ma 251, calculus i, fall 2010 lecture...

MA 251, Calculus I, Fall 2010 Lecture notesLast modified January 19, 2011

1

Contents

1 Review 31.1 Review: Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Limits 92.1 Tangent and velocity problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.3 (and some of section 2.5) Algebraic approach to limits . . . . . . . . . . . . . . . . . . . . 142.5 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.6 Limits at infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.7 Tangents and velocities revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.8 the derivative as a function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3 Rules for Derivatives 343.1 Shortcuts for powers of x, ex, constants, sums, and differences . . . . . . . . . . . . . . . . 343.2 Product and Quotient Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.3 Trigonometric functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.4 The Chain rule!! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.5 Implicit derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453.6 Logarithmic differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.7 Rates of change in the natural sciences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523.9 Related rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4 Applications of Derivatives 614.1 Maximums and minimums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 614.2 Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 624.3 1st and 2cnd derivative tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 664.4 L’Hospital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744.5 (and some of 4.6) Function analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 844.7 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 944.9 Anti-derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

5 The Definite Integral (Compressed Version) 995.2 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 995.3 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1045.4 The Net Change Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1045.5 U -substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

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Chapter 1

Review

1.1 Review: Functions

Rules of exponents

a−b means1

ab

a1/b means b√a

(an)m = anm

anam = an+m

Example 1. Using the above properties, we can simplify the following.

1. (−2)5 = −32

2.x17

x22= x17−22 = x−5 =

1

x5

3. 4−3/2 =1

43/2=

1

(√

4)3=

1

23=

1

8

4.√

36x4 = 6x2

Inverse functions

The function ln(x) is the inverse function of ex. In other words

a = ln(b) means the same thing as ea = b

Another way to put this isln(ex) = x or ln(e2) = 2

where 2 is supposed to represent a box that you would put things inside of.

3

Example 2. Simplify ln(e2x+1).Solution: If we think in terms of boxes, ln(e2) = 2, we have 2x+ 1 inside of the box:

ln(e2x+ 1

) = 2x+ 1 = 2x+ 1

Example 3. We can solveex−1 = 2

using ln(x).Solution: We have

ln(ex−1) = ln(2)

x− 1 = ln(2)

x = ln(2) + 1

Definition. Here’s the most common notation for a function: “f(x)”. In this, f is the name of afunction, x is the input, and whatever f(x) equals is the output.

Example 4. Let f be the function given by f(x) = x2. Then x represents the input, the output isx2. For instance f(2) = 4. Even though this example is so simple, it still helps us illustrate some veryimportant concepts. Make sure you understand what role x plays here: it just stands for the input. Sothe function could have been described this way “f is the function which takes an input and square it.”Why do I care? Because we need to know how to calculate things like f(3x) and f(x + 3). If you gettoo focused on thinking that f squares x , then you might think that f(3x) = 3x2. No! The functionsquares any input, and in the case of f(3x), the input is 3x. So the output is (3x)2.

Now, if you really understand the notation, you should be able to say what f(x + 3) is without amoment’s hesitation. . . . . . . . . . I hope you said (x+ 3)2, but if not, keep practicing!

Example 5. Let f(x) = x2 + x+ 1.

1. Find f(1)

2. Solve f(x) = 1

3. Find f(h)

4. Find f(x+ 1)

Solution:

1. f(1) = 12 + 1 + 1 = 3.

2. We set the output equal to 1 and solve

x2 + x+ 1 = 1

x2 + x = 0

x(x+ 1) = 0

x = 0,−1

3. f(h) = h2 + h+ 1.

4. f(x+ 1) = (x+ 1)2 + (x+ 1) + 1 = x2 + 3x+ 3

4

Trigonometric functions

The trigonometric functions are important, and have all kinds of special properties that we will want toremember.

First, please remember the following:

In Calculus all angles will bemeasured in radians!

The reason for this is that the derivative rules that we will learn, only work in radians.The definition of radians is: the number of radians of θ, as pictured, is given by dividing the length

s by the length r, i.e θ =s

r.

s = arc lengthθ

r = radius

⇒ θ rad =s

r

If you apply this with a circle of radius 1 and use the whole 360◦ you find

θ = 360◦, s = 2π

r = 1

θ = 360◦ = 2π rad

and from this you can easily get a bunch of other angles

180◦ = π, 90◦ = π/2, 60◦ = π/3, 45◦ = π/4, 30◦ = π/6.︸︷︷︸the “standard” angles

Now we’ll calculate sin and cos at basic angles. Recall that sin =opp

hypand cos =

adj

hyp. Now we draw

a 30-60-90 triangle, and recall that the sides have lengths 1, 2 and√

3 (in general such a triangle haslengths with these ratios).

5

√3

12

30◦

60◦

From this we seesin(π/6) = 1/2 and cos(π/6) =

√3/2.

If we look at the 60◦ angle in this triangle, and think about the opposite and adjacent sides, then we canalso figure out

sin(π/3) =√

3/2, cos(π/3) = 1/2.

We can draw a similar picture for 45◦ = π/4, and find the following

1

1

√2

45◦

45◦

⇒ sin(π/4) =1√2, cos(π/4) =

1√2

Now we know how to calculate sin and cos (and also tan since tan(θ) = sin(θ)cos(θ)) at all the basic angles,

between 0 and 90◦, but we need to extend our calculations to all possible angles, in particular all anglesgreater than 90◦.

Given any angle θ, we place θ in a circle of radius r and let α be the reference angle, i.e. the smallestangle between θ and the x-axis, and let (x, y) be the coordinates on the circle determined by θ.

θ

(x, y)α r

Then we have the following definition and result about sin, cos and tan.

sin(θ) =y

r= ± sin(α)

cos(θ) =x

r= ± cos(α)

tan(θ) =y

x= ± tan(α)

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In each line above, the first equality is the definition of how to calculate a trig function for any value ofθ, but the way people actually calculate these values is given using α. In other words, people reduce anyangle θ into an angle α that is between 0 and 90◦. Then they decide whether they should take + or −in the above equations. The + or − is determined by whether y or x is + or −. The results for whichfunctions are positive are given in the following picture, which has various mnemonics.

Positive trig functions in each quadrant

A (All)S (Sine)

T (Tangent) C (Cosine)

Finally, everything we have done so far gives you enough information to fin sin, cos and tan atmultiples of the standard angles. However, if the problem doesn’t give you one angle to work with, butrather asks you to find the angle that solves an equation, then it’s useful to have table listing all thestandard values. Here’s the table

0 π/6 π/4 π/3 π/2

sin(θ) 0 1/2 1/√

2√

3/2 1

cos(θ) 1√

3/2 1/√

2 1/2 0

note this pattern is easy to remember if youwrite the entries with square roots of 0,1,2,3,4on top:

√0/2,

√1/2,

√3/2,

√4/2

Example 6. Solve cos(x) = −1/2.Solution: We break this into to parts: which quadrant should the angle be in to make cosine negative,and how close to the x-axis should the line be to make cos(x) = ±1/2.

We see that cosine is negative in quadrants II and III. In quadrant II, the gap between the line andthe x-axis should have 60◦, this means that x is 120◦, as pictured

x = 120◦

60◦

Another solution can be found in quadrant III. Here, we can have x = 180◦ + 60◦ = 240◦.

Example 7. Solve cos(x+ 5) = 1/2

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Solution: We know already that cos(60) = 1/2 and we want to make cos(x+5) = 1/2. If you comparewhat’s inside of cosine, I hope you will see that we can make this work if x+ 5 = 60. In other words, ifx+ 5 = 60 then cos(x+ 5) = cos(60) = 1/2.

Now we can solve x+ 5 = 60 to get x = 55.

Example 8. Let f(x) be defined by the following formulas, each applying to just one piece of f(x).

f(x) =

x2 if x ≤ 0

−x2 if 0 < x ≤ 3

ex−2 if 3 < x

Find f(−1), f(2) and f(4), and make a graph of f(x).Solution: For f(−1) we plug x = −1 into the first formula. Thus f(−1) = (−1)2 = 1.

For f(2) we use the second formula, f(2) = −22 = −4.For f(3) we use the third formula, f(4) = e2.The graph looks like x2 on the left (i.e. for x ≤ 0); it looks like −x2 in the middle (for 0 < x ≤ 3)

and it looks like ex on the right (for x > 3). Notice that the graph looks “unnatural,” especially at x = 3where it is discontinuous.

K6 K4 K2 0 2 4 6

10

20

30

40

50

8

Chapter 2

Limits

2.1 Tangent and velocity problems

The problems in this section are at the heart of Calculus and lead directly to the main idea in Calculus,limits.

Example 1. The height of a thrown ball is given by the following function:

p(t) = −4.9t2 + 3.5t+ 2.

Find (approximate) the velocity at t = 2.3.Solution: The way we “solve” this problem is by approximation.

Velocity =change in distance

change in time

we will calculate change in distance using the formula for p(t), and plugging in t = 2.3 and a secondvalue close to 2.3.

For t = 2.2 we get

p(2.2) = −14.016, p(2.3) = −15.871 ⇒ vel =−14.016− (−15.871)

2.2− 2.3= −18.55

Watch out: the velocity we have just calculate is the average velocity from t = 2.2 to t = 2.3. It is notthe exact velocity at the instant of time at t = 2.3. Now, we can repeat this calculation for values closerto 2.3. Note, we cannot plug in t = 2.3 directly for our second time value in this velocity fraction, sincethis will give us a division by 0 error. Here are the values my TI-84 gives me:

t p(t)p(t)− p(2.3)

t− 2.32.29 −15.68 −18.992.299 −15.85 −19.042.31 −16.06 −19.092.301 −15.89 −19.04

From these values, it looks pretty clear that the velocity we calculate gets closer and closer to −19.04 aswe use t-values closer and closer to 2.3. So, our best answer appears to be −19.04.

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There are two conclusions we will make from this example (and partly by looking ahead and knowingwhat Calculus is about): Sometimes we can make a sequence of approximations that appear to be gettingcloser and closer to the correct answer; A difference quotient may be an important example of a thingthat we can approximate like this.

In fact, both of these observations are extremely profound and important. The first one becomes, inits finial from, the concept of a mathematical limit. The second one becomes the derivative.

Finally, note that even the partial solution we have of this problem is a monumental step forwardin the history of science. The very idea that you could quantify physical processes would have beenincomprehensible to almost everyone, until Galileo. After Galileo it took another 100 years before peopleunderstood that you could start with one quantified process, position in this case, and do something toit to calculate its rate of change.

Example 2. Find the equation of the tangent line at x = 0.8 for the function f(x) = sin(x), and makea graph showing f(x) and the tangent line.Solution: We always need to know two things to find the equation of a line: this could be two points,or one point and the slope1. In this case, we do know a point that should be on the tangent line, namelythe point x = 0.8, y = sin(0.8) ≈ 0.7173. The hard part, the crucial part, is to find the slope of thetangent line.

As in the velocity problem, we “solve” the problem of finding this slope by making approximations.

slope =rise

run

We will calculate slope by plugging in x = 0.8, and another value close to 0.8.For x = 0.9 we get

sin(0.9) ≈ 0.7833, sin(0.8) ≈ 0.7173 ⇒ slope =0.7833− 0.7173

0.9− 0.8≈ 0.6597

Of course, this is the slope through two points, and we want to figure out the slope at just one point.Now, we can repeat this calculation for values closer to x = 0.8. Note, we cannot plug in x = 0.8 for oursecond x-value in this slope fraction, since this will give us a division by 0 error. Here are the values myTI-84 gives me:

x sin(x)sin(x)− sin(0.8)

x− 0.80.81 0.7424 0.69310.801 0.7181 0.69640.79 0.7104 0.70030.799 0.7167 0.6971

From these values, it looks pretty clear that the slope that we calculate gets closer and closer to 0.697as we use x-values closer and closer to 0.8. So, our best answer appears to be m = 0.697.

Finally, using this, our best tangent line appears to be

y = 0.697(x− 0.8) + 0.7173

Here is the graph, which shows that our line is indeed tangent to sine

1I’ll always use the point-slope form of the equation of a line: y = m(x−x0) + y0, where m is a known slope and (x0, y0)is a known point.

10

0 1 2 30

0.5

1.0

1.5

2.0

2.2 Limits

Recall one of the main conclusions of the previous section: that sometimes we do a calculation, have oneof the inputs of the calculation get closer and closer to a single number, and see that outputs are gettingcloser and closer to a single number. If we take this idea, and make a formal definition out of it, we getthe technical concept of limit.

Definition. We write

limx→a

f(x) = L to mean that

{the y-values of f(x) become infinitely close to L asthe x-values become infinitely close to the numbera.

}

Note two things about this definition. (1) The phrase “infinitely close” is vague. But it can bemade precise using a somewhat technical mathematical description involving finite, but arbitrarily small,distances (δ’s and ε’s). (2) When we have x infinitely close to a, we do not let x equal a.

There are a few of basic variations on the definition of limit.


limx→a+


the y-values of f(x) become infinitely close to L asthe x-values become infinitely close to the numbera, but with x > a.

We write

limx→a−


the y-values of f(x) become infinitely close to L asthe x-values become infinitely close to the numbera, but with x < a.

