 2 � �excos1x 2 dx
ex1�cos1x 2 2 � � 1�cos1x 2 2 # ex dx
uv � �v du
�u dv�exsin1x 2 dx
–INTEGRATION BY PARTS–
149
Calc2e_20_145-150.qxd 11/18/11 1:02 AM Page 149
Practice
Evaluate the following integrals using integration by
parts, substitution, or basic integration.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22. e x dxx cos( )�
cos( ) ln sin( )x x dx⋅ ( )�
sin( ) cos( )x x dx�
e
xdx
x1
2�
e x dxx− cos( ) sin( )�
13x
dx�
15
x xdx
ln( )( )�
cos ( ) sin( )3 x x dx⋅�
xe dxx−�
x x dx − 1�
x dx − 1�
1x
x dx+
ln( )�
( )ln( )x x x dx3 3 1+ − �
ee
dxx
x
3
3 9+�
x e dxx2 13 + �
( sin( ))x x dx2 + �
x x dx2 sin( )�
ln( )x
xdx�
( )cos( )x x dx + 3�
x x dxsin( )2�
x x dxsin( )�
x x dx5 ln( )�
–INTEGRATION BY PARTS–
150
Calc2e_20_145-150.qxd 11/18/11 1:02 AM Page 150
If you have completed all 20 lessons in this book, you are ready to take the posttest to measure your progress.
The posttest has 50 multiple-choice questions covering the topics you studied in this book. Although the
format of the posttest is similar to that of the pretest, the questions are different.
Take as much time as you need to complete the posttest. When you are finished, check your answers with
the answer key that follows the posttest. Along with each answer is a number that tells you which lesson of
this book teaches you about the calculus skills needed for that question. Once you know your score on the
posttest, compare the results with the pretest. If you scored better on the posttest than you did on the pretest,
congratulations! You have profited from your hard work. At this point, you should look at the questions you
missed, if any. Do you know why you missed the question, or do you need to go back to the lesson and review
the concept?
If your score on the posttest doesn’t show much improvement, take a second look at the questions you
missed. Did you miss a question because of an error you made? If you can figure out why you missed the prob-
lem, then you understand the concept and simply need to concentrate more on accuracy when taking a test.
If you missed a question because you did not know how to work the problem, go back to the lesson and spend
more time working that type of problem. Take the time to understand basic calculus thoroughly. You need a
solid foundation in basic calculus if you plan to use this information or progress to a higher level. Whatever
your score on this posttest, keep this book for review and future reference.
POSTTEST
151
Calc2e_21_151-164_Post.qxd 11/18/11 1:03 AM Page 151
Calc2e_21_151-164_Post.qxd 11/18/11 1:03 AM Page 152
–LEARNINGEXPRESS ANSWER SHEET–
153
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
1.2.3.4.5.6.7.8.9.
10.11.12.13.14.15.16.17.
18.19.20.21.22.23.24.25.26.27.28.29.30.31.32.33.34.
35.36.37.38.39.40.41.42.43.44.45.46.47.48.49.50.
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
a b c d
Calc2e_21_151-164_Post.qxd 11/18/11 1:03 AM Page 153
Calc2e_21_151-164_Post.qxd 11/18/11 1:03 AM Page 154
155
Posttest
1. Evaluate when .
a. �12
b. �10
c. �4
d. 4
2. Simplify when .
a.
b.
c.
d.
3. Evaluate (g ° h)(x) when
and .
a.
b.
c.
d.
4. What is the domain of ?
a. all real numbers greater than or equal to –1
b. all real numbers except 0
c. all real numbers except –1 and 0
d. all nonzero real numbers greater than or
equal to –1
Use the following graph for problems 5 through 7.
5. Where does have a local maximum?
a. x � 0
b. x � 1
c. x � 2
d. x � 3
6. On what interval(s) is decreasing?
a. (1,2) and (2,3)
b. (�q,2)
c. (�q,0)
d. (q,1) and (3,q)
g1x 2
g1x 2
f xx
x( )
= + 1
1x2 �
5x
� 1
1x2 � 5x � 1
x � 5 �1x
x xx
2 5 11+ + +
h1x 2 �1x
g1x 2 � x2 � 5x � 1
2x3 � 3x2 � x
2x2 � 3x
4x2 � 2x � 2
4x2 � 6x � 2
f˛1x 2 � x2 � xf˛12x � 1 2
f˛1x 2 � x3 � 2xf˛1�2 2
–POSTTEST–
1
2
3
–1–2 1 2 3 4–1
–2
–3
y
x
y = g(x)
Calc2e_21_151-164_Post.qxd 11/18/11 1:03 AM Page 155
–POSTTEST–
156
7. What is the slope of the line passing through
(2,�4) and (1,7)?
a. –
b. –11
c. 11
d. 3
8. Simplify .
a. 7
b. 12
c. 16
d. 64
9. Simplify .
a. �8
b. 4
c.
d.
10.Solve for x when .
a.
b. 2.5
c.
d.
11. Evaluate .
a. 1
b.
c.
d.
12. Evaluate .
a. –
b.
c. –
d.
13. Evaluate .
a. 0
b. 1
c. –2
d. undefined
limxS2
x2 � 5x � 6x2 � 2x � 3
2
2
2
2
3
2
3
2
sin a4p3b
3
2
2
2
12
cos ap
4b
ln a52b
104
ln110 2
ln14 2
4x � 10
116
14
16�12
43
111
Calc2e_21_151-164_Post.qxd 11/18/11 1:03 AM Page 156
14. Evaluate .
a. –2
b.
c. 1
d. undefined
15. Evaluate .
a. 5
b. �qc. qd. –4
16. What is the slope of the tangent line to f(x) = x2
at x � 3?
a. 2
b. 6
c. 9
d. 2x
17. What is the slope of the tangent line to
at x � 2?
a. �7
b. �3
c. 1
d. 4
18. What is the derivative of
?
a. g′(x) = 0
b. g′(x) = – 1
c. g′(x) =
d. g′(x) =
19. Suppose that after t seconds, a falling rock is
feet off the ground.
How fast is the rock traveling after 2 seconds?
a. 10 feet per second
b. –32 feet per second
c. 59 feet per second
d. 146 feet per second
20. Differentiate y = + 4sin(x).
a.
b. + 4cos(x)
c. + 4sin(x)
d.1
24
xx − cos( )
dydx
=
x
2
dydx
=
xdydx
=
1
24
xx + cos( )
dydx
=
x
s1t 2 � �16t2 � 5t � 200
32x3 � 30x2 � 3
32x3 � 10x2 � 3x � x
8x4 � 10x3 � 3x
g1x 2 � 8x4 � 10x3 � 3x � 1
y � 4x � 7
limx
xx→ +
+−1
412
14
limxS3
x2 � 5x � 6x2 � 2x � 3
–POSTTEST–
157
Calc2e_21_151-164_Post.qxd 11/18/11 1:03 AM Page 157
21. What is the derivative of ?
a. f ′(x) =
b. f ′(x) =
c. f ′(x) =
d. f ′(x) =
22. Differentiate .
a.
b.
c.
d.
23. Differentiate .
a. g′(x) =
b. g′(x) =
c. g′(x) =
d. g′(x) =
24. What is the derivative of ?
a. f ′(x) =
b. f ′(x) =
c. f ′(x) =
d. f ′(x) = 0
25. What is the slope of the line that is tangent to
at x � 2?
a. 8
b. 12
c. 24
d. 48
26. Differentiate y = .
a.
b.
c.
d. –2x2cos(x2) + sin(x2)
27. Find when .
a.
b.
c.
d. 11x y y+( )sec( ) tan( )
dydx
=
x
y1 2 + sec ( )
dydx
=
1
1 2x y( sec ( )) +dydx
=
1x
� sec21x 2dydx
=
tan1y 2 � y � ln1x 2 � 1dy
dx
dydx
=
2x2cos1x2 2dydx
=
2x2cos1x2 2 � sin1x2 2dydx
=
xcos1x2 2 � sin1x2 2dydx
=
xsin1x2 2
y � 1x2 � 2 23
sec1x 2tan1x 2
tan21x 2
sec1x 2
f˛1x 2 � sec1x 2
�1x2 � 5x 2sin1x 2 � 12x � 5 2cos1x 2
1x2 � 5x 22
12x � 5 2cos1x 2 � 1x2 � 5 2sin1x 2
1x2 � 5 22
�sin1x 2
2x � 5
sin1x 2
2x � 5
g1x 2 �cos1x 2
x2 � 5x
xex�1dydx
=
1x � 1 2exdydx
=
xexdydx
=
exdydx
=
y � xex
5ex � 2x
5xex�1 � 2x
5xex�1 �2x
5ex �2x
f˛1x 2 � 5ex � 2ln1x 2
–POSTTEST–
158
Calc2e_21_151-164_Post.qxd 11/18/11 1:03 AM Page 158
28. Find when .
a.
b.
c.
d.
29. What is the slope of the curve
at (1,2)?
a.
b.
c.
d. 6
30. The volume of a sphere is . If the
radius increases by 3 meters per second, how
fast is the volume changing when
meters?
a.
b.
c.
d.