We write

limx→a

f(x) =∞ to mean that

{the y-values of f(x) become infinitely large as thex-values become infinitely close to the number a.

}

We can also combine these concepts to have

limx→a+

f(x) =∞, limx→a−

f(x) =∞, limx→a

f(x) = −∞, limx→a+

f(x) = −∞, limx→a−

f(x) = −∞.

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Example 1. Example 1 from section 2.1 involved a limit. Can you figure out what we took the limit ofand what it equalled?Solution: Although we started with a function p(t), the height of a ball, this wasn’t what we took thelimit of. If you don’t see the limit right away, think about this: what sort of t-values were we pluggingin? Were they getting closer and closer to something? Hopefully you see that we took lim

t→2.3of something.

What were we plugging these t-values into? Not just p(t), although that was part of it. We were plugging

them intop(t)− p(2.3)

t− 2.3. Putting all this together, we have the following

Example 1 in 2.1 showed that limt→2.3

p(t)− p(2.3)

t− 2.3≈ −19.04.

Example 2. Example 2 from section 2.1 involved a limit. Can you figure out what we took the limit ofand what it equalled?Solution: Although we started with a function sin(x), the height of a ball, this wasn’t what wetook the limit of. If you don’t see the limit right away, think about this: what sort of x-values were weplugging in? Were they getting closer and closer to something? Hopefully you see that we took lim

x→0.8of

something. What were we plugging these x-values into? Not just sin(x), although that was part of it.

We were plugging them intosin(x)− sin(0.8)

x− 0.8. Putting all this together, we have the following

In Example 2 from section 2.1 we showed that limx→0.8

sin(x)− sin(0.8)

x− 0.8≈ 0.7.

Example 3. Find limx→3

sin(x− 3)

x− 3.

Solution: At this point, the only way we can find this limit, is to use our calculators. We can eithermake a table of numbers, or we can graph the function. Since we made a table of numbers for ourexamples in section 2.1, this time we will look at the graph.

1 2 3 4 5

0.5

0.6

0.7

0.8

0.9

1.0

To read the limit from this graph you have to look at two things at the same time: the y-values on thegraph and the x-values near 3. If you do this, I hope you see that the y-values are getting infinitely closeto y = 1 as the x-values get infinitely close to x = 3. Thus, the limit is 1

limx→3

sin(x− 3)

x− 3= 1

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Example 4. Find the following limit, if it exists.

limx→1

sin

(1

x− 1

)

Solution: Again, we graph the function, although this time the graph is perhaps confusing.

0.5 1.0 1.5 2.0

K1.0

K0.5

0

0.5

1.0

This is because the graph oscillates up and down between 1 and −1 faster and faster as x approaches 1.Therefore, there is not a single number that all the y-values get closer and closer to, as x gets closer to1. Therefore, this limit does not exist

limx→1

sin

(1

x− 1

)= DNE

Example 5. (Section 2.2#7, from the book)Solution: The key here is to look at the y-values on the graph, and see what they do as x gets closeto 0 or 2 or 4. In the case of 0 and 2, the left side and the right side are different. For instance, if x isclose to 0, but just a little to the left, then the y-values are close to −1. In other words, lim

x→0−f(x) = −1.

Similarly, limx→0+

f(x) = −2. Etc.

Example 6. (Section 2.2, #15, from the book) Make up a graph of a function f(x) that has the followingproperties: (a) lim

x→3+f(x) = 4, (b) lim

x→3−f(x) = 2, (c) lim

x→−2f(x) = 2, (d) f(3) = 3, (e) f(−2) = 1.

Solution: The key is to make up pieces of a function that have the appropriate property. You cannothave a single, basic formula, that has all these properties. But the concept of limit only describes what’shappening very close to a certain point, so the conditions we’ve been given describe what happens closeto the point x = 3 for instance. We can make up whatever we want for x-values away from these certainpoints. Here are two solutions.

13

K4 K2 0 2 4 6

1

2

3

4

K4 K2 0 2 4 6

K5

5

10


1

(x− 2)2.

Solution: We graph the function, with two different windows, in case there is a peak that we can’tsee in the first window.

K2 K1 0 1 2 3 4 5 6

K1

1

2

3

4

5

K2 K1 0 1 2 3 4 5 6

10

20

30

40

50

From these graphs we see that the function has a vertical asymptote at x = 2. The best way to describethis is to write

limx→2

1

(x− 2)2=∞

Note: people also sometimes say that this limit does not exist. It is correct to say this, because the limitis not a real number, however it is better to write ∞ as we have, because it provides more information.If you like, you can write both DNE and ∞.

2.3 (and some of section 2.5) Algebraic approach to limits

Now we start to learn how to find limits algebraically. This starts with the simplest possible limits, andthen builds these up to more complicated examples.

14

Fact 1. If C is a constant, then limx→a

C = C.

Fact 2. limx→a

x = a

Fact 3. If f(x) and g(x) are any functions with limx→a

f(x) and limx→a

g(x) both existing then we have

1. limx→a

[f(x)± g(x)] = limx→a

f(x)± limx→a

g(x)

2. limx→a

[f(x)g(x)] = limx→a

f(x) limx→a

g(x)

3. limx→a

f(x)

g(x)=

limx→a

f(x)

limx→a

g(x)(as long as the limit on the bottom is not zero).

Using these facts we can already do a simple limit.


(3x2 + 2x+ 5), algebraically, using the limit laws, showing all possible steps.

Solution:

limx→4

(3x2 + 2x+ 5) Next use Fact 3.1

= limx→4

3x2 + limx→4

2x+ limx→4

5 now use use Fact 3.2

= limx→4

3 · limx→4

x · limx→4

x+ limx→4

2 · limx→4

x+ limx→4

5 now use Facts 1 and 2

= 3 · 4 · 4 + 2 · 4 + 5

= 61

You can generalize the same argument as in Example 1, to any polynomial

Fact 4. If p(x) is any polynomial then limx→a

p(x) = p(a)

In other words JUST PLUG IT IN!

Theorem 1. If f(x) is any combination (i.e. sum, product, fraction, composition, etc.) of basic functions(i.e. powers of x, exponentials, trig functions, inverses, etc.) and x = a is in the domain of f(x), then

limx→a

f(x) = f(a)

In other words JUST PLUG IT IN! (as long as it’s defined).


5x2 + 2x

sin(π2x) .

Solution: Using Theorem 1 we just plug in x = 3 and get5 · 32 + 2 · 3sin(3π/2)

= −51.

If we can’t just plug it in then we use the following result.

Theorem 2 (Squeeze Theorem). Let g(x) ≤ f(x) ≤ h(x) for all x near x = a (except possibly at x = a).Suppose

limx→a

g(x) = limx→a

h(x) = L.

Thenlimx→a

f(x) = L

15

Proof. We have that the y-values of g(x), and the y-values of h(x) are becoming infinitely close to L.But since f(x) is between g(x) and h(x), the y-values of f(x) must also become infinitely close to L.

Example 3. Make up graphs that illustrate the Squeeze Theorem.Solution: We start by making up two graphs that will be the top and bottom, but that squeezetogether at one point. Then, between these curves, we try to draw any curve we feel like, but with therequirement that it stay between the top and bottom. The curve in the middle will have no choice butto go through the squeeze point.


x2 sin(1/x) + 1 using the Squeeze Theorem and graph the results.

Solution: Recall that sin(?) ≤ 1, where “?” represents anything. Thus

x2 sin(1/x) + 1 ≤ x2(1) + 1

where we replaced sin(1/x) with 1. Similarly, −1 ≤ sin(?) and so

x2(−1) + 1 ≤ x2 sin(1/x) + 1.

Combining these inequalities we have

x2(−1) + 1︸︷︷︸g

≤ x2 sin(1/x) + 1︸︷︷︸f

≤ x2(1) + 1︸︷︷︸h

Now we are half-way to using the Squeeze Theorem, but we need to verify that the limits of g and hare equal. Here we can use the JUST PLUG IT IN results:

limx→0

g(x) = 0(−1) + 1 = 1 limx→0

h(x) = 0(1) + 1 = 1

Since these are equal, we have

therefore, by Sq. Thm, limx→0

f(x) = 1

Here is the graph:

16

Example 5. (Section 2.3#38) Find limx→0+

√x esin(π/x).

Solution: Using −1 ≤ sin( ) ≤ 1 we have

√x e−1 ≤ √x esin(π/x) ≤ √x e1

Let g(x) be the first function and h(x) the last.limx→0+

g(x) = limx→0+

√x e−1 =

√0e−1 = 0

limx→0+

h(x) = limx→0+

√x e1 =

√0e1 = 0.

Therefore these limits are the same.Therefore, by the Squeeze Theorem, we have lim

x→0+f(x) = 0.

This example also has a nice picture that illustrates the Squeeze Theorem.

0 0.2 0.4 0.6 0.8 1.00

0.5

1.0

1.5

2.0

2.5

Corollary (Corollary of Squeeze Theorem). If h(x) = f(x) for all x near x = a (except possibly atx = a) and lim

x→ah(x) = L, then lim

x→af(x) = L.

17

Proof. We apply the squeeze theorem. Let g(x) = h(x). Then we have g(x) ≤ f(x) ≤ h(x) for all xnear x = a, as needed in the Squeeze Theorem. Then lim

x→ag(x) = lim

x→ah(x) = L, and so lim

x→af(x) = L as

well.


x2 + x− 6

x− 2.

Solution:

limx→2

x2 + x− 6

x− 2now factor

= limx→2

(x− 2)(x+ 3)

x− 2now cancel x− 2

= limx→2

x+ 3 now “just plug in”

= 2 + 3 = 5

The crucial theoretical step is where we cancel x− 2. Note that the functions x− 2 andx2 + x− 6

x− 2are

not exactly the same function. They are the same everywhere except x = 2, where the first is definedand the second is not. By the Corollary of the Squeeze Theorem these two functions do have the samelimit.

Example 7. Find limx→−4

x2 + 5x+ 4

x2 + 3x− 4.

Solution:

limx→−4

x2 + 5x+ 4

x2 + 3x− 4now factor

= limx→−4

=(x+ 1)(x+ 4)

(x+ 4)(x− 1)now cancel the factor x+ 4

= limx→−4

x+ 1

x− 1now “just plug in”

=−4 + 1

−4− 1=−3

−5=

3

5

Again, the crucial theoretical step is that it’s ok to cancel.

Here are some rules of thumb for algebraic manipulation of limits:

• If the limit off(x)

g(x)involves division by 0, then factor f and/or g, cancel the factor that gives the

0.

• If the limit off(x)

g(x)− h(x)

p(x)involves division by 0, then combine the fractions using a greatest

common denominator, then factor things and cancel the part giving division by 0.

• If the limit of

√f(x)−

√g(x)

h(x)involves division by 0, then rationalize the numerator:

√f(x)−

√g(x)

h(x)·√f(x) +

√g(x)√

f(x) +√g(x)

=f(x)− g(x)

h(x)(√f(x) +

√g(x))

then factor things and cancel the part giving division by 0

18

Example 8. Find limh→0

√1 + h− 1

h.

Solution:

limh→0

√1 + h− 1

hrationalize the numerator

= limh→0

√1 + h− 1

h

√1 + h+ 1√1 + h+ 1

finish rationalizing

= limh→0

(1 + h)− 1

h(√

1 + h+ 1)cancel 1 and −1

= limh→0

h

h(√

1 + h+ 1)cancel the h

= limh→0

1√1 + h+ 1

“Just plug in”

=1√

1 + 0 + 1=

1

2

Example 9. Find limt→0

(1

t− 1

t2 + t

)

Solution:

limt→0

(1

t− 1

t2 + t

)Get a common denominator

= limt→0

(1

t

t+ 1

t+ 1− 1

t2 + t

)finish getting common denominator

= limt→0

(t+ 1

t2 + t− 1

t2 + t

)combine fractions

= limt→0

t+ 1− 1

t2 + tcancel 1 and −1

= limt→0

t

t2 + tfactor bottom

= limt→0

t

t(t+ 1)cancel t

= limt→0

1

t+ 1“Just plug in”

=1

0 + 1= 1


sin(x)

xSolution: Numerically, or graphically, it appears that this limit should be 1. However, we wish toprove this (pretty) rigorously. Suppose the angle θ is between 0 and π/2. If we draw θ in a unit circle,the areas pictured below can be compared as shown

19

area≤

area≤

Labelling these quantities we have

1

cos(θ)

sin(θ)

1

2cos(θ) sin(θ)

area≤

1 θ

θ

2π· π · 12 =

θ

2

area≤ 1

tan(θ)

1

2· 1 · tan(θ)

We arrange these by size and calculate

1

2cos(θ) sin(θ) ≤ θ

2≤ 1

2tan(θ) now mult by 2, and replace tan(θ)

cos(θ) sin(θ) ≤ θ ≤ sin(θ)

cos(θ)now divide by sin(θ)

cos(θ) ≤ θ

sin(θ)≤ 1

cos(θ)now take inverses

1

cos(θ)≥ sin(θ)

θ≥ cos(θ)

Note that limx→0+

1

cos(x)= lim

x→0+cos(x) = 1.

Therefore, by the Squeeze Theorem,

limx→0+

sin(x)

x= 1

Finally, since sin(x) is an odd function, and x is an odd function, we have thatsin(x)

xis an even function.