31. If a 10-foot ladder slides down a wall at 2 feet
per minute (see the figure that follows), how
fast does the bottom slide when the top is 6 feet
up?
a. foot per minute
b. feet per minute
c. 2 feet per minute
d. 12 feet per minute
32. Evaluate .
a. 2
b. 3
c.
d. ∞
33. Where does have a vertical asymptote?
a. y � 0
b. x � 1
c. x � e
d. no vertical asymptote
y � ex
75
limxSq
3x2 � 7x � 2x2 � 5x � 1
32
2
3
1 200, π m
sec
3
4 000, π m
sec
3
4 000
3
, πm
sec
3
2403
π msec
r � 10
V r= 4
33π
306
3
2
311
y3 � y � 3x � 3
y2 � 2xy
x2 � 2xydydx
=
y2 � 2xy
x2 � 2xydydx
=
y � 2x
x � 2ydydx
=
1y
dydx
=
x2y � xy2dy
dx
–POSTTEST–
159
ladder
10 feet
wall
y
x ground
Calc2e_21_151-164_Post.qxd 11/18/11 1:03 AM Page 159
34. Evaluate .
a. 5
b. 0
c. qd. –3x3
35. On what interval(s) is f(x) = x3 + 6x2 – 15x + 2
decreasing?
a. (4,5)
b. (�5,1)
c. (2,6) and (15,q)
d. (�q,�5) and (1,q)
36. Which of the following is the graph of
?
a.
b.
c.
d.
y � 3x � x3
limxSq
x5 � 3x3
ex � 1
–POSTTEST–
160
1
2
3
–1
–2–3 2 3
y
x
–2
–3
1–1
1
2
3
–1
–2–3 2 3
y
x
–2
–3
1–1
1
2
3
–1
–2–3 2 3
y
x
–2
–3
1–1
1
2
3
–1
–2–3 2 3
y
x
–2
–3
1–1
Calc2e_21_151-164_Post.qxd 11/18/11 1:03 AM Page 160
37. If up to 30 apple trees are planted on an acre,
each will produce 400 apples a year. For every
tree over 30 on the acre, each tree will produce
10 apples less each year. How many trees per
acre will maximize the annual yield?
a. 5 trees
b. 32 trees
c. 35 trees
d. 40 trees
38. An enclosure will be built, as depicted, with
100 feet of fencing. What dimensions will
maximize the area?
a. x = 20 ft, y = 20 ft
b. x = 25 ft, y = 10 ft
c. x = 25 ft, y = 25 ft
d. x = 50 ft, y = 10 ft
39. If and , then
what is ?
a. 6
b. 9
c. 10
d. 16
40. What is ?
a. –4
b. 0
c. 4
d. 6
�4
0
f˛1x 2 dx
�10
1
f˛1x 2 dx
�10
7
f˛1x 2 dx � 8�7
1
f˛1x 2 dx � 2
–POSTTEST–
161
y
x
1
2
3
1 2 3 4
y
x
y = f(x)
–1
–2
Calc2e_21_151-164_Post.qxd 11/18/11 1:03 AM Page 161
41. If , then what is ?
a. g ′(x) =
b. g ′(x) =
c. g ′(x) =
d. g ′(x) =
42. Evaluate .
a. 6
b. 24
c. 36
d. 40
43. Evaluate .
a.
b. �c
c.
d.
44. Evaluate
a.
b.
c.
d.
45. Evaluate .
a.
b.
c.
d.
46. Evaluate .
a.
b.
c.
d.121 ln1x 2 22 � c
1 ln1x 2 22 � c
12
x2 � c
1x
� c
� ln1x 2
x dx
3ex�1
x � 1� sin1x2 2 � c
31
22e x cx + + cos( )
3ex � cos12x 2 � c
3ex � cos12x 2 � c
� 13ex � sin12x 2 2 dx
cos1x 2 � c
�cos1x 2 � c
sin1x 2 � c
�sin1x 2 � c
�cos1x 2 dx
5x4 �2x3 � c
5x4 �2x3
16
x6 �1x
16
x6 �3x3 � c
� a x5 �1x2b dx
�3
1
16x2 � 4x 2 dx
3x2
14
x4 � c
14
t4 � c
x3
g¿ 1x 2g1x 2 � �x
0
t3 dt
–POSTTEST–
162
Calc2e_21_151-164_Post.qxd 11/18/11 1:03 AM Page 162
47. Evaluate .
a.
b.
c.
d.
48. Evaluate .
a.
b.
c.
d. 124
49. Evaluate .
a.
b.
c.
d.
50. Evaluate .
a.
b.
c.
d. xex � ex � c
xex � ex � c
12
x2ex � c
12
xex � c
�xex dx
xln1x 2 � x � c
121 ln1x 2 22 � c
1x
� c
ln11 2 � c
� ln1x 2 dx
1256
2483
2
3
�6
0
24x � 1 dx
12
x2e1x22 � c
12
xe1x22 � c
12
e1x22 � c
e1x22 � c
�xe1x22
dx
–POSTTEST–
163
Calc2e_21_151-164_Post.qxd 11/18/11 1:03 AM Page 163
Answers
1. c. Lesson 1
2. a. Lesson 1
3. d. Lesson 1
4. d. Lesson 1
5. b. Lesson 2
6. a. Lesson 2
7. b. Lesson 2
8. d. Lesson 3
9. c. Lesson 3
10. a. Lesson 3
11. c. Lesson 4
12. a. Lesson 4
13. a. Lesson 5
14. b. Lesson 5
15. c. Lesson 5
16. b. Lessons 6, 7
17. d. Lessons 6, 7
18. d. Lesson 7
19. c. Lesson 8
20. a. Lesson 8
21. a. Lesson 8
22. c. Lesson 9
23. d. Lesson 9
24. c. Lesson 9
25. d. Lesson 10
26. b. Lessons 9, 10
27. b. Lesson 11
28. c. Lesson 11
29. a. Lesson 11
30. d. Lesson 12
31. b. Lesson 12
32. b. Lesson 13
33. d. Lesson 13
34. b. Lesson 13
35. b. Lesson 14
36. d. Lesson 14
37. c. Lesson 15
38. b. Lesson 15
39. c. Lesson 16
40. c. Lesson 16
41. a. Lesson 17
42. c. Lesson 18
43. b. Lesson 18
44. b. Lesson 18
45. c. Lessons 18, 19
46. d. Lesson 19
47. b. Lesson 19
48. a. Lesson 19
49. d. Lesson 20
50. c. Lesson 20
–POSTTEST–
164
Calc2e_21_151-164_Post.qxd 11/18/11 1:03 AM Page 164
Lesson 1
1.
2.
3.
4. . Because there is no x in the
description of f, the 7 never gets used. This is
called a constant function because it is always
equal to the constant 2.
5.
6.
7. The rock is feet high after
3 seconds.
8. The profit on 100 cookies is P(100) = $39.
9.
10.
11.
=
=
12.
13.
14.
=
=− − = − + = − +2 2
22ax a
aa x a
ax a
( )( )
− + +( ) − − +( )( )x a x
a
2 25 5
− + +( ) − − +( )( )x a x
a
2 25 5
− + +( ) − − +( )( )x a x
a
2 25 5
g12x 2 � g1x 2 �
g1x2 � 2x 2 �8
x2 � 2x� 61x2 � 2x 2
−+1
2x x h( )
−+
hhx x h2 ( )
x x hhx x h
− −+
=2 ( )
x x hx x h
h
− ++
( )( )2
12
12( )x h x
h+
−=
f x h f xh
( ) ( )+ − =
f˛1x � a 2 � x2 � 2xa � a2 � 3x � 3a � 1
f˛1y 2 � y2 � 3y � 1
s13 2 � 16
h164 2 � 4
m −
= − −
= − ⋅ −
⋅ −
⋅ −
=
15
515
515
15
15
125
3
f˛17 2 � 2
h1
21
=
g1�3 2 � �20
f˛15 2 � 9
SOLUTION KEY
165
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 165
15.
16.
=
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27. (Note that t cannot equal –5.)
28.
29.
30.
31. and
32. (Note that this single interval isenough because it excludes by and –3.
Lesson 2
− 4
3
− ∞
4
3,
−( ]8 2,( , )−∞ −8
−∞ ∞( ),
−∞ ∞( ),
−∞ ∞( ),
− ∞( )5,
− ∞[ )1,
−∞ −( ) −( ) ∞( ), , , , ,3 3 5 5and
( )( )f h f x f hx
x x
o o 21
21
12
1
2
=
=−
1x3 � 2x2 � 1
� B 1x3 � 2x2 � 1
( )( ) ( ( ( )))h f g x h f g xo o = =
g h g go {( ) = −
= =
=
( ) ( ) ,16 16 16 12 1 4414
h h w w w w wo( ) = −( ) − −( )
f f z
z
zo( ) = =( )11
( )( )f h tt t
o =−1
( )( )g f xx x
o = − +1 21
3 2
( )( )f g xx x
o =− +
12 13 2
x x x xx x
3 2 326 12 8
23 6 4
+ + + − = + +
x x x x( )=
+ + +( ) −2 4 4
2
2 3
g x g x x x( ) ( ) ( )+ − = + −22
22
3 3
h x a h xa
x a xa
x a xa
aa
( ) ( ) ( )+ − =− + +( ) − − +( )
= − − + + −
= − = −
2 1 2 1
2 2 1 2 1
22
–SOLUTION KEY–
166
1
2
3
4
5
6
–1–2–3–4–5–6 1 2 3 4 5 6–1
–2
–3
–4
–5
–6
1.2.
3.4.
5.
6. 7. 8.(0,0)
(0,3)
(3,5)
( , )(–5,0)
(–3,4)
(–1,–5)
(2,–6)
y
x2
__94
__1
1
2
3
4
5
6
–2–3–4–5–6 1 2 3 4 5 6–1
y
x
7
8
9
10
12. (–2,13)
11. (0,5)
10. (1,4)
9. (3,8)
11
12
13
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 166
–SOLUTION KEY–
167
13. There is a discontinuity at x � 0. The graph of
f is decreasing on (�q,0) and on (0,q). The
graph of f is concave down on (�q,0) and
concave up on (0,q). There are no points of
inflection, no local maxima, and no local
minima. There is a vertical asymptote at x � 0
and a horizontal asymptote at y � 0.
14. There are no discontinuities. The function
increases on (�q,�3) and on (0,3), and it
decreases on (�3,0) and on (3,q). There are
local maxima at (�3,4) and (3,4), and there is
a local minimum at (0,2). The graph is concave
up on (�1,1) and concave down on (�q,�1)
and on (1,q). There are points of inflection at
(1,3) and (–1,3). There are no asymptotes.
15. There is a discontinuity at x � 1. The function
increases on (�1,1) and on (1,q), and
decreases on (�q,�1). The function is
concave up on (�q,1) and on (1,q). There is
a local minimum at (–1,�2). There are no
asymptotes, nor any points of inflection.
16. The discontinuities occur at x � �2 and x � 2.
The function increases on (0,1), (1,2) and
(2,q), and decreases on (�q,�2), (–2,–1) on
(�1,0). The point (0,2) is a local minimum.
The graph is concave up on (�2,–1), (–1,1),
and (1,2) and concave down on (�q,�2) and
(2, q). There are no points of inflection. There
are vertical asymptotes at x � �2 and x � 2,
and a horizontal asymptote at y � 0.
17. There is a discontinuity at x � �1. The
function increases on (�q,�1) and on
(�1,2), and it decreases on (2,q). There are
local maxima at (�1,3) and (2,3). The graph is
concave up on (�1,0) and concave down on
(0,q), so there is a point of inflection at (0,2).
Because the line is straight before , it
does not bow upward or downward, and thus
has no concavity. There are no asymptotes.
18. There are no discontinuities. The graph
decreases on (�q,q), is concave down on
(�q,0), and is concave up on (0, q). There is
a point of inflection at (0,0). There are
horizontal asymptotes at y � �2 and y � 2.
19. There are no discontinuities. The graph
increases on (0,2), has a local maximum at
(2,5), and decreases on (2,q). The graph is
concave down on (0,3) and concave up on
(3,q) with a point of inflection at (3,3). There
is a vertical asymptote at x � 0 and a
horizontal asymptote at y � 1.
20. The discontinuities occur at x � 2, x � 5, x � 6,
and x � 7. The function increases on (�q,1),
(4,5), and on (5,q). The function decreases on
(1,2), on (2,4), and on (6,7). There is a local
maximum at (1,2) and at (2,3). The points (4,2)
and (17,4) are a local minima. The graph is
concave up on (–q,1), (1,2), (2,5), (5,6), and
(6,7). Since the line is straight after x = 7, it does
not bow upward or downward and thus has no
concavity. There is a horizontal asymptote at y =
0 and a vertical asymptote at x = 6.
21. 3
22. 0
23.
24.w � 7
3�
7 � w�3
�94
x � �1
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 167
25.
26.
27.
28.
–SOLUTION KEY–
168
1
2
3
4
5
6
–1–2 1 2 3 4 5 6–1
y
x
y = – x + 5
(6,1)
2—3
1
2
3
–11 2 3 4 5–1
–2
–3
y
x
y = x – 2
(–1,–3)
(5,3)
1
2
3
4
56
–1–2–3–4–5–6 1 2 3 4 5 6–1
y
y = 5 (2,5) (6,5)
1
2
3
–11 2 3 4–1
–2
–3
–4
–5
–6
y
x
y = 2x – 4
(1,–2)
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 168
Lesson 3
1.
2.
3.
4.
5. 1
6.
7. 9
8. 50 = 1
9.
10.
11.
12.
13.
14.
15.
16.
17. 1
18. 2
19. 5
20. ln(1) = 0
21.
22.
23.
24. x �
25.
26.
Lesson 4
1.
2.
3.
4.
5.
6. 60°
7. 90°
8. 360°
9. 18°
10. 330°
34π
53π
3p2
p
p
6
log log log ( )
log log ( )
a a a
a a
xy
x y
x y
33
3 2 3 3
2 9 11
= −
= − = − −= + =
ln( ) ln( )
ln( ) ln( ) ln( )
ln( ) ln( ) ln( )
ln( ) ln( )ln( )
ln( )ln( )
3 3 100
3 3 100
3 5 3 100
100 5 33
1003
5
5
5
x
x
x
x
⋅ =
+ =+ =
= − = −
ln110 2
ln12 2
ln ln ln ln( )52
25
52
25
1 0
+
= ⋅
= =
ln a246b � ln14 2
ln17 # 2 2 � ln114 2
e7
e11
3
3
3
33 729
2
4 2
2
86
−
−
−
−= = =( )
44
4 2563
14
− = =
88
18
14 096
2
2 4
−= =
,
51
51
15 6256
6− = =
,
( )3 3 3 9323
3 23 2= = =⋅
12
183
=
34 � 81
6�2 �1
36
104 � 10,000
43 � 64
25 � 32
–SOLUTION KEY–
169
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 169
11. 1
12.
13. 2
14. 2
15.
16.
17.
18.
19.
20.
21.
22. �
23. �1
24. �
25.
26. �1
27. 0
28. undefined
29.
30. �
31.
32.
33. �2
34. �
35.
36.
37.
38.
39.
40. x = π
Lesson 5
1. 1
2. 1
3. 1
4. undefined
5. no
6. �1
7. 4
8. undefined
9. �1
10. yes
11. 1
12. 3
13.
14. q
q
x = 43
53
π π,
32
− 12
2
3
3
2 3
3
3
�12
32
3
2
2
2
2
2
2
2 3
3
3
3
2
2
3
2 3
3 =
3
1
3
3
3 =
3
–SOLUTION KEY–
170
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 170
15. 2
16. �2
17. 16
18. 0
19.
20.
21. 2x � 1
22.
23. �q
24. q
25.
26. �q
27.
28. �q
29. 8
30.
31. 1
32.
33. �q
34.
35. 2x
36.
37.
38. First note that
So,
39.
40. lim( )( )
( )lim ( )
( ) ( )
( ) ( )
ln( )
z
z
z
ze e
ee
e
→ →
− −− −
= − −
= − − = − −=
ln lm2
2
2 2
2
4 2
24
4 2 4
2
lim lim( )( )
( )( )
lim( ) ( )
x x
x
x x xx
x x xx x
x xx
→ →
→
+ −−
= − +− +
= ++
= ++
= =
3
3 2
2 3
3
2 159
3 53 3
53
3 3 53 3
246
4
lim( )
lim( )
lim
lim( )
lim( )
h
h
h
h
h
x h xh
x hx h x h xh
x hx h x h x
h
h x hx hh
x hx h x
→
→
→
→
→
+ −
= + + + −
= + + + −
= + +
= + + =
0
3 3
0
3 2 2 3 3
0
3 2 2 3 3
0
2 2
0
2 2 2
2 2
2 3 3 2
2 6 6 2 2
6 6 2
6 6 2 6
( ) ( ) ( ) ( )( )x h x h x h x hx h x h
x hx h x x h h x h
x hx h x h
+ = + + = + + +
= + + + + +
= + + +
3 2 2 2
3 2 2 2 2 3
3 2 2 3
2
2 2
3 3
02
0−
=
lim
lim
lim
a
a
a
a
x a x
x a x
x a x
a x a x
a
x a x x
→
→
→
+ −⋅ + +
+ +
=+ +( )
= + +( ) =
0
0
02
( )
)( )x
x
x x x→= −
− + +( ) =25
25
25 1 5
1260
lim
(
lim( )( )x
x x
x x x→
−( ) +( )− + +( )25
5 5
2 5
5 1
lim( )( )( )( )x
x xx x→
− −− +
=3
3 12 5
14
lim( )( )x
x
x x→
−− +
=2
2
2 214
72
16
e ee
− − −= =3 2 1212
2 1( )
12p6
�3p
9623
–SOLUTION KEY–
171
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 171
Lesson 6
1.
2.
3.
4.
5.
6.
.
Thus, at x � 2, the slope is .
7.
So, the slope at x = 16 is .
8.
� .
Thus, there is a slope of zero when g′(x) =
2x – 4 = 0. This happens when x � 2.
9.
� .
The slope at (2,�3) is , so
the equation of the tangent line is
.y � �41x � 2 2 � 3 � �4x � 5
h¿ 12 2 � �4
lim( )a
x a x→
− − = −0
2 2
� limaS0
�2xa � a2
a
h¿ 1x 2 � lim
aS0 1 � 1x � a 22 � 11 � x2 2
a
lim( )a
x a x→
+ − = −0
2 4 2 4
� limaS0
2xa � a2 � 4a
a
1 4 12 − − +( )x xa
g¿ 1x 2 � lim
aS0 1x � a 22 � 41x � a 2 �
a
′ =g ( )1638
′ = + −
= + − ⋅ + ++ −
= + −+ −
= // + −
=
→
→
→
→
g xx a x
a
x a xa
x a x
x a x
x a x
a x a x
a
a x a x x
a
a
a
a
( ) lim
lim
lim( )
(
lim( )
0
0
0
0
3 3
3 3 3 3
3 3
9 9
3 3
9
3 3
3
2
f ¿ 12 2 � 13
= + + = +→
lim( )a
x a x0
6 3 1 6 1
� limaS0
6xa � 3a2 � a
a
f ¿ 1x 2 � lim
aS0 31x � a 22 � 1x � a 2 � 13x2 � x 2
a
= + + =→
lim( )a
x xa a x0
2 2 23 3 3
� limaS0
13x2 � 3xa � a2 2a
a
� limaS0
x3 � 3x2a � 3xa2 � a3 � x3
a
k¿ 1x 2 � lim
aS0 1x � a 23 � x3
a
= + −+ +→
lim( )
a
x a x
x a x0
9 9
3 3
3 3
3 3
x a x
x a x
+ ++ +
= + −
→
lima
x a x
a0
3 3
′ = + −→
f xx a x
aa( ) lim
0
3 3
limaS0
0a
� 0� limaS0
10 � 10
a�
g¿ 1x 2 � lim
aS0 g1x � a 2 � g1x 2
a
� limaS0
2x � a � 2x
� limaS0
x2 � 2xa � a2 � 5 � x2 � 5
a
h¿ 1x 2 � limaS0
1x � a 22 � 5 � 1x2 � 5 2
a
� limaS0
8x � 8a � 2 � 8x � 2
a� 8
f ¿ 1x 2 � lim
aS0 81x � a 2 � 2 � 18x � 2 2
a
–SOLUTION KEY–
172
= 3
2 x
=+ +
=+→
lima x a x x x0
9
3 3
9
3 3
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 172
10.