Therefore limx→0−

sin(x)

xequals the same thing as lim

x→0+

sin(x)

x. Since both the left and right hand limits

equal 1, the two sided limit equals 1 as well.

2.5 Continuity

Definition. Let f(x) be any function and x = a a number in the domain of f(x). We say that f(x) iscontinuous at x = a if lim

x→af(x) = f(a).

In this language, the plugging in theorem says that all combinations of our basic functions arecontinuous everywhere they defined.

20

Example 1. Make up two pictures of how a function could fail to be continuous at a point. What doescontinuity mean about holes in the graph?Solution: To not be continuous, we would need to have the y-values of f(x) not getting infinitelyclose to f(a). Thus, there would have to be a hole of some kind at x = a. The picture of a discontinuous(i.e. not continuous) function would have to look like one of the following:

“Removable discontinuity” “Jump discontinuity”

or

So, continuous at x = a means no hole at x = a.

Definition. If f(x) is any function defined on an interval [a, b], we say that f(x) is continuous on theinterval if it is continuous at every point in the interval.

By the discussion above, we can think of this as meaning that f(x) has no holes anywhere in theinterval. From this viewpoint, the plugging in theorem says that all combinations of our basic functionshave no holes on any interval they are defined on.

Example 2. Let g(x) be defined by the graph in section 2.2, #7. List the points where g(x) is notcontinuous.Solution: We looked at the graph and found the holes. They are at x = 0 and x = 2. Another wayto show that the function is discontinuous there is to note that

limx→0

g(x) = DNE 6= f(0)

limx→2

g(x) = DNE 6= f(2)

Example 3. Show2 that the function

f(x) =

2x2 − 5x− 3

x− 3if x 6= 3

6 if x = 3

is not continuous at x = 3.

2The word “show” here, and elsewhere means that you have to provide more than just an answer. You need to record(write) some of your reasons for this answer.

21

Solution: We start with the limit:

limx→3

2x2 − 5x− 3

x− 3= lim

x→3

(x− 3)(2x+ 1)

x− 3= lim

x→32x+ 1 = 2 · 3 + 1 = 7

On the other hand f(3) = 6.Therefore

limx→3

f(x) 6= f(3)

and f(x) is not continuous at x = 3.

Example 4. Find a and b so that the function defined by

f(x) =

x+ 1 if x ≤ −2

ax2 + bx if − 2 < x < 1

−x+ 2 if x ≥ 1

is continuous everywhere, and graph your resultsSolution: I’ll start with a little extra discussion, but then show the brief version. Since basic functionsare continuous everywhere they are defined, we see that x+ 1, ax2 + bx and −x+ 2 are each continuouson their own intervals. Thus, the only place that f(x) could even fail to be continuous are at x = −2and x = 1. Therefore, we need

limx→−2

f(x) = f(−2) and limx→1

f(x) = f(1)

In fact, we need both left hand and right hand limits to be equal at x = −2 and at x = 1, and this givesus the really crucial equations.

limx→−2−

f(x) = limx→−2+

f(x) = f(−2) and limx→1−

f(x) = limx→1+

f(x) = f(1)

Now, how to we find each limit? For limx→−2−

we would use the formula for f(x) when x < −2. Thus, we

would use limx→−2−

x+ 1. Now, to find this limit we would just plug in.

Summarizing the above discussion, here’s how you solve it.For continuity, we need the left and right hand limits to be equal at −2 and 1

limx→−2−

f(x) = limx→−2+

f(x) limx→1−

f(x) = limx→1+

f(x)

limx→−2−

x+ 1 = limx→−2+

ax2 + bx limx→1−

ax2 + bx = limx→1+

−x+ 2

−2 + 1 = a(−2)2 + b(−2) a(1)2 + b(1) = −(1) + 2

−1 = 4a− 2b a+ b = 1

We solve the second equation for b = 1 − a, so the first equation becomes −1 = 4a − 2(1 − a), whencea = 1/6 and b = 5/6.

For entertainment we double check this with a graph

22

K4 K3 K2 K1 0 1 2 3 4

K3

K2

K1

Theorem 3 (IMV: Intermediate Value Theorem). If f(x) is continuous on the interval [a, b], then thegraph of f(x) (for a ≤ x ≤ b) crosses every y-value between f(a) and f(b). In other words, for each Nbetween f(a) and f(b), there exists a solution x, with a ≤ x ≤ b, such that f(x) = N .

Example 5. Make up a graph illustrating the theorem.Solution:

?

The picture shows a function with values above and below some fixed line. Since the function is contin-uous, if you connect the parts of its graph, it has to cross the line.

Example 6. Show that 1071 has 5th root, and find integers that are upper and lower bounds for this5th root.Solution: If x =

5√

1071 then x is a solution of x5 = 1071.The basic idea is that the function f(x) = x5 starts small, but eventually becomes larger than 1071.More precisely:

f(1) = 15 = 1 ⇒ f(1) < 1071f(10) = 105 = 100, 000 ⇒ f(10) > 1071

Therefore, the IMV shows that there is some solution x, with 1 ≤ x ≤ 10, such that x5 = 1071.

23

Example 7. Show that the equation x+ sin(x) = e has a solution and find upper and lower bounds.Solution: Let f(x) = x+ sin(x).

f(0) = 0, which is too small.f(π) = π + 0, which is too big.Therefore, there is a solution of x+ sin(x) = e for some value of x with 0 ≤ x ≤ π.

2.6 Limits at infinity


limx→∞


{the y-values of f(x) become infinitely close to Las the x-values become infinitely large.

}

This limit goes by another name: f(x) has a horizontal asymptote, on the right, of y = L.We have obvious variations

limx→−∞

f(x) = L, limx→∞

f(x) =∞, limx→∞

f(x) = −∞, limx→−∞

f(x) = −∞

Example 1. Find limx→∞

1

xand discuss why this limit makes sense.

Solution: We approach this in three ways. Graphically, we look at the graph of1

x, and track the

y-values as x moves to the right. The limit of these y-values, also called the horizontal asymptote, is

y = 0. Thus, limx→∞

1

x= 0.

Alternatively, let’s think about fractions of the form 1/x for large values of x. What are your odds

of winning the lottery? One in a million, or1

1, 000, 000. What are your odds of winning the lottery and

being struck by lightning on the same day? One in a trillion, or1

1, 000, 000, 000, 000. What are your

odds of winning the lottery, being struck by lightning, and having a meteorite landing on your house, all

on the same day? One in a septillion, or1

1024. Now, what nonnegative number is less than all of these

odds, and all similar sorts of odds? The only such number is 0.

24

Finally, we take a somewhat algebraic approach3. Let ε be any positive number. Now let x be any

number greater than 1/ε. In other words x >1

ε. Now if we take inverses and reverse the inequality we

have1

x< ε. In other words, no matter how small ε is, we can make

1

xeven smaller. Thus, we can make

1

xinfinitely close to 0, and so the limit is 0.

Fact 1. Here are all the basic functions you know with horizontal asymptotes (as well as two functionsthat don’t have them).

• limx→±∞

1

xp= 0 for any real number p > 0.

• limx→−∞

ex = 0

• limx→∞

tan−1(x) = π/2, limx→−∞

tan−1(x) = −π/2

• limx→∞

ln(x) =∞, limx→∞

ex =∞, limx→∞

√x =∞

Now we generalize two of the above facts

Fact 2. If limx→a

f(x) = L (with L 6= ±∞) and limx→a

g(x) = ±∞, then

limx→a

f(x)

g(x)= 0

Note: the same result holds if we replace x→ a with x→ a+, x→ a−, x→∞ or x→ −∞.

Shorthand mnemonic: “L

±∞ = 0”. Be careful here, we are not treating ∞ as a real number, but

merely writing something which is shorthand for the correct statement involving limits.

Example 2. Find the horizontal asymptote of3x8 − 100x3 + 17.304

7x8 − x+ 100, 000.

Solution:

limx→∞

3x8 − 100x3 + 17.304

7x8 − x+ 100, 000now divide top and bottom by x8

= limx→∞

1x8

1x8

3x8 − 100x3 + 17.304

7x8 − x+ 100, 000now distribute

= limx→∞

3x8 1x8− 100x3 1

x8+ 17.304 1

x8

7x8 1x8− x 1

x8+ 100, 000 1

x8

now simplify

= limx→∞

3− 100x5

+ 17.304x8

7− 1x7

+ 100,000x8

now apply basic facts

=3− 100

∞ + 17.304∞

7− 1∞ + 100,000

∞now apply basic facts

=3− 0 + 0

7− 0 + 0=

3

7

3By “algebraic” I mean “notation, with rules”

25

Fact 3 (Horizontal Asymptotes of Rational Functions).

limx→∞

anxn + . . . (a poly. of degree n)

bmxm + . . . (a poly. of degree m)=

0 if n < manbm

if n = m

∞ if n > m

Rule 1. To find

limx→∞

sum of powers of x

sum of powers of x

we divide the top and the bottom by the biggest (simplified) power of x that is on the bottom. (“Sim-

plified” means we have to take into account things like√x2 + . . ..)


3x2 + x+ 1/x√9x4 + 1/x

Solution: The biggest “simplified” power on the bottom is x2, since we have x4 inside of a square root.

We divide the top and bottom by x2. At one step we have to move1

x2inside of a square root. Recall

that√a2b = a

√b, so to move something inside a square root, it comes in squared. Thus

1

x2√a =

√1

x4a

limx→∞

1x2

(3x2 + x+ 1/x)1x2

√9x4 + 1/x

distribute1

x2and move it inside the square root

= limx→∞

3 + 1x + 1

x3√1x4

(9x4 + 1x)

simplify

= limx→∞

3 + 1x + 1

x3√9 + 1

x5

apply basic facts

=3 + 1

∞ + 1∞3√

9 + 1∞5

=3 + 0 + 0√

9 + 0=

3

3= 1


x2 − 7x√x5 + 1

.

Solution: We start by dividing the top and bottom by x5/2

limx→∞

1x5/2

(x2 − 7x)1

x5/2

√x5 + 1

distribute1

x5/2

= limx→∞

x2−5/2 − 7x1−5/2√1

(x5/2)2(x5 + 1)

simplify

= limx→∞

1x1/2− 7

x3/2√1x5

(x5 + 1)simplify more

26

= limx→∞

1x1/2− 7

x3/2√1 + 1

x5

apply basic facts

=1∞ − 7

∞√1 + 1

∞

=0− 0√1 + 0

= 0


cos(x/π).

Solution: We look at the graph of f(x)

20 40 60 80 100

K1.0

K0.5

0

0.5

1.0

From this graph, it is clear that no horizontal asymptote exists, and thus the limit does not exist.

limx→∞

cos(x/π) = DNE


(√

4x2 + x+ 1− 2x).

Solution: The trick here is to first rewrite the function as a fraction, and then rationalize thenumerator.

limx→∞

(√

4x2 + x+ 1− 2x) force it into a fraction

= limx→∞

√4x2 + x+ 1− 2x

1rationalize numerator

= limx→∞

√4x2 + x+ 1− 2x

1·√

4x2 + x+ 1 + 2x√4x2 + x+ 1 + 2x

finish rationalizing

= limx→∞

(4x2 + x+ 1)− (2x)2√4x2 + x+ 1 + 2x

simplify

= limx→∞

x+ 1√4x2 + x+ 1 + 2x

divide top and bottom by x

27

= limx→∞

1x1x

x+ 1√4x2 + x+ 1 + 2x

distribute1

x

= limx→∞

1x(x+ 1)

1x(√

4x2 + x+ 1 + 2x)bring

1

xinside square root

= limx→∞

1 + 1x√

1x2

(4x2 + x+ 1) + 2distribute

1

x2inside square root

= limx→∞

1 + 1x√

4 + 1x + 1

x2+ 2

apply basic facts

=1 + 1

∞√4 + 1

∞ + 1∞ + 2

=1 + 0√

4 + 0 + 0 + 2=

1

4


(−28x11 + 1000x10 + 1).

Solution: This function is a polynomial, and its graph will not have a horizontal asymptote, norwill it oscillate up and down like sine and cosine. Rather, the limit will either be +∞ or −∞, and theproblem is to determine which of these is the answer.

As before, we will factor out powers of x. As before, we focus on the highest power of x.

limx→∞

(−28x11 + 1000x10 + 1)

= limx→∞

x11(−28 + 10001

x+

1

x11)

=∞(−28 + 0 + 0) = −∞

In fact, this example generalizes to a familiar rule: the end behavior of a polynomial depends only onthe leading term, i.e. the term with the largest power of x.

2.7 Tangents and velocities revisited

Definition. The derivative of f(x) at the point x = a is a number denoted by f ′(a) and defined asa limit

f ′(a) = limx→a

f(x)− f(a)

x− a = limh→0

f(a+ h)− f(a)

h

note that the two limits are equal: just let h = x − a. The fractions used here are called differencequotients, and they equal the slope of a secant line through the points (a, f(a)) and (x, f(x)) (or(a+ h, f(a+ h)) ), as pictured

28

f(x)

m =f(x)− f(a)

x− a =f(a+ h)− f(a)

h

a x = a+ h

h = x− a

Definition. The tangent line to a function f(x) at a point x = a is given by

y = f ′(a)(x− a) + f(a).