� =
10x �2. The y-value at x � 1 is .
The slope at x � 1 is . The
equation of the tangent line is
.
Lesson 7
1.
2. = 21x 20
3.
4.
5. = 12t11
6.
7.
8.
9.
10. , so
11. , so
12. , so
13. , so
14. .
So, .
15. . So, .
16. . So, .
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32. , so ′ = + −f x x x( ) 6
152
352f x x x( ) = − −3
254
32
dy
dxx x
x x
= = 2 32 31
223
23
− −+ +
dy
du� 2u � 2u�3 � 2u �
2u3
dy
dx� �2x�2 � 2x�3 � �
2x2 �
2x3
h¿ 1u 2 � 5u4 � 16u3 � 21u2 � 4u � 8
g˛¿ 1x 2 �35
x�45 � 15x2 �
3
5x 45
� 15x2
F¿ 1x 2 � 600x99 � 500x49 � 100x24 � 20x9
′ = + +s t t et( ) 3 2 32π
dydx
x xx x
= − − − − = − + −− −4 2 3 14 6
13 43 4
( )
f ¿ 1x 2 � 24x2 � 6x
dydt
t t = −12 62
k¿ 1x 2 � �2x
g˛¿ 1t 2 �485
t3
V¿ 1r 2 � 4p r2
f ¿ 1x 2 � 30x�11 �30x11
dydx
x = 42 6
dydt
tt
= =−112
1
12
1112
1112
yt
tt t= = =−
13
14
13
14
112
′ =h t t( ) 7 6h t t( ) = 7
′ = − = −−g x x
x( )
32
3
2
52
52
g xx x x
x( ) =⋅
= = −1 112
32
32
f ¿ 1x 2 � �12
x�32 � �
1
2x˛
˛
32
f˛1x 2 �1
x˛˛
12
� x�12
dy
dx� �x�2 � �
1x2y � x�1
dy
duu
u =
1
2
1
2
12
− =y � u12
k¿ 1x 2 �14
x�34 �
1
4x˛
34
k1x 2 � x˛
14
g¿ 1x 2 � �45
x�95 � �
4
5x 95
f ¿ 1t 2 � 0
f ¿ 1x 2 � 100x99
dy
dx�
75
# x 25
dy
dt
h¿ 1x 2 � 0
g¿ 1u 2 � �5u�6 ��5u6
dy
dx
f ¿ 1x 2 � 5x4
y � 121x � 1 2 � 7 � 12x � 5
k¿ 11 2 � 12
k˛11 2 � 7
lima
x a→
+ +( )0
10 5 2 limaS0
10xa � 5a2 � 2a
a
limaS0
51x � a 22 � 21x � a 2 � 15x2 � 2x 2
a�
k¿ 1x 2 �
–SOLUTION KEY–
173
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 173
33. , ,
,
and
34.
35.
36. , , and
Lesson 8
1. pay rate in dollars per hour
2. fuel economy in miles per gallon
3. baby’s growth rate in pounds per month
4. rate at which the radius shrinks in inches per
hour
5. , so . So, it is increasing at
the rate of 1 foot per year.
6. , so . So, it is
decreasing at the rate of 6 feet per day.
7. , so
. So, the profit is increasing at
the rate of $3,750 per car sold.
8. , so the cost would
increase at a rate of $2.13 per inch at the
instant when x = 3 inches.
9. After 3 seconds, it is at meters from
the start. Note that ν(t) = 3t 2 + 4t + 10 and
a(t) = 6t + 4. At that moment, it is traveling at
meters per second and accelerating
at meters per second per second.
10. The position function is ,
the velocity function is , and the
acceleration is a constant . Since
s(2) = 0, it will hit the ground when t � 2
seconds and be traveling at feet
per second (downward) at that instant.
11. The position function is
and the velocity function is .
The bullet will stop in the air when the velocity
is zero. This happens at t � 25 seconds, when
the bullet is feet in the air.
12. The position function is
, so after 4 seconds, it is feet
off the ground. It has therefore fallen 296 feet
by this moment. Since ν(t) = –32t – 10, it is
traveling feet per second
(downward) at this moment.
13.
14.
15.
16.
17. because is a constant
18. Because
, the point is . Since
f ′(x) = cos(x) – sin(x), the slope is
, so the equation is
.� �x �p
2� 1� a x �
p
2b � 1
y �f ¿ ap2b � �1
ap
2,1b1 � 0 � 1
sin ap
2b � cos a
p
2b �f a
p
2b �
cos15 2h¿ 1x 2 � �sin1x 2
r¿ 1u 2 �12
cos1u 2 �12
sin1u 2
′ = − − +
= − − +
− −g x x x x
x xx
( ) sin( )
sin( )
653
6 5
3
343
3 43
π
π
f ¿ 1t 2 � 3cos1t 2 �2t2
dy
dx� 20x4 � 10sin1x 2
v14 2 � �138
s14 2 � 7041,000
s1t 2 � �16t2 � 10t �
s125 2 � 10,000
v1t 2 � �32t � 800
s1t 2 � �16t2 � 800t
v12 2 � �64
a1t 2 � �32
v1t 2 � �32t
s1t 2 � �16t2 � 64
a13 2 � 22
v13 2 � 49
s13 2 � 75
′ = − ≈C ( ) . .3 4 883
2 13
′ =P ( ) ,50 3 750
′ = − +P x x x( ) ,3
10120 9 0002
′ = −L ( )7 6′ = − +L t t( ) 2 8
′ =h ( )5 1′ =h tt
( )25
2
d3y
dt3 �209
t�83
d2y
dt2 � �43
t�53
dy
dt� 2t�2
3
d y
dxx x x
3
36 5 4180 48 6= − + −− − −
s– 1t 2 � �32
f 1x 2 � 24x�5
f ‡ 1x 2 � �6x�4
f – 1x 2 � 2x�3f ¿ 1x 2 � �x�2
–SOLUTION KEY–
174
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 174
19.
20.
21.
22. . (Note that e–3 is a
constant.)
23. . (Note that ln(π) is a
constant.)
24. , so
25.
26.
Lesson 9
1. = x(2cos(x) – x
sin(x))
2.
3.
4.
= x(6ln(x) + 3 – 20x2)
5.
6.
7.
8.
9.
10.
11.
12.
�
13.So, ,
so the slope is 1.
14.
15.
16.
17.
18.
19.
14et � 1 2 1t3 � 2t � 1 2 � 13t2 � 2 2 14et � t 2
1t3 � 2t � 1 22
dy
dt�
x x
x
4
2
5 1ln( )
ln( )
−( )( )
dy
dx�
5x4ln1x 2 � 1x# x5
1 ln1x 2 22�
5x4ln1x 2 � x4
1 ln1x 2 22
f ¿ 1x 2 �11 � 1
x 2 1ex � 1 2 � ex1x � ln1x 2 2
1ex � 1 22
2x1x2 � 1 2 � 2x1x2 � 1 2
1x2 � 1 22�
4x1x2 � 1 22
f ¿ 1x 2 �
16x � 5 2 1x3 � 10x � 7 2
13x2 � 5x � 2 22
13x2 � 10 2 13x2 � 5x � 2 2
13x2 � 5x � 2 22�h¿ 1x 2 �
y � �p1x � p 2 � �px � p2
f ¿ 10 2 � 210 2e0 � e010 22 � 1 � 1
′ = + +f x xe e xx x( ) .2 12
3x3cos1x 2 � 3x4sin1x 2ln1x 2
12x3ln1x 2cos1x 2 �
a1x
cos1x 2 � sin1x 2ln1x 2b # 3x4
g¿ 1x 2 � 12x3ln1x 2cos1x 2 �
dy
dx� exsin1x 2 � 1exsin1x 2 � cos1x 2ex 2 # x
� 2cos1x 2sin1x 2
cos1x 2sin1x 2 � cos1x 2sin1x 2f ¿ 1x 2 �
dy
dxx x
xx
x x
5 1 ln( ) + 1
= − ⋅ ⋅
= − −
1
15 1
2
2 ln( )
′
= −( ) + +( ) +( )h t
t t t t t t
( )
sin( ) cos( ) cos( ) sin( )3 42 2
dy
dx�
8sin1x 2
x� 8ln1x 2cos1x 2 � sin1x 2
′ = ⋅ − ⋅
− ⋅ − ⋅
−
− −
k x x x x x
x x x x
( ) cos( ) sin( )
cos( ) sin( )
12
14
12
12
14
54
′ = + + +
= + +
h u e u u u e
e u u
u u
u
( ) ( ) ( )
( )
2
2
3 2 3
5 3
g˛¿ 1x 2 � 6xln1x 2 �1x13x2 2 � 20x3
dy
dxx x = −π(cos ( ) sin ( ))2 2
dy
dt� 24t˛
2et � et # 8t˛
3 � 8t˛
2et13 � t 2
f ¿ 1x 2 � 2xcos1x 2 � sin1x 2 # x2
f ¿ 110 2 �1
10
g˛
11002 1x 2 � 3ex
f – 1x 2 � e˛
x �1x2f ¿ 1x 2 � ex �
1x
k¿ 1u 2 �152
x 32 � 5ex
h¿ 1x 2 �1
22x�
8x
dy
dx� �sin1x 2 � 10e˛
x � 8
g˛¿ 1t 2 �12t
� 2t
f ¿ 1x 2 � 1 � 2x � 3x2 � ex
–SOLUTION KEY–
175
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 175
20. g ′t
21.