Example 1. Find the equation of the tangent line at x = 5 of f(x) = 2x2 − x+ 3.Solution: We have a = 5 and f(5) = 48. To find the slope, we find the limit of the differencequotient.

f ′(5) = limx→5

f(x)− f(5)

x− 5

= limx→5

2x2 − x+ 3− (2(52)− 5 + 3)

x− 5

= limx→5

2x2 − x+ 3− 48

x− 5

= limx→5

2x2 − x− 45

x− 5

= limx→5

(2x+ 9)(x− 5)

x− 5

= limx→5

2x+ 9

= 19

Thus, we havey = 19(x− 5) + 48

Definition. If f(t) equals position as a function of time, then the velocity at t = a (i.e. the instantaneousvelocity) is given by

v(a) = limt→a

f(t)− f(a)

t− a = limh→0

f(a+ h)− f(a)

h

note that the two limits are equal: just let h = t− a.

Example 2. Repeat the calculations of Example 1, Section 2.1, using limits and algebra.Solution: The function was p(t) = −4.9t2 + 3.5t+ 2 and we had a = 2.3.

(I’m always somewhat surprised that we can factor this function exactly since it has decimals!)We find v(2.3) = −19.04 m/s

29

Interpretations:

• If we are looking at a graph of f(x) then f ′(a) equals the slope of the tangent line at x = a.

• If f(t) is a position function with t = time, then f ′(a) equals the velocity at t = a.

• If f(x) equals any real quantity whatsoever with, then f ′(a) equals the rate of change of this

quantity, with units given byunits of f(t)

units of x.

Example 3. (Stewart 2.7#49) The quantity of oxygen that can dissolve in water depends on the tem-perature of the water. (So thermal pollution influences the oxygen content of water.) The graph showshow oxygen solubility S varies as a function of the water temperature T .

1. What is the meaning of the derivative S′(T )? What are its units?

2. Estimate the value of S′(16) and interpret it.

Solution: (a) The derivative measures the rate of change of oxygen as temperature changes. It’s units

aremg/L◦C

(b) We draw the tangent line on the graph, and use two numbers on the line, as far apart as possible.It appears that the line contains the points (0, 14) and (40, 4) (this depends on how you draw the line).

Thus, the slope is approximately −1

4

mg/L◦C

= −0.227mg/L◦C

. An “interpretation” should be a restatement

of this fact, using words that someone who doesn’t know Calculus would understand. Here’s a commoninterpretation: at T = 16, each one degree increase in temperature would result in 0.227 mg/L lessdissolved oxygen.

Example 4. Find the derivative of f(x) =1

xat x = 4.

30

Solution: This problem is asking for f ′(4). There are two ways to set this up: f ′(4) = limh→0

f(4 + h)− f(4)

h

and f ′(4) = limx→4

f(x)− f(4)

x− 4. We’ll use the first approach here, but the student may want to also practice

with the second.

f ′(4) = limh→0

f(4 + h)− f(4)

h

= limh→0

14+h − 1

4

h

= limh→0

44(4+h) −

(4+h)(4+h)4

h

= limh→0

4−(4+h)4(4+h)

h

= limh→0

−h4(4 + h)

· 1

h

= limh→0

−1

4(4 + h)

=−1

4(4 + 0)=−1

16

Example 5. Find a formula for the derivative of f(x) =√x at x = a where a is an arbitrary unknown

constant.Solution: This problem is asking for f ′(a) where a is an unknown constant. There are two ways to

set this up: f ′(a) = limh→0

f(a+ h)− f(4)

hand f ′(a) = lim

x→a

f(x)− f(a)

x− a . We’ll use the second approach

here, but the student may want to also practice with the first.

f ′(a) = limx→a

f(x)− f(a)

x− a= lim

x→a

√x−√ax− a

= limx→a

√x−√ax− a ·

√x+√a√

x+√a

= limx→a

x− a(x− a)(

√x+√a)

= limx→a

1√x+√a

=1√

a+√a

=1

2√a

31

2.8 the derivative as a function

Definition. The derivative of f(x) is the function f ′(x) defined as follows

f ′(x) = limh→0

f(x+ h)− f(x)

h.

Note: this differs from the definition in section 2.7 in that we don’t have the phrase “at the point x = a”.As a result, in this definition f ′(x) is a function, not a number.

We interpret f ′(x) as giving a formula for the slope of the tangent line, the velocity, etc.

Example 1. Find f ′(x) for f(x) =1

x2.

Solution:

f ′(x) = limh→0

1(x+h)2

− 1x2

h

= limh→0

x2

x2(x+ h)2− (x+ h)2

x2(x+ h)2· 1

h

= limh→0

x2 − (x+ h)2

x2(x+ h)2· 1

h

= limh→0

x2 − (x2 + 2xh+ h2)

x2(x+ h)2· 1

h

= limh→0

−2xh− h2x2(x+ h)2

· 1

h

= limh→0

−2x− h1x2(x+ h)2

· 1

1

=−2x− 0

x2(x+ 0)2

=−2x

x4

=−2

x3

Notation for derivative: f ′(x) or y′ (with y = f(x)) ord

dxf(x) or

df

dxor

dy

dx(with y = f(x)).

Last three = Leibniz notation.Leibniz notation advantages:

• You don’t have to name f(x),

• This notation suggests a ratio,

• this notation makes explicit the role of x in taking the derivative,

• this notation makes the chain rule (coming later) look nice,

32

• this notation reminds us thatd

dxis an operator (i.e. it is a thing that we apply to functions),

• this notation reads as “take the derivative (with respect to x) of . . . ”,

• this notation is easier to see than ′.

Example 1 in different notation: f ′(x) = − 2

x3, or

d

dx

1

x2= − 2

x3or

dy

dx= − 2

x3, etc.

At this point we turn to graphical derivatives.

Example 2. Do problem #6 in 2.8.Solution: Here we looked at a function given by a graph, and estimated what the graph of itsderivative would be.

When we do this, all we can estimate from the graph is when the derivative is 0, when it is positive,when is negative, and a little bit of when it is bigger or smaller. But this is enough to make a roughgraph.

In particular, we conclude that f(x) is flat at x = 0, and so the f ′(x) must have a y-value of 0 atx = 0. Similarly, f(x) is going down a little just to the right of it’s peak, so f ′(x) must have a negativevalue just to the right of x = 0. Farther to the right, f(x) is still going down, but not as steeply, so f ′(x)must still be negative, but not as negative. Similar reasoning applies on the left, in fact this is just amirror image. So, the graph of f ′(x) will start out positive on the left, become a little more positive asit gets close to the y-axis, plunge down through the origin, become most negative a little to the right ofthe y-axis, and then start coming back towards the x-axis.

Example 3. Do problem #41 in 2.8Solution: Here we had three graphs which were labeled a, b and c, and which represented a functionf , as well as f ′ and f ′′. We had to figure out which graph was which.

When we do this, we can only verify the sorts of properties we had in the previous example: if a = b′,does the graph of a equal 0 when the graph of b is flat? Is the graph of a positive when the graph of b isgoing up? Etc. But this is enough to tell which graph must be the derivative of which other graph.

In particular, I think that b = a′. To check this note that when a is flat, we have b = 0. When a isincreasing, we have b is positive. When a is decreasing, we have that b is negative.

Now I think that c = b′. To check this note that when b is flat, we have c = 0. When b is increasing,we have c is positive. When b is decreasing, we have that c is negative.

33

Chapter 3

Rules for Derivatives

3.1 Shortcuts for powers of x, ex, constants, sums, and differences

Now we start to fill in shortcuts. Recall that we already knowd

dx

1

x2= − 2

x3.

Example 1. Findd

dxx.

Solution:

f ′(x) = limh→0

(x+ h)− xh

= limh→0

h

h

= limh→0

1

= 1

Example 2. Findd

dxx2.

Solution:

f ′(x) = limh→0

(x+ h)2 − x2h

= limh→0

x2 + 2xh+ h2 − x2h

= limh→0

2xh+ h2

h

= limh→0

2x+ h1

1

= 2x+ 0

Example 3. Findd

dx

√x

34

Solution:

f ′(x) = limh→0

√x+ h−√x

h

= limh→0

√x+ h−√x

h·√x+ h+

√x√

x+ h+√x

= limh→0

(x+ h)− xh(√x+ h+

√x)

= limh→0

h

h(√x+ h+

√x)

= limh→0

1

1(√x+ h+

√x)

=1√

x+ 0 +√x

=1

2√x

We generalize the previous examples in the following rule:

Power rule:d

dxxn = nxn−1 for any real number n

We’ve prove this so far for n = 1, 2, 1/2,−1/2 (and in the homework probably you’ve done n = 3). Wewill prove it for all n when we do logarithmic differentiation.

Example 4. Find the equation of the tangent line at x = 1 of f(x) = x3.7.

Solution: We used

dxxn = nxn−1 with n = 3.7. Thus, f ′(x) = 3.7x2.7. Then m = f ′(1) = 3.7 and our

equation is y = 3.7(x− 1) + 1.

Constant multiple rule:d

dxC · f(x) = C · f ′(x), where C is a constant

Proof.

d

dxC · f(x) = lim

h→0

C · f(x+ h)− C · f(x)

h

= limh→0

C

1· f(x+ h)− ·f(x)

h

= C limh→0

f(x+ h)− ·f(x)

h

= Cf ′(x)

35

The next rule follows from the previous one and a special case of the Power Rule:

Constant rule:d

dxC = 0, where C is a constant

Sum and Difference rule:d

dx[f(x)± g(x)] = f ′(x)± g′(x)

Proof. We prove this just for f(x) + g(x), and note that the result for f(x) − g(x) has essentially thesame proof.

d

dx[f(x) + g(x)] = lim

h→0

f(x+ h) + g(x+ h)− (f(x) + g(x))

h

= limh→0

f(x+ h)− f(x) + g(x+ h)− g(x)

h

= limh→0

(f(x+ h)− f(x)

h+g(x+ h)− g(x)

h

)

= limh→0

f(x+ h)− f(x)

h+ limh→0

g(x+ h)− g(x)

h

= f ′(x) + g′(x)

Example 5. Find when f(x) = x3 − 20x2 + 4x has a horizontal tangent line.Solution: Saying that a line is horizontal is exactly the same as saying that it has a slope of 0. Thus,we need to find out when the slope of a tangent line is 0. Thus, we take the derivative, set it equal to 0,and solve this

f ′(x) = 3x2 − 20x+ 4 = 0

Apply the quadratic formula, and get

x =40±

√1552

6

Exponential rule:d

dxex = ex

Proof. First we will need to know the following limit:

limh→0

eh − 1

h

We look at a graph of this function, and find that the limit equals 1

36

ehK 1h

hK1.0 K0.5 0.0 0.5 1.0

0.8

1.0

1.2

1.4

1.6

Now we will use this limit in the following calculation.

d

dxex = lim

h→0

ex+h − exh

= limh→0

ex · eh − exh

= limh→0

ex

1· e

h − 1

h

=ex

1· limh→0

eh − 1

h

=ex

1· 1

Example 6. Find the equation of the tangent line at x = 0 where f(x) = 3x3/2 − 5ex.Solution: We have a = 0, y0 = f(0) = −5. For the slope we calculate first the derivative

f ′(x) = 3 · 3

2x1/2 − 5ex =

9

2

√x − 5ex

and plug in x = 0f ′(0) = 0− 5 = −5

Thus, the tangent line is y = −5x− 5.

Example 7. Find the derivative of5√

32x+ 5√x5.

Solution: First we simplify the function f(x) =5√

32x1/5 +5x5/2 = 2x1/5 +5x5/2. Then the derivativeis

f ′(x) =2

5x−4/5 +

25

2x3/2

37

3.2 Product and Quotient Rules

Product rule:(f · g)′ = f ′ · g + f · g′

There is another more verbose way to state this

d

dx[f(x)g(x)] =

[d

dxf(x)

]g(x) + f(x)

[d

dxg(x)

]

Figure 3.1 shows one way you can understand this rule.

Proof. Now we do the real proof of the product rule:

d

dx[f(x)g(x)] = lim

h→0

f(x+ h)g(x+ h)− f(x)g(x)

h

= limh→0

f(x+ h)g(x+ h)− f(x)g(x+ h) + f(x)g(x+ h)− f(x)g(x)

h

= limh→0

f(x+ h)g(x+ h)− f(x)g(x+ h)

h+f(x)g(x+ h)− f(x)g(x)

h

= limh→0

f(x+ h)− f(x)

h

g(x+ h)

1+f(x)

1

g(x+ h)− g(x)

h

Note that when you take the limit the first fraction becomes f ′(x), the second fraction becomes g(x),the third becomes f(x) and the last becomes g′(x), so we get

f ′(x) · g(x) + f(x) · g′(x).

Example 1. Findd

dx(3x+ ex)(x2 − 4ex).

Solution: Let f(x) = 3x+ ex and g(x) = x2 − 4ex. Then the derivative is

f ′(x)g(x)+f(x)g′(x) = (3x+ex)′ ·(x2−4ex)+(3x+ex)·(x2−4ex)′ = (3+ex)(x2−4ex)+(3x+ex)(2x−4ex).

Quotient Rule:

(f

g

)′=f ′ · g − f · g′

g2

Here is the more verbose way to state this

d

dx

f(x)

g(x)=f ′(x)g(x)− f(x)g′(x)

(g(x))2

Example 2. Findd

dx

x

ex.

Solution: Let f(x) = x and g(x) = ex. Then the derivative is

f ′(x)g(x)− f(x)g′(x)

(g(x))2=

(x)′ex − x(ex)′

(ex)2=ex − xex

(ex)2=ex(1− x)

(ex)2=

1− xex

.

38

Figure 3.1: Intuitive, graphical, understanding of product rule

f

g

Picture fg as a rectangle with sides f and g,let d(fg) represent an infinitely small changein fg, pictured in red (well, a small change ispicture in red).

f

g

df

dg

Now, the red part, the change in fg, not onlyrepresents d(fg), but can also be connected tochanges in sides f and g.

f

g

df

dg dgdf

g

f

Using this, we can break the red part up intotwo rectangles and a square with areas df · g,g · df , and df · dg.

So, the red part is d(fg), and this equals thesum of two rectangles and a square

d(fg) = df ·g + f ·dg + df ·dg

Now, since df and dg are infinitely small, wehave df · dg ≈ 0. This gives us the crucial ideaof the whole picture/proof

d(fg) = df · g + f · dgChange in area equals df · g + f · dg

Now, we “divide” by dx to get

d

dxfg =

df

dx· g + f

dg

dx

39

Example 3. Find the derivative of2t+ 1

t

2 +√t.

Solution: Let f(x) = 2t+1

tand g(x) = 2 +

√t. Then the derivative is

f ′(x)g(x)− f(x)g′(x)

(g(x))2=

(2t+ 1

t

)′ · (2 +√t)−

(2t+ 1

t

)· (2 +

√t)′

(2 +√t)2

=(2− t−2)(2 +

√t)− (2t+ 1

t )(12 t−1/2)

(2 +√t)2

Example 4. Suppose h(x) is a function such that h(5) = 3 and h′(5) = −2. Find the derivative ofexh(x) at x = 5.Solution: Let f(x) = ex. Then the derivative is

f ′(5)h(5) + f(5)h′(5)

Since f ′(x) = ex this becomese5 · 3 + e5 · (−2) = e5.

Today we did a worksheet that verified the short cut rules for ex, sin(x), cos(x) and tan(x).

3.3 Trigonometric functions

Sin Rule:d

dxsin(x) = cos(x)

Note: this formula is only correct in radians.

Proof. We need two limits

limh→0

sin(h)

h= 1 and lim

h→0

cos(h)− 1

h= 0

(The first of these we proved using the Squeeze theorem, back in section 2.3, the second follows from thefirst if you multiply on top and bottom by cos(x) + 1.)

We also need one identity

sin(x+ h) = sin(x) cos(h) + sin(h) cos(x)

Now we apply the definition

d

dxsin(x) = lim

h→0

sin(x+ h)− sin(x)

h

= limh→0

sin(x) cos(h) + sin(h) cos(x)− sin(x)

h

= limh→0

sin(x) cos(h)− sin(x)

h+

sin(h) cos(x)

h

= limh→0

sin(x)

1· cos(h)− 1

h+

sin(h)

h· cos(x)

1

= sin(x) · 0 + 1 · cos(x)

= cos(x)

40

Cos Rule:d

dxcos(x) = − sin(x)

The proof is left as homework.

Tan Rule:d

dxtan(x) = sec2(x) or

1

cos2(x)

Proof. We apply the identity tan(x) =sin(x)

cos(x)and the quotient rule:

d

dxtan(x) =

d

dx

sin(x)

cos(x)

=(sin(x))′ · cos(x)− sin(x) · (cos(x))′

(cos(x))2

=cos(x) · cos(x)− sin(x) · (− sin(x))

(cos(x))2

=cos2(x) + sin2(x)

(cos(x))2

=1

(cos(x))2

Example 1. Find the equation of the tangent line at x = π/6 for sin(x).Solution: Use y = m(x− x0) + y0.

x0 = π/6,y0 = f(π/6) = sin(π/6) = 1/2.f ′(x) = cos(x), m = f ′(π/6) = cos(π/6) =

√3/2.

y =

√3

2(x− π/6) + 1/2.

Example 2. Findd

dx(x sin(x))

Solution: Let f(x) = x, g(x) = sin(x).f ′(x) = 1, g′(x) = cos(x).The product rule gives f ′(x)g(x) + f(x)g′(x) which is 1 · sin(x) + x cos(x).

Example 3. Findd

dx

1 + sin(x)

x+ cos(x), and simplify.

Solution: f(x) = 1 + sin(x)g(x) = x+ cos(x)f ′(x) = cos(x)g′(x) = 1− sin(x)

41

f ′g − fg′(g)2

=cos(x)(x+ cos(x))− (1 + sin(x))(1− sin(x))

(x+ cos(x))2

=x cos(x) + cos2(x)− (1− sin2(x))

(x+ cos(x))2

=x cos(x) + cos2(x)− 1 + sin2(x)

(x+ cos(x))2

=x cos(x)

(x+ cos(x))2

We state for future reference the derivatives of the other trigonometric functions

d

dxcot(x) = − csc2(x),

d

dxsec(x) = sec(x) tan(x),

d

dxcsc(x) = − csc(x) cot(x)

3.4 The Chain rule!!

Chain rule:d

dxf(g(x)) = f ′(g(x)) · g′(x)

In wordsd

dxf(g(x)) = the derivative of the outside function (don’t

change the inside) times the derivative of theinside function.

In Leibniz notation

Chain rule: If

{y = y(u)u = u(x)

}then

dy

dx=dy

du· dudx

42

Here’s how a lot of people really remember it: function by function

ordinary rule −−−−−−→ chain rule version

d

dxxn = nxn−1 −−−−−−→ d

dx(2)n = n(2)n−1 · (2)′

d

dxex = ex −−−−−−→ d

dxe2 = e2 · (2)′

d

dxsin(x) = cos(x) −−−−−−→ d

dxsin(2) = cos(2) · (2)′

d

dxcos(x) = − sin(x) −−−−−−→ d

dxcos(2) = − sin(2) · (2)′

d

dxtan(x) = sec2(x) −−−−−−→ d

dxtan(2) = sec2(2) · (2)′

For each formula, put whatever you want in each 2, as long as you put the same thing in each 2.

Example 1. Findd

dxsin(x2 + 1).

Solution: We’ll do this one using a few different kinds of notation:

Box notation:d

dxsin(2) = cos(2) ·2′. Into 2 we put x2 + 1 so we get

d

dxsin(x2 + 1

)= cos

(x2 + 1

)· x2 + 1

′= cos(x2 + 1) · 2x.

Leibniz notation: u = x2 + 1, y = sin(x2 + 1) = sin(u),dy

du= cos(u),

du

dx= 2x, so we get

dy

dx=dy

du· dudx

= cos(u) · 2x = cos(x2 + 1)2x.

Word notation:

d

dxsin(x2 + 1) = cos(x2 + 1) · 2x

deriv. ofoutside don’t

changeinside

deriv. ofinside

Example 2. Findd

dx

√4x2 + x.

43

Solution: We apply the chain rule form of the derivative of√2 and get

d

dx

√2 =

1

2√2 ·2′

d

dx

√4x2 + x =

1

2

√4x2 + 1

· 4x2 + x′

=1

2√

4x2 + 1· (8x+ 1)

Here’s another way to write it:

d

dx

√4x2 + x =

1

2√

4x2 + x· (8x+ 1)


changeinside

deriv. ofinside

Example 3. Findd

dt

(sin(t)

t2 + t+ 1

)15

Solution: We combine the chain rule form of the derivative of (2)15 and use the quotient rule onthe inside to get

d

dt(2)15 = 15(2)14 ·2′

d

dt

(sin(t)

t2 + t+ 1

)15

= 15

(sin(t)

t2 + t+ 1

)14

· sin(t)

t2 + t+ 1

′

= 15

(sin(t)

t2 + t+ 1

)14

· cos(t)(t2 + t+ 1)− sin(t)(2t+ 1)

(t2 + t+ 1)2

Here’s another way to write it:

d

dt

(sin(t)

t2 + t+ 1

)15

= 15

(sin(t)

t2 + t+ 1

)14

· cos(t)(t2 + t+ 1)− sin(t)(2t+ 1)

(t2 + t+ 1)2


changeinside

deriv. ofinside

Example 4. Findd

dx(3x2 − 5

√x)esin(

√tan(x))

44

Solution: We apply the product rule, and then the chain rule with four “layers”.

d

dx(3x2 − 5

√x)esin(

√tan(x)) = (3x2 − 5

√x)′· esin(

√tan(x)) + (3x2 − 5

√x)(esin(

√tan(x)))

′

(esin(√

tan(x)))′

= esin(√

tan(x)) · sin(√

tan(x))

′

= esin(√

tan(x)) · cos(√

tan(x)) ·√

tan(x)

′

= esin(√

tan(x)) · cos(√

tan(x)) · 1

2√

tan(x)· tan(x)

′

= esin(√

tan(x)) · cos(√

tan(x)) · 1

2√

tan(x)· sec2(x)

Example 5. Find F ′(−11) where F (x) = f(g(x)) and

f(−3) = 5 g(−3) =3

2

f ′(−3) = 2 g′(−3) = −1

7f(−11) = 7 g(−11) = −3

f ′(−11) = 2 g′(−11) = 5

Solution: The chain rule says that F ′(x) = f ′(g(x)) ·g′(x). Therefore, we have F ′(−11) = f ′(g(−11)) ·g′(−11) = f ′(−3) · 5 = 2 · 5 = 10.

General exponential rule:d

dxax = ax ln(a), for a > 0

Proof. Let a = eln(a). Then

d

dxax =

d

dx(eln(a))x =

d

dxex ln(a) = ex ln(a) · x ln(a)

′= ex ln(a) · ln(a) = ax ln(a)

3.5 Implicit derivatives

Example 1. Find the slope at x = 1/2 on the circle x2 + y2 = 1, by solving explicitly for y.

Solution: y =√

1− x2.y′ =

1

2(1− x2)−1/2 · 2x.

m = y′(1/2) =1

2(1− (1/2)2)−1/2 · 2 · 1

2=

1

2√

1− 1/4· 1 =

1

2√

3/4= −1/

√3.

Example 2. Find the slope at

(1

2,

√3

2

)on the circle x2 + y2 = 1, by working implicitly: remember, y

is a function of x.

45

Solution: We take the derivative of both sides of x2 + y2 = 1,d

dx(x2 + y2) =

d

dx1

2x+d

dxy2 = 0

To figure outd

dxy2 we use the chain rule (after all, y is somehow a function of x, so we have one

function, inside another function, the outside one is the squaring function).

d

dx22 = 22 ·2 ′

We plug y into this,d

dxy 2 = 2 y · y

′= 2y · y′

Thus, the above equation becomes 2x+ 2y · y′ = 02y · y′ = −2x

y′ = −xy

At the point

(1

2,

√3

2

)we have

y′ = − 1/2√3/2

= −1/√

3

Note, the implicit formula shows us something nice about the slope of a point on the circle: if the point

is (x, y), then the slope of the radius will bey

x. Note that each tangent line is perpendicular to the radius

at that point. Therefore, the slope of a tangent line will be − 1

y/x, which is what we found.

How to take implicit derivatives: You start with an equation involving x and y. Take the

derivative,d

dxof both sides. Anytime you see an x, you just take the derivative like normal. Anytime

you see a y, you treat it like a hidden function and use the chain rule (and possibly the product andquotient rules). Here’s a simpler way to say this: take the derivative of anything involving y just the

same as you would if it were x, but then multiply by y′ ordy

dx. 1 Finally, you solve for y′.

1It is often confusing to students why we treat y this way, I’ll try to offer a little explanation here. The crucial differenceis between a function and an equation. When we write f(x) = x2 + y2 we are defining a function f(x). We can plug in anynumber we want for x, and get the output from the formula x2 + y2. For example, f(1) = 1 + y2. Now, in this case, y hasno relation to x. We could have that y = 3 or y = 5, or just treat it as an unknown constant, or as an unrelated variable.

In this case, if we take the derivatived

dxof f(x) we get that

d

dxy = 0, and so

d

dxf(x) = 2x. This makes sense because the

graph of f(x) is a parabola, shifted either up or down due to the value of y, and such a parabola should have slope givenby 2x, just like f(x) = x2.

Now, suppose x2 + y2 = 1. Here, the fact that we have an equation means that y cannot be a number unrelated to x.If I plug in x = 1, then I must have y = 0. There is no choice about y, it is determined by x. Thus, y is a function of x.Looking at it the other way, we are not free to plug in different values of y and still have a function f(x). We cannot havey = 3 and then graph f(x). There is no f(x) here and there is no value of x that corresponds to y = 3. Again, y is notunrelated to x, it is an implicit function of x.

Now, when we take the derivative of this equation, we have to do it in a way that (1) is correct given that y is a function

of x, and (2) is correct given that we could in fact replace y with√

1− x2. So, if we have y = g(x), then y2 = (g(x))2 and

46

Example 3. Find the derivative at x = 2/3, y = 4/3 for the shape defined by x3+y3 = 3xy. Remember:y is a function of x.