22.
23.
24.
25.
26.
27.
28. , so
.
29.
30.
31.
32.
33.
34.
Lesson 10
1.
2.
3.
4.
5.
�2x � 9
22x2 � 9x � 1
121x2 � 9x � 1 2�
12 # 12x � 9 2g¿ 1x 2 �
dy
du�
721u5 � 3u4 � 7 2
52 # 15u4 � 12u3 2
18t7 � 27t2 � 3 2
101t8 � 9t3 � 3t � 2 2 #h¿ 1t 2 �
dy
dx� 31x2 � 8x � 9 22 # 12x � 8 2
f ¿ 1x 2 � 418x3 � 7 23 # 124x2 2
′ =
+
− ⋅ +
− −
j x
x x x x x x x
x
( )
sec( ) sec( ) tan( )
sec ( )
13
14
23
34
13
14
2
′ = ⋅ + ⋅
+
⋅ ⋅
h t e tt
e t t
e t
t t
t
( ) ln( ) tan( ) sec ( )
ln( )
1 2
′ = +
= +( )g x e x x x e
e x x
x x
x
( ) sec( ) sec( ) tan( )
sec( ) tan( )1
f ¿ 1x 2 � tan1x 2 � x # sec21x 2
�csc21x 2��1
sin21x 2�
��sin21x 2 � cos21x 2
sin21x 2
ddxa
cos1x 2
sin1x 2b
ddx1cot1x 2 2 �
= − = − ⋅
= −
cos( )
sin ( ) sin( )cos( )sin( )
csc( )cot( )
x
x xxx
x x
2
1
ddxa
1sin1x 2
bddx1csc1x 2 2 �
′ = + = + ==
f e e e e e e e( ) ln( )2 2 31
{
′ = ⋅ +f x x x x( ) ln( )2
dydx
e e x e e x e
d y
dxe x e e x
x x x
x x x
= ⋅ + ⋅ + = ⋅ + +
= ⋅ + + ⋅ = ⋅ +
1 1
1 1 22
2
( )
( ) ( )
f˛¿ 1x 2 �12xex � ex # x2 2cos1x 2 � sin1x 2 # x2ex
cos21x 2
�ln1x 2 � 1 � xln1x 2
ex
1 ln1x 2 � 1 2ex � xex˛ln1x 2
ex # ex
dy
dx�
′ =
+
⋅ −
⋅ + ⋅
⋅ +( )=
h t
tt
t t t t t t
t
t t
( )
sin ( )
cos( ) sin( ) cos( ) sin( ) ln( )
sin ( )
sin( )cos( )
11 2
2
4
6 744444 844444
dydx
x x x x
x x=
− ⋅ −( ) − ⋅ +( )( )+( ) −( )
02 2
cos( ) cos( ) sin( ) sin( )
sin( ) cos( )
π π
π π
′ = ⋅ − − − ⋅
−( )g u
u u u u u
u u( )
cos( ) ( ) ( ) sin( )3 2
32
3 3 2
3
dydx
x x x x
x
x
x
=− +
− −
+
= −
+
32
132
1
1
3
1
12
32
12
32
32
12
32
2
2
= −
= −
3
3
2 3
2 2
2 3
2
π ππ
π
t t t t
t
t t t t
t
sin( ) cos( )
sin ( )
sin( ) cos( )
sin ( )
–SOLUTION KEY–
176
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 176
6.
7.
8.
9.
10. h′(x) =
11.
12.
13. , so
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
Lesson 11
1. , so
2. , so
3.
4.
5.
6.
7.dy
dx�
�sin1x 2
sec21y 2� �sin1x 2cos21y 2
dy
dxe ee e
x x
y y = +
+22
2
2
dy
dx�
12x3 � 12y � 8
dy
dx�
1x
1 � 121y
�22y
2x2y � x
dy
dx�
4cos1y 2
� 4sec1y 2
dy
dx�
cos1x 2
3y2 � 13y2 # dy
dx�
dy
dx� cos1x 2
dy
dx�
4x3 � 831y � 1 22
31y � 1 22 # dy
dx� 4x3 � 8
′ = −−
⋅ − −
−
⋅⋅ ⋅ −( ) + ⋅ −
−( )
h xe
e
e
e
e x e e e
e
x
x
x
x
x x x x
x
( ) cos sin21
1
1
1
8 1 1
1
4 4
4 4
2
2 2
2 2
�3u
sec1 ln18u3 2 2tan1 ln18u3 2 2
sec1 ln18u3 2 2tan1 ln18u3 2 2 # 18u3
# 24u2k¿ 1u 2 �
dy
dx� cos1sin1sin1x 2 2 2 # cos1sin1x 2 2 # cos1x 2
g¿ 1t 2 �sec21et � 1 2 # et
tan1et � 1 2
dy
dx� 41e9x2�2x�1 23 # 1e9x2�2x�1 2 # 118x � 2 2
� �24cos218x 2sin18x 2
1�sin18x 2 2 # 8f ¿ 1x 2 � 3cos218x 2 #
dydx
x x= ( ) ⋅ −( )sec cos( ) sin( )2
′ = +( ) ⋅ −( )s u u u u u( ) sin( ) cos( ) cos( ) sin( )32
dydx x x
x x x= ⋅ + ⋅( )1sin( )
sin( ) cos( )
dy
dxe x exx e = ⋅ + −2 2 12 2
sec110x2 � ex 2tan110x2 � ex 2 # 120x � ex 2
f ¿ 1x 2 �
dy
dx� e2x � 2xe2x
= −2 2 22
θ θ θθ
cos( ) sin( )
cos12u 2 # 2 # u � 1 # sin12u 2
u2f ¿ 1u 2 �
g˛¿ 1x 2 �121ex � e�x 2g1x 2 �
121ex � e�x 2
f ¿ 1x 2 � ex � 2e2x � 3e3x
dy
dx�
51 ln1x 2 24
x
π πcos( )x
dy
dt�
33t � 5
′ = ( ) ⋅−
g x x x( ) tan( ) sec ( )12
12 2
′ = ( ) ⋅f x xx
( ) sec2 1
2
dy
dx�
131ex � 1 2�
23 # ex �
ex
31ex � 1 223
–SOLUTION KEY–
177
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 177
8. , so
9.
10.
11.
12.
13. , so
14.
15. , so at
(�3,1), the tangent slope is .
16. , so at (1,�2) the tangent slope
is .
17. , so at (4,2) the tangent slope is
1.
18. , so at
the tangent slope is .
19. at , so the tangent
equation is .
20. , so at (0,0) the tangent slope is
. As such, the equation of the tangent line is
y = x.
Lesson 12
1.
2.
3.
4.
5.
6. 2A # dAdt
� 2B # dBdt
� 2C # dCdt
dzdt
�45
x # dxdt
�45
y # dy
dt�
35x2
# dxdt
1y
# dy
dt� ex # dx
dt� 2x # dx
dt# y2 � 2y # dy
dt# x2
1
22x# dx
dt�
1
22y# dy
dt� 30x2 # dx
dt� 7 # dx
dt
4y3dy
dt� 6x
dxdt
� �sin1y 2dy
dt
dy
dt� 51x3 � x � 1 24 # a 3x2dx
dt�
dxdtb
− 13
− 13
dydx
ee ee
e e
x
x y
x
x y
=−
+
++ 1
y x = 3
2 3 1
2 6−
+ π
12 6
,π
dy
dx =
3
2 3
− 4 2π
22 4
,π
dydx
y y
= −
⋅
1
2 2 2 2
π πcos( ) sin sin( )
dydx
yy
= −
−−
13
3 52
− 49
dydx
xy
= − −−
3 13 3
2
2
dy
dx� 1
3y2 # dy
dx� 2x � 2y # dy
dx� 5 # dy
dx
dy
dx�
3x2cos1y 2
sec1y 2tan1y 2 � 9 � x3sin1y 2
dy
dx�
1 � y �1y
�xy
2 � x � 1�
y2 � y3 � y
�x � xy2 � y2
y � x # dydx
y2 � y �dy
dx# x � 1 �
dy
dx
dy
dx
x y xy
x y x = −
−4 2
4
3 4
2 3 4
dy
dx�
521y � x2 23
� 2x
dydx x y y
= −− −
1
3 2 22( )
dydx
x x yy x y
= − + −− + −
cos( ) cos( )cos( ) cos( )
dy
dx
x y
x yx y
=
1
2 +
1 1
2 +
1
2 + −=
− 1
dy
dx x y
dy
dx =
1
2 + 1 +
–SOLUTION KEY–
178
Though tempting, don’t cancel the terms cos(x – y).
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 178
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17. Because , A is increasing at the rate
of 172 square feet per minute when I � 20.
18. , so R is increasing at the rate of
per hour at this instant.
19. , so the area is decreasing at the
rate of 65 square feet per minute.
20. , so the area shrinks by
320p square inches per hour.
21. , so the radius grows at
feet per hour.
22. , so each side is shrinking at the rate of
4 inches per minute.
23. , so each side is growing at the rate of
inch per second.
24. , so the base increases at the rate of
7 inches per hour.
25. If the height is y and the base is x, then
and . After 6
seconds, y � 6 and , so x � 8.