Solution:d

dx(x3 + y3) =

d

dx(3xy)

3x2 +d

dxy3 = 3

d

dx(xy)

Ford

dxy3 we use the chain rule:

d

dxy 3 = 3 y 2 · y

′= 3y2 · y′

Ford

dx(xy) we use the product rule

x′ · y + x · y′ = y + x · y′

Thus we have3x2 + 3y2 · y′ = 3(y + x · y′)

Now we solve for y′:

3x2 + 3y2 · y′ = 3y + 3x · y′

3y2 · y′ − 3x · y′ = 3y − 3x2

(3y2 − 3x)y′ = 3y − 3x2

y′ =3y − 3x2

3y2 − 3x

y′ =y − x2y2 − x

Plug in x = 2/3 and y = 4/3 to get m = 4/5.

Example 4. Find y′ for ey cos(x) = 1 + sin(xy)Solution: ey · y′ cos(x)− ey · sin(x) = cos(xy) · (1 · y + x · y′).

ey · y′ cos(x)− cos(xy)x · y′ = cos(xy)y + ey sin(x)(ey cos(x)− cos(xy)x)y′ = cos(xy)y + ey sin(x)

y′ =ey sin(x) + cos(xy)y

ey cos(x)− cos(xy)x

the derivative becomes 2g(x) · g′(x). This is the same as we found (but in different notation), 2y · y′. This would be trueno matter what function y equals. We will have other examples where y is a different function of x, but we will still have

that the derivative of y2 is 2y · y′. Finally, given the fact that y =√

1− x2 we should get the correct thing when we take

the derivative of (√

1− x2)2. On the one hand, we can simplify this first and take the derivative of 1−x2 and get −2x. Onthe other hand, we should be able to apply the chain rule and get

2(√

1− x2) ·√

1− x2

′

= 2√

1− x2 · 1

2√

1− x2· (−2x) = −2x.

The point is, all of these different ways of looking at x2 + y2 = 1 have to give the same result, the one we got in Example 1,and it can’t be the same as taking the derivative of f(x) = x2 + y2.

47

Inverse trig functions

Recall that if we know one trig function, then we know them all2.

Example 5. Suppose sin(θ) =5

7(and 0 < θ < π/2). Find cos(θ) and tan(θ).

Solution: From sin(θ) =5

7we can draw a triangle and label two sides as follows

?

57

θ

Then the missing side is√

72 − 52 =√

24 = 2√

6 and so cos(θ) = 6√

2/7, tan(θ) =5

6√

2.

Example 6. Suppose sin(y) = x, find cos(y) and tan(y).Solution: From sin(y) = x we can draw a triangle and label two sides as follows

?

x1

y

Then the missing side is√

1− x2 and so cos(y) =√

1− x2 and tan(y) =x√

1− x2.

Example 7. Findd

dxsin−1(x).

Solution: Let y = sin−1(x),sin(y) = x,d

dxsin(y) =

d

dxx

cos(y) · y′ = 1

y′ =1

cos(y)

Example 6 shows sin(y) = x⇒ cos(y) =√

1− x2.y′ =

1√1− x2

2As I like to say in this case, You might say, “one trig function to rule them all, one trig function to find them, one trigfunction to bring them all and in the darkness bind them.”

48

Sin inverse:d

dxsin−1(x) =

1√1− x2

Similar rules

Cos inverse:d

dxcos−1(x) =

−1√1− x2

Tan inverse:d

dxtan−1(x) =

1

1 + x2

We started the class by doing two quick tests: one for 3.4, the chain rule, and one for 3.5, implicitderivatives.

3.6 Logarithmic differentiation

Example 1. Findd

dxln(x)

Solution: y = ln(x)

ey

= xd

dxey

=d

dxx

eyy′ = 1

y′ =1

ey

y′ =1

x

ln:d

dxln(x) =

1

x

Example 2. Find the derivative of y =3

√x4 + 1

x4 − 1

Solution: ln(y) = ln

(3

√x4 + 1

x4 − 1

)

ln(y) =1

3

(ln(x4 + 1)− ln(x4 − 1)

)

d

dxln(y) =

d

dx

1

3

(ln(x4 + 1)− ln(x4 − 1)

)

1

yy′ =

1

3

(1

x4 + 1(4x3)− 1

x4 − 14x3)

y′ = y1

3

(1

x4 + 1(4x3)− 1

x4 − 14x3)

y′ =3

√x4 + 1

x4 − 1

1

3

(1

x4 + 1(4x3)− 1

x4 − 14x3)

Procedure for logarithmic differentiation: Start with y =something.Take ln( ) of both sides, and use the properties of ln( ) to simplify the right hand side.Take implicit derivatives, solve for y′ and replace y with the formula you started with.

49

Example 3. Find the derivative of y = xx.Solution: ln(y) = ln(xx)

ln(y) = x ln(x)d

dxln(y) =

d

dx(x ln(x))

1

y· y′ = ln(x) + x

1

xy′ = y(ln(x) + 1)y′ = xx(ln(x) + 1)

Example 4. Find a general formula for the derivative of f(x)g(x).Solution: y = f(x)g(x)

ln(y) = g(x) ln(f(x))d

dxln(y) =

d

dx(g(x) ln(f(x)))

1

y· y′ = g′(x) ln(f(x)) + g(x)

1

f(x)· f ′(x)

y′ = y(g′(x) ln(f(x)) + g(x)1

f(x)· f ′(x))

y′ = f(x)g(x)(g′(x) ln(f(x)) + g(x)

1

f(x)· f ′(x)

)

Example 5. Prove the ruled

dxxn = nxn−1 for any real number n.

Note that we have proven this rule for x2, 1/x,√x, and maybe one or two other values of n. But

now we are giving a proof for an infinite number of values of n, including values that would be reallyhard to do using lim

h→0. . . , such as n = 1.79, n = π, n =

√2, etc.

Let y = xn.Then ln(y) = ln(xn).Then ln(x) = n ln(x).Taking the derivative of both sides we get

d

dxln(y) =

d

dxn ln(x)

1

y· dydx

= n · 1

x

dy

dx= yn · 1

xdy

dx= xnn · 1

x= nxn−1

50

Derivatives

Here all the derivative rules and techniques that we have learned this semester. With them, you can take thederivative of any function that you have ever seen! (Note: there are other functions in the world, that you don’tknow how to take the derivative of, but these involve definitions and formulas that you have probably never seen.)

Formulas marked with a “*” should definitely be memorized.

Basic functions

* d

dxxn = nxn−1 for all real numbers n

The derivatives of x2, x3,√x = x1/2 were found in

2.8, using the definition of derivative. The derivativesof 1/x = x−1, 1/x2 = x−2 were found in 3.1, again usingthe definition. The derivative of xn, for all real numbersn, was found in 3.6 by using logarithmic differentiation.

* d

dxsin(x) = cos(x)

d

dxsec(x) = sec(x) tan(x)

* d

dxcos(x) = − sin(x)

d

dxcsc(x) = − csc(x) cot(x)

* d

dxtan(x) = sec2(x)

d

dxcot(x) = − csc2(x)

The derivatives of sin(x) and cos(x) were found in 3.3,using the definition, trig identities, and the special lim-

its limh→0

sin(h)

h= 1 and lim

h→0

cos(h)− 1

h= 0. The rest

of these derivatives were found in 3.3 using the quotientrule.

d

dxsin−1(x) =

1√1− x2

d

dxsec−1(x) =

1

x√x2 − 1

d

dxcos−1(x) =

−1√1− x2

d

dxcsc−1(x) =

−1

x√x2 − 1

d

dxtan−1(x) =

1

1 + x2

d

dxcot−1(x) =

−1

1 + x2

All of these derivatives were found in 3.5, by using im-plicit derivatives to obtain a relationship between eachof these functions and the noninverse function.

* d

dxex = ex and * d

dxln |x| = 1

x

The derivative of ex was found in 3.1 using the special

limit limh→0

eh − 1

h= 1. The derivative of ln(x) was found

in 3.6 using implicit derivatives.

Combinations

*(f + g)′ = f ′ + g′ *(f · g)′ = f ′ · g + f · g′

*(f − g)′ = f ′ − g′ *(f/g)′ =f ′ · g − f · g′

(g)2

*(f(g(x)))′ = f ′(g(x)) · g′(x)

(the combination f(x)g(x) is described below

)

Here is how some of our basic functions look when com-bined with the chain rule:

* d

dx�n = n�n−1 ·�′ * d

dxsin(�) = cos(�) ·�′

* d

dxe� = e� ·�′ * d

dxcos(�) = − sin(�) ·�′

* d

dxln(�) =

1

� ·�′ * d

dxtan(�) = sec2(�) ·�′

You should imagine putting other functions inside theboxes. Of course, the exact same stuff should go insideeach box in one of these equations.

The derivatives of f ± g were proven in 3.1 usingbasic definitions. The product and quotient rules wereproven in 3.2 using the basic definition and some tricks.The chain rule was proven in 3.4, also using the defintiionand some tricks.

Techniques

Implicit derivatives. (1) You start with an equationinvolving x’s, y’s, and/or numbers. (2) You take thederivative of both sides of the equation. When you dothis you view y as a function of x and use the chain rule,product rule, etc. as needed. (For example d

dxy2 = 2yy′,

ddx

sin(y) = cos(y)y′, ddx

xy = y + xy′, etc..) (3) You solvethe new equation for y′.

Logarithmic differentiation. (1) You start with afunction, often of the form y = f(x)g(x). (2) Take ln ofboth sides, bring the exponent down in front. (3) Takethe implicit derivative. (4) Solve for y′.

3.7 Rates of change in the natural sciences

derivative = rate of change. The main difficulty in this section is just seeing different ways of interpretingthe rate of change, and extracting from each problem just what the function is, and what the variablesare. Since I do not know all the applications that are in this book, nor an equal number of applicationsthat are not in the book, I will follow the examples in the book more closely than I usually do.

Example 1. Identify the two functions in Example 1 from the book, and what their derivatives mean.Solution:

f(t) = positiondf

dt=

change in position

change in time= velocity

andv(t) = velocitydv

dt=

change in velocity

change in time= acceleration

Example 2. Identify what the function is in example 2 from the book, and what its derivative means.Solution:

f(x) = mass in first x meters of roddf

dx=

change in mass

change in length= linear density

Note: towards the end of the problem, the book assumes that f(x) =√x and then finds f ′(1), i.e. the

linear density at x = 1 meter.


Q(t) = charge in regiondQ

dt=

change in charge

change in time= current

Example 4. Identify what the function is in example 4 in the book, and what its derivative means.Solution:

[A] = concentration of chemical Ad[A]

dt=

change in concentration

change in time= instantaneous rate of reaction


V (P ) = volume as a function of pressure PdV

dP=

change in volume

change in pressure

compressibility = − 1

V

dV

dP

Note: towards the end of this example, the book uses a gas equation of the form V =5.3

Pand calculates

the compressibility at P = 50 kPa, by taking the derivative.


P (t) = population at time tdP

dt=

change in population

change in time= instantaneous rate of growth

52

Note: towards the end of this example, the book considers a population of bacteria with the number ofbacteria given by n02

t where n0 is the initial population. Then the book calculates the growth rate att = 4 by taking the derivative.


v(r) = velocity, v, of blood flow as a function of r, the distance fromcenter of blood vessel

dv

dt=

change in velocity

change in distance from center= velocity gradient

Note: in this example the book introduces the equation

v =P

4η`(R2 − r2)

where P , η, `, and R are constants that depend on the pressure, viscosity, length of vessel, and radius of

vessel. Then they take the derivativedv

drand plug in .002 cm to find the velocity gradient.


C(x) = cost as a function of x, the number of units madedC

dx=

change in cost

change in number of units= marginal cost

Note: towards the end of the example, the book uses C(x) = 10000 + 5x+ .01x2 and finds the marginalcost at 500.

Example 9. The graphs in #5 from the book, show the velocity of two different particles. When is eachparticle speeding up? When is it slowing down?Solution: The particle is speeding up, if its graph of velocity is increasing. For graph (a), this is for0 < t < 1. For graph (b), this is for 1 < t < 2.

Example 10. In #11 in the book, a company is making square computer chips. We analyze the functionthere, and its derivative.Solution: We have

A(x) = area of square

dA

dx=

change in area

change in length of side

(a) A(x) = x2 mm2

A′(x) = 2xmm2/m

A′(15) = 30 mm2/m

At x = 15, each 1 mm increase in side length will produce a 30 mm2 increase in area.

(b) A′(x) = 2x and the perimeter is 4x, so A′(x) is half the perimeter.

Geometrically, this is because a small change in side length ∆x, produces a small change in area∆A, that is (approximately) 2∆x · x.

53

x

x

∆x

∆x

Thus,∆A

∆x≈ 2x.

Example 11. In #13(b),(c) in the book, we calculate the change in area of a circle as a function ofradius.Solution:

(b) A(r) = πr2 units2

A′(r) = 2πr units2/unit

A′(2) = 4π units2/unit

(c) A′(r) = 2πr and the circumference is 2πr, so these are equal.

Geometrically, this is because a small change in radius ∆r, will produce a small change in area ∆A,that is approximately 2πr∆r.

r

∆r

2πr∆r

Therefore,∆A

∆r≈ 2πr.

Example 12. In #17 in the book, we analyze the change in mass of a rod, i.e. linear density.Solution:

m(x) = mass of rod between position 0 and position x

= 3x2 kg

m′(x) =change in mass

change in length of rod

m′(x) = 6x kg/mm′(1) = 6 kg/mm′(2) = 12 kg/mm′(3) = 18 kg/m

54

Homework comments

For the homework, you may want to have a skeleton key guide to what the functions are in the problems.Here it is

• #7, you are given position. Take the derivative to get velocity. Take the derivative again to getacceleration.

• #8, you are given position. Take the derivative to get velocity.

• #12, start with V (x) = x3. CalculatedV

dxat x = 3 (and interpret).

• #15, you are given surface area S. CalculatedS

dr(and interpret).

• #16, you are given volume V . CalculatedV

dr(not for part (a) ) (and interpret).

• #17, you are given mass M = 3x2. CalculatedM

dx.

• #20, you are given force F , where G, m and M are constants. CalculatedF

dr(and interpret).

• #23, you need to start by finding a formula for the population P , as in example 6 from the book.

Then takedP

dt.

• #24, you are given a formula for the population f , where a, b are constants. Finddf

dtand use this

(with f) to solve for the constants a and b.

• #29, you are given cost C. TakedC

dxto find the marginal cost.

• #30, you are given cost C. TakedC

dxto find the marginal cost.

3.9 Related rates

The basic idea here is that we have an equation, and every variable in it is changing (or could be changing)with time. If the variable is changing with time, that means that it is a function of time, although we do

not usually know what formula it has. Thus, if we take the time derivative,d

dt, of the equation, we treat

each variable as an implicit function of t, and so we take the derivative in the same way as we did withimplicit derivatives. The only difference now is that this approach applies to every variable, not just y,and we’re taking the derivative with respect to t, not x.

There are two parts of every related rates problem. The pre-calculus part involves finding the formulas

to use, possibly more than one. The calculus part involves, taking the derivatived

dt, plugging in numbers

and solving for the remaining quantity. Part of what students find confusing is combining these steps,so I want to de-confuse things and separate them. We’ll start with three examples that involve onlythe pre-calculus side, then do one example that involves only taking derivatives, and finally do someexamples that combine the steps.

55

Example 1. Imagine that a 10 m long ladder is leaning against a wall and starts to slip, so that the topof the ladder is going down, and the bottom of the ladder is moving out away from the wall.

1. Find an equation that relates the positions x and y of each end the ladder.

2. Discuss what happens to x and y as the ladder slides.

3. What do can you say, without actually taking derivatives, aboutdx

dtand

dy

dt?

Solution: x = the position of the bottom of the ladder.y = the position of the top of the ladder.x2 + y2 = 10.Note that x and y change as the ladder moves. Thus, x and y are functions of time, though we don’t

know exactly what formula these functions have. We do know thatdy

dtwill be negative, and

dx

dtwill be

positive.

Example 2. Imagine that a balloon is being blown up.

1. Find an equation that relates the volume of the balloon V , to the radius r.

2. Discuss what happens to V and r as the balloon is blown up.

3. What can you say, without actually taking derivatives, aboutdV

dtand

dr

dt?

Solution: Volume of a sphere: V =4

3πr3 where r is radius.

Note that r and V change as the balloon is blown up. Thus, again, V and r are functions of time,though we don’t know exactly what formula these functions have.

Example 3. Imagine that gravel is moving along a conveyor belt, up a ramp, and falling into a pile thatmakes a cone.

1. Find an equation that relates the volume of the pile, V , to the radius of the pile, r, and the heightof the pile, h.

2. Discuss what happens to V , r and h as the gravel is added.

3. What can you say, without actually taking derivatives, aboutdV

dt,dh

dtand

dr

dt?

Solution: Volume of a cone: V =1

3πr2h, where r is the radius (at the widest part) and h is the

height.Note that r, h and V change as more gravel is added. Thus, again, r, h and V are functions of time,

though we don’t know exactly what formula these functions have.

Example 4. In each of the following equations, treat each variable as a function of time, and taked

dtof

both sides of the equation.

56

1. Volume of a cube: V = x3.

2. Area of a circle: A = πr2.

3. Area of a square: A = x2.

4. Area of a rectangle: A = l · w.

5. Volume of a cylinder: V = πr2h.

6. Volume of a sphere: V =4

3πr3.

7. Lengths in a right triangle: z2 = x2 + y2.

8. Surface area of sphere: S = 4πr2.

9. Volume of a cone: V =1

3πr2h.

10. Angle made by tracking an object:

tan(θ) =y

x.

Solution:

1.d

dtV =

d

dtx3

dV

dt= 3x2

dx

dt

2.d

dtA =

d

dtπr2

dA

dt= π2r

dr

dt

3.d

dtA =

d

dtx2

dA

dt= 2x

dx

dt

4.d

dtA =

d

dtl · w

dA

dt=dl

dt· w + l · dw

dt

5.d

dtV =

d

dtπr2h

dV

dt= π

(2rdr

dt· h+ r2 · dh

dt

)

6.d

dtV =

d

dt

4

3πr3

dV

dt=

4

3π3r2

dr

dt

7.d

dtz2 =

d

dt(x2 + y2)

2zdz

dt= 2x

dx

dt+ 2y

dy

dt

8.d

dtS =

d

dt4πr2

dS

dt= 4π2r

dr

dt

9.d

dtV =

d

dt

1

3πr2h

dV

dt=

1

3π

(2rdr

dt· h+ r2 · dh

dt

)

10.d

dttan(θ) =

d

dt

y

x

sec2(θ)dθ

dt=

dydt · x− y · dxdt

x2

Example 5. A ladder that is 10 m long is leaning against a wall. The bottom end is 2 m out from thewall and starts to slide farther out at a speed of 1 m/s. How fast is the top sliding down?Solution:

57

Example 6. A balloon is being blown up with air at a rate of 9.3 in3/s. Find how fast the radius isincreasing when the volume is 1000 in3.

Solution:d

dtV =

d

dt

4

3πr3

dV

dt=

4

3π · 3r2dr

dtPlug in V = 1000 and solve for r

1000 =4

3πr3 ⇒ r = 3

√750/π

Plug indV

dt= 9.3, r = 3

√750/π and solve for

dr

dt

dr

dt=

1

4· 9.3

π(750/π)2/3.

Example 7. (c.f. Stewart, 3.9#4) A lecturer is positioning a projector in a large room. They start withthe projector close to the screen and move it back. As they move it back, the rectangle of light thatit makes grows larger. Suppose at a certain point in time, the length of the rectangle is 72 inches, theheight is 48 inches, the length is increasing by 2 in/s and the height is increasing by 1.3333 in/s. How fastis the area increasing at this moment?Solution:

`

w ⇒ A = ` · w

d

dtA =

d

dt(` · w)

d

dtA =

d`

dt· w + ` · dw

dt.

d`

dt= 2 cm/s

dw

dt= 1.3333 cm/s

` = 72 cmw = 48 cmdA

dt= 2 · 48 + 72 · 1.3333 = 192 in2/s

Example 8. (c.f. Stewart 3.9#20) A boat is being pulled into a dock by a rope connected to the frontof the boat. The point where the rope connects to the boat is 2 m higher than the point where the ropeconnects to the dock. The rope is being pulled in at a rate of 0.5 m/s. How fast is the boat approachingthe dock, when the distance between the boat and the dock is 10 m?Solution: Here is a crude sketch of the situation

boat

dock

rope

The crucial parts of the picture are the rope, the distance between the boat and the dock, and the relativeheights of the boat and the dock. Thus, we have a simplified picture

58

zy

x

From this picture, we can find an equation relating x, y and z:

x2 + y2 = z2.

Is it time to plug numbers in yet? The only number you can plug in, if you want to, is y = 2, becausethis number is the only one that doesn’t change as the boat is moving. (If you don’t plug this number

in yet, that’s ok too: in another of couple of steps you’ll plug indy

dt= 0.) So, we have

x2 + 22 = z2.

We taked

dtof both sides:

d

dt(x2 + 22) =

d

dtz2

2x · dxdt

+ 0 = 2z · dzdt

dz

dt= −0.5 m/s

x = 10z =

√x2 + y2 =

√102 + 22 =

√104.

2 · 10 · dxdt

= 2 ·√

104 · (−0.5)

dx

dt= −0.05

√104

Example 9. (Stewart 3.9#16)A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight

toward the building at a speed of 1.6 m/s, how fast is the length of his shadow on the building decreasingwhen he is 4 m away from the building?Solution: The hardest part of this problem is getting the first equation: the one that relates all thevariables. Start with a picture

h

(spotlight)

2 (man)

x 12− x12

From this picture we can see that the crucial variables are x and h. We need some equation that relatesx, h, 2, 12 and the triangles. I don’t think you should use the Pythagorean theorem, because that wouldintroduce another variable, which would make it more complicated. You could use a trig function, with

59

the same angle for the smaller triangle and the bigger one, but why not eliminate the trig function andthe angle? This leaves you with similar triangles. Similar triangles have the same ratio of sides (this isactually how trigonometry works: right triangles with the same angles are similar, and so they have thesame ratios of sides, and so they have the same trigonometric values).

The smaller triangle is similar to the first. In each triangle, we will take the ratio of the vertical sideto the horizontal side:

2

12− x =h

12

You can take a time derivative of this, or you can rewrite the equation first and then take the timederivative.

d

dt2(12− x)−1 =

d

dt

1

12h

−2(12− x)−2 · (−1)dx

dt=

1

12

dh

dt

Plug in x = 4,dx

dt= −1.6 m/s and solve for

dh

dt

dh

dt= 12(−2)

1

(12− 4)2(−1)(−1.6) = −0.6 m/s

60

Chapter 4

Applications of Derivatives

4.1 Maximums and minimums

Definition. Let x = c be in the domain of f(x).x = c is an absolute maximum if f(x) ≤ f(c) for all x in the domain.x = c is a local maximum if f(x) ≤ f(c) for all x near c.

(c cannot be an endpoint)x = c is an absolute minimum if f(x) ≥ f(c) for all x in the domain.x = c is a local minimum if f(x) ≥ f(c) for all x near c.

(c cannot be an endpoint)

We first learn to understand this definition by looking at graphs.

Example 1. Identify the local/absolute max/mins in the graph in textbook 4.1 figure 1 (p. 271).(Assume that the domain of the function is everything that appears in the graph, and nothing else.)Solution: x = a is an absolute max (because the y-value there is ≤ all other y-values on the graph).

x = b is an absolute maximum (because the y-value there is ≥ all other y-values on the graph).x = b and x = e are both local mins (because the y-values at x = b are ≤ all y-values near x = b, and

similarly for x = e).x = b and x = d are both local maxs.

Theorem 4 (Fermat). If x = c is a local min/max then f ′(c) = 0 or f ′(c) is undefined.

Proof. A rigorous proof of this result uses the definition of the derivative, and limits, etc. We will becontent to understand why it should be true by looking at pictures. (Draw four pictures.)

f ′(c) = 0, local max f ′(c) DNE, local max f ′(c) = 0, local min f ′(c) DNE, local min

61

Definition. If f ′(c) = 0 or f ′(c) is undefined we call c a critical point.

Fermat’s Theorem justifies our approach to finding mins and max’s, which always starts with findingthe critical points.

Theorem 5 (Absolute max/min test (aka “closed interval method”)). To find the absolute max/min ofa function f(x) on an interval [a, b], do the following.

1. Find the critical points of f(x) in the interval [a, b] (i.e. find f ′(x), solve f ′(x) = 0 and identifywhere f ′(x) is undefined).

2. List the f(x)-values (i.e. y-values) at a, b and at each critical point. The absolute max is the biggesty-value in the list. The absolute min is the smallest y-value in the list.

Example 2. Find the absolute max/mins of sin(x) +x

2on the interval [π/2, 5π/2].

Solution: f ′(x) = cos(x) +1

2

cos(x) +1

2= 0

cos(x) = −1

2x = 2π/3 + n2π, 4π/3 + n2π,In the interval we have only 2π/3, 4π/3

x f(x)

π/2 sin(π/2) + π/4 ≈ 1.8

2π/3 sin(2π/3) +2π

6≈ 1.9

4π/3 sin(4π/3) + 4π/6 ≈ 1.2 abs min5π/2 sin(5π/2) + 5π/4 ≈ 4.9 abs max

Example 3. Find the absolute max/mins of x3 − 10x2 + 5x+ 10 on the interval [−2, 10].Solution: f ′(x) = 3x2 − 20x+ 5

3x2 − 20x+ 5 = 0

x =20±

√400− 4 · 3 · 5

6=

20±√

340

6=

10±√

85

3≈ 0.2, 6.4

x f(x)

−2 ≈ −470.2 ≈ 106.4 ≈ −105 abs min10 ≈ 60 abs max

4.2 Mean Value Theorem

We “know” that the derivative of a function tells us about the function itself right? Well, how do weknow this? Is it obvious that if f ′(x) > 0 then f(x) is increasing? It turns out that all of our knowledgeabout how f ′(x) affects f(x) depends entirely upon the following result.