Because , , so
. The end of the board is moving at
the rate of of a foot each hour along the
ground.
26. If the base is x and the hypotenuse (length of
the string) is s, then . Using
this, when s � 260, x must be 240. Because
, we can calculate that .
Thus, the string is being let out at 12 feet
per second.
Lesson 13
1. 0
2.
3.
4.q
52
45
dsdt
� 12dxdt
� 13
x2 � 1002 � s2
34
dxdt
� �34
218 2dxdt
� 216 2 11 2 � 0dy
dt� 1
x2 � 62 � 100
2xdxdt
� 2ydy
dt� 0x2 � y2 � 102
dbdt
� 7
1240
dedt
= − 1240
dsdt
� 4
52p
� 0.796drdt
�5
2p
dAdt
in
hour= − 320
2
π
dAdt
= − 65
2764
dRdt
= 2764
dAdt
� 172
dBdt
� 28
dKdt
� 24
dy
dt� �
43
dxdt
� �3512
dDdt x y
xdxdt
ydydt
xdxdt
ydydt
x y
=+
⋅ ⋅ + ⋅
=⋅ + ⋅
+
1
22 2
2 2
2 2
dSdt
ededt
= ⋅12
= ⋅ + ⋅
12
dbdt
hdhdt
b
dAdt
�12
# dbdt
# h �dhdt
# 12
b
dCdt
� 2p # drdt
dAdt
� 8p r # drdt
dVdt
� 4p r2 # drdt
–SOLUTION KEY–
179
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 179
5. 0
6.
7. �q
8. q
9. 1
10. 0
11. 3
12. q
13. �q
14. q
15. 0
16. 0
17. q
18. 0
19. 0
20.
21. vertical asymptote at x � 4, horizontal
asymptote at y � 1, sign diagram:
22. vertical asymptotes at x � 2 and x � �2,
horizontal asymptote at y � 0, sign diagram:
23. vertical asymptote at x � �3, horizontal
asymptote at y � 1, sign diagram:
24. vertical asymptotes at x � 1 and x � 3, hori-
zontal asymptote at y � 0, sign diagram:
25. vertical asymptote at x = –5, horizontal asymp-
tote at y = 0, sign diagram:
26. vertical asymptotes at x = –2 and x = –1, no
horizontal asymptote, sign diagram:
27. q
28. q
29. q
30. �q
31. �q
32. �q
–1–2 0
m(x)+- +-
–5
j(x) + -
�47
�89
–SOLUTION KEY–
180
++ -f (x)
–2 4
++ --g(x)
–2 2 3
+++ -h(x)
–3 –1 1
++ --k(x)
1 31__2
–
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 180
Lesson 14
1. has no asymptotes. ,
thus there is a local minimum at (15,�215).
Because , the graph is always
concave up.
2. has no asymptotes.
, so there
is a local maximum at (�2,4). Because
, the graph is always concave
down.
g– 1x 2 � �2
g¿ 1x 2 � �4 � 2x � �212 � x 2
g1x 2
f – 1x 2 � 2
f˛¿ 1x 2 � 2x � 30f˛1x 2
–SOLUTION KEY–
181
1
2
3
–1
–2–3 1 2 3–1
y
x
–2
g(x) = –4x – x2
increasingdecreasing
concavity
4(–2,4)
(0,0)
–2
y
x
f(x) = x2 – 30x + 10
15increasingdecreasing
concavity
5 10 15 20
25
–25
–50
–75
–100
–125
–150
–175
–200
–225
–250
(0,10)
(15,–215)
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 181
3. has no asymptotes.
,
so there is a local maximum at (�2,49) and a
local minimum at (3,�76). Because
, there is a point of
inflection at .
4. has no asymptotes.
, so there
is a local minimum at (�1,�2) and a local
maximum at (1,2). , so there is a
point of inflection at (0,0).
k– 1x 2 � �6x
k¿ 1x 2 � 3 � 3x2 � 311 � x 2 11 � x 2
k1x 2
1
2
27
2,−
12 a x �12b12x � 6 �
h– 1x 2 �
h¿ 1x 2 � 6x2 � 6x � 36 � 61x � 3 2 1x � 2 2
h1x 2
–SOLUTION KEY–
182
1
2
3
–1
–2–3
1
2 3
–1
y
x
–2
–3
k(x) = 3x – x3
0
increasingdecreasing
concavity
(–1,–2)
(1,2)
(0,0)–1 125
50
75
–25
–2–3 1 2 3–1
y
x
–50
–75
h(x) = 2x3 – 3x2 – 36x + 5
increasingdecreasing
concavity
–4
(–2,49)
,
(3, –76)
–2 3
1—2
1—2–27—2
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 182
5. has no asymptotes.
, so there
is a local minimum at (6,�427).
, so there
are points of inflection at (0,�5) and at
(4,�251).
6. has a vertical asymptote at x � �2 and a
horizontal asymptote at y � 1. The first
derivative is , and the
second is .g– 1x 2 ��4
1x � 2 23
g¿ 1x 2 �2
1x � 2 22
g1x 2
f – 1x 2 � 12x2 � 48x � 12x1x � 4 2
f ¿ 1x 2 � 4x3 � 24x2 � 4x21x � 6 2
f˛1x 2
–SOLUTION KEY–
183
–3 1 2 3
y
x
f(x) =x4 – 8x3 + 5
increasingdecreasing
concavity
4 5 6 7 8 9–2 –1
50
–50
–100
–150
–200
–250
–300
–350
–400
–450
(0,5)
(4,–251)
(6,–427)
6
0 4
______
1
2
3
–1–2–3 1 2 3–1
y
x
–2
–3
g(x) = xx + 2
increasingdecreasing
concavity
–4
4
4 5–4–5
–2
–2
(–3,3)
(–1, –1)
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 183
7. has vertical
asymptotes at x � �3 and x � 3, and a
horizontal asymptote
at y � 0. Because
, there is a local maximum at
. The second derivative is
.
8. has vertical
asymptotes at x � 1 and x � �1, and a
horizontal asymptote at y � 0.
The first derivative is
,
and the second derivative is
. There is a point of
inflection at (0,0).
k– 1x 2 �2x1x2 � 3 2
1x � 1 231x � 1 23
k¿ 1x 2 � �x2 � 11x2 � 1 22
� �x2 � 1
1x � 1 221x � 1 22
k1x 2 �x
1x � 1 2 1x � 1 2
6x2 � 181x � 3 231x � 3 23
h– 1x 2 �6x2 � 181x2 � 9 23
�
a 0,�19b
�2x1x � 3 221x � 3 22
h¿ 1x 2 ��2x1x2 � 9 22
�
h1x 2 �1
1x � 3 2 1x � 3 2
–SOLUTION KEY–
184
–2–3 1 2 3–1
y
x
k(x) = –——— xx2 – 1
4 5–4–5
1
2
–1
–2
0
–1 1
–1 1
(2, )2—3
(–2, ) 2 – — 3–2
–3 1
2
3–1
y
x
h(x) = 1______x2 – 9
4 5–4–5
1
2
–1
–2
–3 30
–3 3
0, 1—9–
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 184
9. has a vertical asymptote at x � 0 but no
horizontal asymptote. Because
, there is a
local maximum at (�1,�2) and a local
minimum at (1,2). The second derivative is
.
10. has a horizontal asymptote at y � 0 but
no vertical asymptotes. Because
, there is
a local minimum at and a local
maximum at . Because
, there are points
of inflection at , (0,0), and
.33
4,
33
4,
= − ++
2 3 3
12 3
x x x
x
( )( )
( )
f – 1x 2 �2x1x2 � 3 2
1x2 � 1 23
a ˇˇˇˇ1,12b
a ˇˇˇˇ�1,�12b
f˛¿ 1x 2 �1 � x2
1x2 � 1 22�11 � x 2 11 � x 2
1x2 � 1 22
f˛1x 2
j– 1x 2 �2x3
j¿ 1x 2 �x2 � 1
x2 �1x � 1 2 1x � 1 2
x2
j1x 2
–SOLUTION KEY–
185
(– , –
–2–3 1 2 3–1
y
x
–
f(x) = x2+1
______x
0
1–1
–
1,
–1,–
3
3
3 ___4 )
( ), 3___4
33
1—2
1—2
1—2
1—2
–2–3 1 2 3–1
y
x
j(x) = x______x2+ 1
4 5–4–5
1
2
–1
–2
0
3
4
–3
–4
(1,2)
(–1,–2)
–1 10
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 185
Lesson 15
1. when x � 10.
Using the first derivative test, is
positive, so the function increases to x � 10
and , so the
function decreases afterward, thus x � 10
maximizes the profit.
2. If x is the number of trees beyond 30 that are
planted on the acre, then the number of
oranges produced will be:
O(x) = (number of trees) (yield per tree)
� (30 + x)(500 � 10x) � 15,000 + 200x � 10x2.
The derivative is zero
when x � 10. Using the second derivative test,
is negative, so this is maximal.
Thus, x � 10 more than 30 trees should be
planted, for a total of 40 trees per acre.
3. The total sales will be figured as follows:
Sales = (number of copies)(price per copy),
so S(x) = (20 + x)(100 – x) = 2,000 + 80x – x2
where x is the number of copies beyond 20.
The derivative S′(x) = 80 – 2x is zero when x
� 40. Because S″(x) = –2, this is maximal by
the second derivative test. Thus, the artist
should make x � 40 more than 20 paintings,
for a total of 60 paintings in order to maximize
sales.
4. After x days, there will be pounds of
watermelon, which will valued at cents
per pound. Thus, the price after x days will be
P(x) = (200 + 5x)(90 – x) = 18,000 + 250x – 5x2
cents. The derivative is P(x) = 250 – 10x, which
is zero when x � 25. Because P(x) is clearly
positive when x is less than 25 and negative
afterward, this is maximal by the first
derivative test. Thus, the watermelons will
fetch the highest price in 25 days.
5. The area is and the total fencing is
. Thus, , so the
area function can be written as follows:
A(y) = xy = (200 – 26) ⋅ y = 200y – 2y 2. The
derivative A′(y) = 200 – 4y is zero when y � 50.
Because the second derivative is A″(y) = –4, this
is an absolute maximum. Thus, the optimal
dimensions for the pen are y � 50 feet and
feet.
6. Here, and the total fencing is
. Because , the
area function can be written as follows:
A(y) = (150 – 5y)y = 150y – 5y2. The
derivative A′(y) = 150 – 10y is zero when y �
15. Because A″(y) = –10, this is an absolute
maximum. Thus, the optimal dimensions are y
� 15 feet and therefore
feet.
7. Because Volume = πr2h = 16π, it follows
that . Thus, the surface area function is
A(r) =
The derivative is zero
when , so . Thus, the only
point of slope zero is when r � 2. The second
derivative is , which is
positive when r � 2. Thus, by the second
derivative test, r � 2 is the absolute minimum.
Thus, a radius of r � 2 inches and a height of
inches will minimize the surface
area.
h �16r2 � 4
′′ = +A ( )rr
464
3π π
r3 � 8432
2π π
rr
=
′ −V x x( ) = 150 3
22
2 216
2322
22π π π π
r rr
rr
+
= +
h �16r2
x � 150 � 5115 2 � 75
x � 150 � 5y5y � x � 150
Area � xy
x � 200 � 2y � 200 � 2150 2 � 100
x � 200 � 2y4y � 2x � 400
Area � xy
90 � x
200 � 5x
′′ = −O ( )x 20
′ = −O ( )x x200 20
P¿ 1100 2 � �1
10�
1100
� �9
100
P¿ 11 2 � 9,000
P¿ 1x 2 � �1,000
x2 �10,000
x3 � 0
–SOLUTION KEY–
186
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 186
8. Because the box has a square bottom, let x be
both its length and its width, and let y be the
height. Thus, the volume is
and the surface area is
(the top, the four sides, and the bottom). And
because , the height
. Thus, the volume
is
.
The derivative is zero
when . Negative lengths are
impossible, therefore this is zero only when
x � 10. By the second derivative test,
is negative when x � 10, so
this is a maximum. The corresponding height
is feet,
so the largest box is a cube with all sides of
length 10 feet.
9. Because the box has a square bottom, let x be
both the length and width, and y be the height.
The area of each side is thus xy, so the cost to
build four of them at ten cents a square foot is
0.10(4xy) = 0.4xy dollars. The area of the top is
, so it will cost dollars to build.
Similarly, it will cost dollars to build the
base. The total cost of the box is therefore
Cost = 0.4xy + 8x2. Because the volume is
, we know . Thus, the
cost function can be written:
.
The derivative:
is zero only when or x � 10.
By the second derivative test,
is positive when
x � 10, so this is the absolute minimum. The
cheapest box will be built when x � 10 feet and
y � 400 feet.
′′ =
+C 6( ),
xx
32 0001
3
x3 � 1,000
′ = −
+C 6( ),
xx
x16 000
12
�16,000
x� 8x2
C . ( ),
x xx
x=
+0 440 000
82
2
y �40,000
x2x2y � 40,000
7x2
x2x # x � x2
y �15010
�102
� 10
′′ = −V ( )x x3
x2 � 100
′ −V x x( ) = 150 3
22
V ( )x xx
xx x= −
= −2 31502
15012
y �600 � 2x2
4x�
150x
�x2
Area � 2x2 � 4xy � 600
Area � x2 � 4xy � x2
Volume � x2y
–SOLUTION KEY–
187
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 187
10. Inside the margins, the area is:
.
The total area of the page is , so
. Therefore, the area is
.
The derivative is zero
when , thus when x � 12 ignore
negative lengths. The second derivative
is negative when x � 12,
so this is the absolute maximum. Thus, the
dimensions that maximize the printed area are
x � 12 inches and y � 8 inches.
Lesson 16
1. 2
2.
3.
4. 2p
5. 3
6. 3 � 2p
7.
8. 0
9.
10. 4
11. 0
12.
13. 16
14. 6
15. 0
16.
17. 35
18. 48
19. 5
20.21. 7
22. 7
23.24.25. 8
26.27. 25
28. 11
29. –1
30. 1
Lesson 17
1. 0 (Note that , for any constant a.)
2.
3. 8
4.
5. 28
6.
7. 0
8. 7
9. 14
852
332
52
g x dxa
a( ) =∫ 0
�17
�1
�11
�3
212
�12
�32
�32
72
32
′′ −A = 576
3( )x
x
x2 � 144
′ − +A = 288
2( )x
x2
� 102 � 2x �288
x
A = ( )x xx
xx
962 3
966
− −
+
y �96x
xy � 96
� 1x � 3 2 1y � 2 2 � xy � 2x � 3y � 6
Area = x y−
−232
2( )
–SOLUTION KEY–
188
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 188
10. 21
11. 28
12. 35
13. 10
14.
15. 36
16. 20
17.
18.
19.
20.
21.
22.
23.
24. 1
Lesson 18
1.
2.
3.
4.
5. 0
6.
7.
8.
9.
10.
11.
12. 40
13.
14.
15.
16.
17. 48
18. 11
19.
20.
21.
22. 58
23. 208
24.
25.
26. 3ex �12
x4 � c
13
x3 � 5sin1x 2 � c
913
x133 �
5611
x117 � c
− + − +− −12
43
12
4 3 2t t t c
�6
14
t˛
12 � 3t˛
3 �1˛
2t˛
2 � c
2x3 � 5x2 � 5x � c
34
23
43
32x x c− +
83
u3 � c
95
x5 � c
5x � c
34
u 43 � c
45
x 54 � c
38
x˛
˛
83 � c
−[ ] = −
=−
−
−
−
−
tt
1
4
1
4
11 3
4
1�2
t�2 � c � �1
2t2 � c
13
723
0
6
x
=
17
u7 � c
113
x13 � c
15
5x c+
− 22
12
22
2,0003
383
163
23
�4
–SOLUTION KEY–
189
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 189
27.
28.
29.
30.
31.
32.
Lesson 19
1. , by substituting u = x 5 + 1
2. , by substituting u = 4x + 3
3. , by substituting u = x 3 + 1
4.
5. , by substituting u = x 2 – 1
6. , by substituting u = 2x + 1
7. , by substituting u = 1 – x
8. ln 3x3 – 5x, by substituting u = 3x3 – 5x
9. , by substituting u = x 4
10. , by substituting
u = 3x 4 – 2x + 1
11. , by substituting
u = 4x 2 + 5x – 1
12. , by substituting
u = 4x 2 + 5
13. , by substituting
u = 4x + 10
14. Using , the solution is
. Using , the solution
is . Because
, these solutions will
be the same if the second is greater
than the first one.
15. , by substituting u = sin(x)
16. , by substituting u = 4x
17.
18. , by substituting
u = 7x – 2
19. cos(ex) + c, by substituting u = ex (the one
inside sin(ex))
20. , by substituting u = ln(x)
21. , by substituting u = ln(x)
22. , by substituting (the one
occurring in the term )
23. , by substituting u = cos(x)
24. , by substituting u = 1 + e2x12
1 2ln( )+ +e cx
− +lncos( )x c
e x
u x=2e cx +
ln ln( )x c +
141 ln1x 2 24 � c
�17
cos17x � 2 2 � c
4sin1x 2 � c
14
sin14x 2 � c
13
sin31x 2 � c
12
�c
sin21x 2 � cos21x 2 � 1
�12
cos21x 2 � c
u � cos1x 212
sin21x 2 � c
u � sin1x 2
12
4 10ln x c + +
�1
1614x2 � 5 22� c
1414x2 � 5x � 1 24 � c
23x4 � 2x � 1 � c
12
sin1x4 2 � c
56
13
2 132( )x c+ +
38
1243( )x c− +
14
x4 �92
x2 � 4x � c
115
14414x � 3 211 � c
1401x5 � 1 28 � c
8 06
56sin( )x[ ] =π
π
4 4 4 4 3 4 2 42
33 2e e ex[ ] = − = − =
ln( )
ln( )ln( ) ln( ) ( ) ( )
a12
� e1b � 10 � e0 2 � e �12
− + +5 2cos( )x e cx
12u2 � 2cos1u 2 � c
2 2 2
4 2 4 2 2
22ln ln ln
ln( ) ln( )
u e e
e ee
e[ ] = −
= − = − =
–SOLUTION KEY–
190
Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 190
Lesson 20
1. , done by parts with
2. , by parts with
3. , by the substitution
4. , by parts with
5. , by substituting
6. , using
parts twice, first with u = x2, and then with
u = 2x
7. , by basic integration
8. , by substituting
9. , by substituting u = e3x + 9
10. ,
by parts with
11. , evaluating
by parts with
12. , substituting
13. , by parts
with
14. , by parts with
15. , by substituting u = cos(x)
16. , by substituting u = ln(x)
17. , by basic integration
18. e–cos(x) + c, by substituting u = –cos(x)
19. , by substituting
20. , by substituting
21. First, let u = six(x).
Then, .
From an earlier example,
. So, we get
sin(x) ⋅ ln(sin(x)) – sin(x) + c.
22. , by parts twice,
plus the trick from the last example in the
lesson.
121e
xsin1x 2 � e xcos1x 2 2 � c
ln( ) ln( )u du u u u c= − +�
ln( )u du�cos( )ln sin( )x x dx( ) =�
u � cos1x 2�231cos1x 2 2
32 � c
u �1x
�e 1x � c
13
ln x c +
−( )
+1
44
ln( )xc
− +14
4cos ( )x c
u � x�xe�x � e�x � c
u � x
23
x˛1x � 1 232 �
4151x � 1 2
52 � c
u � x � 1231x � 1 2
32 � c
u � ln1x 2ln( )x dx�ln ln( )x x x x c + − +
u � ln1x 2
a14
x4 �32
x2 � xb ln1x 2 �1
16x4 �
34
x2 � x � c
13
93ln( )e cx + +
u � x3 � 113
ex3�1 � c
13
x3 � cos1x 2 � c
�x2cos1x 2 � 2xsin1x 2 � 2cos1x 2 � c
u � ln1x 2121 ln1x 2 22 � c
u � x � 3
1x � 3 2sin1x 2 � cos1x 2 � c
u � x2�12
cos1x2 2 � c
u � x�xcos1x 2 � sin1x 2 � c
u � ln1x 2
16
x6ln1x 2 �1
36x6 � c
–SOLUTION KEY–
191
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Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 192
acceleration the rate at which the velocity of a
moving object is increasing or decreasing
additive rule parts of a function added together
can be differentiated separately:
antiderivative given a function , the
anti-derivative is a function such that
.
asymptote a line that a graph flattens out
toward. It occurs as either an infinite limit or a
limit as .
Chain Rule
closed interval the set of all the real numbers
between and including two endpoints, like all x
such that
composition the process of plugging one func-
tion into another. The composition of functions
f and g is .
concave down when the graph of a function
bows downward, like a frown
concave up when the graph of a function bows
upward, like a smile
concavity the way a graph curves either upward
or downward
Constant Coefficient Rule a constant c multi-
plied in front of a function is unaffected by dif-
ferentiation:
Constant Rule the derivative of a constant is
zero.
continuous function one whose graph can be
drawn without picking up the pencil
cosecant abbreviated csc; see trigonometry
cosine abbreviated cos; see trigonometry
cotangent abbreviated cot; see trigonometry
critical points points of slope zero, points where
the derivative is undefined, and endpoints of the
domain
decreasing when the graph of function goes
down from left to right
definite integral the area between a graph
and the x-axis from x � a to x � b
where area below the x-axis counts as negative,
written
degrees measure the size of angles in such a way
that a complete circle is 360°
�a
b
f˛1x 2dx
y � f˛1x 2
ddx1c # f˛1x 2 2 � c # f ¿ 1x 2
( )( ) ( ( ))f g x f g xo =
a � x � b
ddx1f˛1g1x 2 2 2 � f ¿ 1g1x 2 2 # g¿ 1x 2
x → ±∞
g¿ 1x 2 � f˛1x 2
g1x 2
f˛1x 2
ddx1f1x 2 � g1x 2 2 � f ¿ 1x 2 � g¿ 1x 2
GLOSSARY
193
Calc2e_23_193-196_GLOS 11/18/11 1:07 AM Page 193
derivative the derivative of function is
,
which is the slope of the tangent line at point
.
discontinuity a break in a graph
domain the set of all the real numbers at which a
function can be evaluated
e a transcendental number approximately equal to
2.71828 . . .
explicit a function is explicit if its formula is
known exactly.
exponent an exponent says how many times a
factor is multiplied by itself. In the case of roots,
the exponent is a fraction.
First Derivative Test if a function increases to a
point and then decreases afterward, then that
point is a local maximum. If the function
decreases to a point and then increases afterward,
then the point is a local minimum.
function a mathematical object that assigns one
number in its range to every number in its
domain
Fundamental Theorem of Calculus if
, then . Thus,
where .
graph a visual depiction of a function where the
height of each point is the value assigned to the
number on the horizontal axis
horizontal asymptote a horizontal line toward
which the graph flattens out as or
implicit a function is implicit if it was defined in
an indirect manner so that its exact formula is
unknown.
implicit differentiation the process of taking a
derivative of both sides of an equation and using
the Chain Rule with , ,
, and so on
increasing when the graph of a function goes up
from left to right
indefinite integral represents the antiderivative:
if and only if
integral see either definite integral or indefinite
integral
L’Hôpital’s Rule If and
, then .
The same is true when .
limit the limit means that the values
of get very close to L as x gets close to a.
limit from the left means that the
values of are close to L when x is close to,
and less than, a.
limit from the right means that the
values of are close to L when x is close to,
and greater than, a.
limits at infinity means that
the values of get close to as x gets
really big. If large negative values of x are used and
gets close to , then .
limits of integration the limits of the integral
are a and b.
local maximum the highest point on a graph in
that immediate area, like a hilltop
�b
a
f˛1x 2dx
lim ( )x
f x L→−∞
= y � Ly � f˛1x 2
y � Ly � f˛1x 2
lim ( )x
f x L→∞
=
f˛1x 2
limxSa�
f˛1x 2 � L
f˛1x 2
limxSa�
f˛1x 2 � L
f˛1x 2
limxSa
f˛1x 2 � L
limx→−∞
lim( )
( )
( )
( )x x
f x
g x
f x
g x→∞ →∞= ′
′ limlim ( )
xg x
→∞= ± ∞
lim ( )x
f x→∞
= ± ∞
g¿ 1x 2 � f˛1x 2� f˛1x 2dx � g1x 2 � c
ddt1x 2 �
dxdt
ddx1y 2 �
dy
dxddx1x 2 � 1
x → −∞x → ∞
g¿ 1x 2 � f˛1x 2�b
a
f˛1x 2dx � g1b 2 � g1a 2
g¿ 1x 2 � f˛1x 2g1x 2 � �x
0
f˛1t 2dt
1x,f˛1x 2 2
dy
dx� f ¿ 1x 2 � lim
aS0
f˛1x � a 2 � f˛1x 2
a
y � f˛1x 2
–GLOSSARY–
194
Calc2e_23_193-196_GLOS 11/18/11 1:07 AM Page 194
local minimum the lowest point on a graph in that
immediate area, like a valley
natural logarithm the inverse of the expo-
nential function . Thus, if and only if
.
oscillate to repeatedly go back and forth across a
range of values
point of inflection a point on a graph where the
concavity changes
point-slope formula the equation of a straight
line through with slope m is
.
polynomial the sum of powers of a variable,
complete with constant coefficients. For example,
and are
both polynomials.
position function gives the mark on a line where a
moving object is at a given time
Power Rule
Product Rule
Pythagorean theorem the squares of the legs of a
right triangle add up to the square of the
hypotenuse.
Quotient Rule
radians measure the size of angles in such a way
that a complete circle is 2p radians
range the set of all the numbers that can be the
value of a function
rate of change how fast a quantity is increasing or
decreasing
secant abbreviated sec; see trigonometry
rational function a rational function is the quo-
tient of two polynomials. For example,
is a rational function.
second derivative the derivative of the first derivative
Second Derivative Test a point of slope zero is the
maximum if the second derivative is negative at
the point, and a minimum if the second derivative
is positive at the point.
sign diagram tells where an expression is positive
and negative
sine abbreviated sin; see trigonometry
slope a measure of steepness of a straight line. It is
the amount the y-value goes up or down with
each step to the right.
slope-intercept formula the equation of a straight
line with slope m that crosses the y-axis at y � b is
.
Squeeze Theorem if and
, then .
substitution an integration technique used to
reverse the Chain Rule
tangent abbreviated tan; see trigonometry
tangent line a straight line that indicates the direc-
tion of a curve at a given point
third derivative the derivative of the second
derivative
trigonometric identities ,
, ,
, and sin2 1x 2 � cos2 1x 2 � 1cot1x 2 �cos1x 2
sin1x 2
csc1x 2 �1
sin1x 2sec1x 2 �
1cos1x 2
tan1x 2 �sin1x 2
cos1x 2
limxSa
g1x 2 � LlimxSa
f˛1x 2 � L � limxSa
h1x 2
f˛1x 2 � g1x 2 � h1x 2
y � mx � b
8x3 � 10x � 45x � 2
ddxa
f˛1x 2
g1x 2b �
f ¿ 1x 2 # g1x 2 � g¿ 1x 2 # f˛1x 2
1g1x 2 22
ddx1f˛1x 2 # g1x 2 2 � f ¿ 1x 2 # g1x 2 � g¿ 1x 2 # f˛1x 2
ddx1xn 2 � n # xn�1
10x7 � 12x5 � 4x2 � xx2 � 3x � 5
y � m1x � x1 2 � y1
1x1,y1 2
ey � x
y � ln1x 2ex
ln1x 2
–GLOSSARY–
195
Calc2e_23_193-196_GLOS 11/18/11 1:07 AM Page 195
trigonometry the study of functions formed by
dividing one side of a right triangle by another.
When the right triangle has angle x, the
hypotenuse has length H, the side adjacent to x
has length A, and the side opposite x has length O,
then
The derivatives are:
unit circle the circle of radius 1 centered at the
origin
velocity the rate of change of a moving object at a
particular time
vertical asymptote a vertical line x = a that a graph
more closely resembles as the inputs x approach a
ddx1cot1x 2 2 � �csc2 1x 2
ddx1tan1x 2 2 � sec2 1x 2
ddx1csc1x 2 2 � �csc1x 2cot1x 2
ddx1sec1x 2 2 � sec1x 2tan1x 2
ddx1cos1x 2 2 � �sin1x 2
ddx1sin1x 2 2 � cos1x 2
cot1x 2 �AO
tan1x 2 �OA
csc1x 2 �HO
sec1x 2 �HA
cos1x 2 �AH
sin1x 2 �OH
–GLOSSARY–
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