62

Theorem 6 (Mean Value Theorem). Suppose f(x) is continuous on the interval [a, b] and differentiable

on the interval (a, b). Then there exists an x-value c such that f ′(c) =f(b)− f(a)

b− a .

• In other words, there is a point c where the slope f ′(c) equals the average slope from a to b.

• Geometrically, we have this picture

c

m=f '(c)

(a,f(a))

(b,f(b))

• Here’s a physical interpretation of this theorem. Suppose a cop pulls you over for speeding and saysthat you were doing 70 mph. You argue with him and say that at no point did your speedometerread 70 mph. He disagrees and says that he’s sure that at some point in the last five minutes yourspeedometer did say 70 mph. You ask him how he could know this since he wasn’t in your car.

He says that they’ve been timing you, and that you went 5 miles in 4.25 minutes. He pulls out acalculator, and shows you that 5 divided by 4.25 minutes is 1.176 miles per minute, or 70.59 mph.He then, correctly, tells you that if your average speed was 70.59 mph, then at some point yourspeedometer said, 70.59 mph. That’s the Mean Value Theorem, and there’s no point in arguingabout that!

Example 1. Let f(x) = x3 − x2 − 6x + 2. Solve for c so that f ′(c) equals the average slope on theinterval [0, 2].

Solution: The average slope isf(2)− f(1)

2− 1= −4.

f ′(x) = 3x2 − 2x− 6.3x2 − 2x− 6 = −4

x =1±√

7

3We take the positive solution since it’s the one in [0, 2]

x =1 +√

7

3.

Example 2. Suppose you know that f(3) = 7 and that f ′(x) is always between 1 and 2. Find upperand lower bounds for f(4.5) and give a geometric interpretation of your reasoning.Solution: Graphically, we have the following picture, where the straight lines come from the slopesof 1 and 2, and give upper and lower bounds for f(x) in general, and f(4.5) in particular.

63

f(x)

2 4 6 8 10 12 14

5

10

15

20

25

Although the picture is pretty convincing, we want an algebraic approach to finding these upper andlower bounds, and show that indeed f(4.5) lies between them.

We apply the MVT with the interval [3, 4.5],

f ′(x) =f(4.5)− f(3)

4.5− 3.

We know something about f(3), we know something about f ′(x) and we want to find something outabout f(4.5). Let’s start with the upper bound

2 > f ′(x) =f(4.5)− 7

1.5

which we can easily solve for f(4.5)

2 >f(4.5)− 7

1.5⇒ f(4.5) < 10

Now we get the lower bound

1 < f ′(x) =f(4.5)− 7

1.5⇒ f(4.5) > 8.5

Example 3 (optional). Suppose that f ′(x) > 0 for all x in the interval [a, b]. Prove that f(x) isincreasing.Solution: Let x1 and x2 be two x-values in the interval [a, b] with x1 < x2. We want to show thatf(x1) < f(x2).

Apply the Mean Value Theorem to f on the interval [x1, x2]. Then there is some number c such that

f ′(c) =f(x2)− f(x1)

x2 − x1.

64

But, since f ′(x) is always positive, this means that f ′(c) > 0 and so

f(x2)− f(x1)

x2 − x1= pos#

f(x2)− f(x1) = pos# · (x2 − x1) = pos# · pos#

f(x2)− f(x1) > 0

f(x2) > f(x1)

Example 4 (optional). Show that x3 + x− 1 has exactly one root.Solution: Let f(x) = x3 + x− 1.

f(x) is continuous everywhere.f(0) = −1 which is < 0.f(1) = 1 which is > 0.Therefore, by the Intermediate Value Theorem, there exists a number a, between 0 and 1 such that

f(a) = 0.Now we show that f(x) has only one root. Suppose b is a second number such that f(b) = 0. Then

f(b)− f(a)

a− b =0− 0

a− b = 0

If this were true, then Mean Value Theorem would show that there is a number c such that

f ′(c) = 0

But, f ′(x) = 3x2 + 1, which can never equal 0.Therefore, there is no such c, and therefore there is no such b, and therefore a is the only root.

Now we mention one important application of the MVT that will later be crucial to studying anti-derivatives.

Example 5 (optional). Use the MVT to show the following:

1. If f ′(x) = 0 for all x, then f(x) is a constant function.

2. If f(x) and g(x) have the same derivative, then f(x) = g(x) + C for some constant C.

Solution:

1. Let C = f(x0). Then if x is any other x-value, we have

f(x)− f(x0)

x− x0= f ′(c)

For some c. Since f ′(x) is always 0, we have f ′(c) = 0, and so

f(x)− f(x0)

x− x0= 0⇒ f(x)− f(x0) = 0⇒ f(x) = C

2. Let F (x) = f(x) − g(x). Then F ′(x) = f ′(x) − g′(x) and F ′(x) = 0 since we have assume thatf ′(x) = g′(x) for all x. Then by the first part we have F (x) ≡ C (i.e. F (x) equals C for all x).Then f(x)− g(x) = C and so f(x) = g(x) + C.

65

4.3 1st and 2cnd derivative tests

Definition. If f ′(x) > 0 we say that f(x) is increasing.If f ′(x) < 0 we say that f(x) is decreasing.

f ′(x) > 0 f ′(x) < 0

Theorem 7 (First derivative test). To find the local max/mins of a function f(x) do the following.

1. First find the critical points.

2. Figure out whether f ′(x) is + or − on each side of each critical point (four cases, lots of pictures):

f ′(x) = +,−left of c right of c outcome

+ − ⇒ x = c local max− + ⇒ x = c local min+ + ⇒ x = c neither− − ⇒ x = ctextneither

local max,f ′(c) = 0,f ′(x) = +−

local max,f ′(c) DNE,f ′(x) = +−

local min,f ′(c) = 0,f ′(x) = −+

local min,f ′(c) DNE,f ′(x) = −+

66

neither,f ′(c) = 0,f ′(x) = ++

neither,f ′(c) = 0,f ′(x) = −−

neither,f ′(c) DNE (∞),f ′(x) = ++

neither,f ′(c) DNE (cusp),f ′(x) = ++

neither,f ′(c) DNE (∞),f ′(x) = −−

neither,f ′(c) DNE(cusp),f ′(x) = −−

Finding when a function changes sign.

To use the first derivative test we need to be able to take a function f ′(x) and say when it is positiveand when it is negative. Here’s the most general way to do this, for any function g(x) (which may bef ′(x), or f ′′(x), or f(x), etc.)

1. Solve for when g(x) = 0 or is undefined. These are the only places when g(x) can change signs.(By the Intermediate Value Theorem! Yay! You thought you could forget about this. Not!)

2. Test a single value of x between each pair of numbers you found in step 1 (including a value to theright of all the numbers and a value to the left of all the numbers)

Example 1. Find the local max/mins of f(x) =ln(x)√x

.

Solution: f ′(x) =1x ·√x− ln(x)12x

−1/2

(√x)2

Simplify f ′(x) =x−1/2 − 1

2 ln(x)x−1/2

x=

1− 12 ln(x)

x1/2x=

1− 12 ln(x)

x3/2Note: f ′(x) is undefined at x = 0, but so is the original f(x), so x = 0 is not a critical point.

Solve 0 =1− 1

2 ln(x)

x3/2

0 = 1− 1

2ln(x)

67

ln(x) = 2x = e2.1st deriv test: plug in x = e and x = e3

x = e2

f ′(e) =1− 1

2 ln(e)

e3/2= + f ′(e3) =

1− 12 ln(e3)

e3/2= −

so x = e2 is a local max.

Example 2. Find the local max and mins of f(x) = x2/3e−3x.

Solution: f ′(x) =2

3x−1/3e−3x + x2/3e−3x(−3)

f ′(x) =2

3x1/3e−3x − 3x2/3e−3x

f ′(x) DNE at x = 0, so x = 0 is a critical point.

Solve 0 =2

3x1/3e−3x − 3x2/3e−3x

×3x1/3 ⇒ 0 = 2e−3x − 9xe−3x

0 = 2e−3x(2− 9x)2− 9x = 0

x =2

9Now we test a f ′(x)-value to either side of each critical value. It’s easy to plug in x = −1 and x = 1.

We also need a number between 0 and2

9. I’ll use

1

8because I know how to find the cube root of 8.

f ′(−1) =2

3(−1)1/3e3 − 3(−1)2/3e3 = e3

(−2

3 − 3)

= pos · neg = neg

f ′(1/8) =2

3(1/8)1/3e−24 − 3(1/8)2/3e−24 = e−24

(2

3·1/2 − 3 · 1

4

)= e−24

(4

3− 3

4

)= pos · pos = pos

f ′(1) =2

3(1)1/3e3 − 3(1)2/3e3 = e3

(23 − 3

)= pos · neg = neg

Here’s another way to figure out when f ′(x) is positive and negative. Factor f ′(x)

f ′(x) = e−3x(

2

3

1

x1/3− 3x2/3

)

= e−3x1

3

(2

1

x1/3− 9x2/3

)

= e−3x1

3

1

x1/3(2− 9x)

Note that e−3x is always positive (so is1

3). Also, x1/3 is negative for x < 0 and positive for x > 0.

Finally, 2− 9x is positive for x < 2/9 and positive for x > 2/9. Thus, we have

e−3x1

3· 1

x1/3·(2− 9x)=f ′(x)

x < 0 + · − · + = −0 < x < 2/9 + · + · + = +

2/9 < x + · + · − = −

68

We summarizex = 0 x = 1/3

f(−1) = − f(1/8) = + f(1) = −So x = 0 is a local min, and x = 1 is a local max.

Definition. If f ′′(x) > 0 we say that f(x) is concave up.If f ′′(x) < 0 we say that f(x) is concave down.

f ′(x) > 0, f ′′(x) < 0incr., conc. down

f ′(x) < 0, f ′′(x) < 0decr., conc. down

f ′(x) > 0, f ′′(x) > 0incr., conc. up

f ′(x) < 0, f ′′(x) > 0decr., conc. up

Interpretations of concavity:

• Concavity is not related to whether the graph is increasing or decreasing.

• Concave up means that it is curving more upwards or less downwards.

• Concave up means that it lies above each tangent line.

• Concave down means that it is curving more downwards or more upwards.

• Concave down means that it lies below each tangent line.

Theorem 8 (Second derivative test). To find the local max/mins of a function f(x) try the following.

1. First find the critical points.

2. Figure out whether f ′′(c) is + or − (three cases):

f ′′(c) outcome

+ local min− local max

0 or DNE test says nothing

f ′′(c) = − f ′′(c) = + f ′′(c) = 0, max

69

f ′′(c) = 0, min f ′′(c) = 0, neither f ′′(c) DNE(cusp), min

f ′′(c) DNE(cusp), max f ′′(c) DNE(cusp),neither

f ′′(c) DNE(∞), nei-ther

Example 3. Test the critical point x = e2 from Example 1, using the second derivative test.

Solution: We already know that f ′(x) =1− 1

2 ln(x)

x3/2, and that the only critical point is x = e2.

We take the second derivative,

f ′′(x) =(0− 1

2 · 1x)x3/2 − (1− 12 ln(x))32x

1/2

(x3/2)2

We plug in e2:

f ′′(e2) =−#−#

+#= −

So again x = e2 is a local max.

Definition. For any function f(x) we define1. the intervals of increase/decrease: a list of the intervals where f(x) is increasing, and a list of

the intervals where f(x) is decreasing.

2. the intervals of concavity: a list of the intervals where f(x) is concave up, and a list of theintervals where f(x) is concave down.

3. an inflection point: a number x = c where the concavity changes from sign (i.e. the concavity ispositive on one side and negative on the other).

Example 4. Find the intervals of increase/decrease, intervals of concavity, local max/mins, and inflectionpoints for f(x) = x+

√1− x. Would the second derivative test have worked for the local max/mins?

Solution: f ′(x) = 1− 1

2√

1− x

70

f ′(x) DNE at x = 1. However, this is an end point, and so we do not count it as a local max/min.Solve f ′(x) = 0

0 = 1− 1

2√

1− x1

2√

1− x = 1

2√

1− x = 1√1− x =

1

2

1− x =1

4

x = 1− 1

4=

3

4The domain of f(x) is (−∞, 1].

Test f ′(0) = 1− 1

2= +

Test f ′(15/16) = 1− 1

2√

1/16= −

Intervals of increase/decrease

f(x) ↑: (−∞, 3/4)

f(x) ↓: (3/4, 1]

This shows that x = 3/4 is a local max, since f ′(x) changes from + to − at this point.

f ′′(x) = 0− 1

2· (−1

2)(1− x)−3/2 · (−1) = −1

4(1− x)−3/2

f ′′(x) DNE at x = 1, but this doesn’t matter.

Solve 0 =1

4

1

(1− x)3/2

This has no solution.Test f ′′(0) = −Intervals of concavity

f(x) concave down (−∞, 1]

f(x) has no inflection point since f ′′(x) is always negative.By the way, this also verifies that x = 3/4 is a local max, since f ′′(3/4) = −.

Example 5 (4.3#28). Sketch the graph of a function that has the following properties:

f ′(x) > 0 if |x| < 2, f ′(x) < 0 if |x| > 2f ′′(x) < 0 if 0 < x < 3, f ′′(x) > 0 if x > 3f ′(2) = 0, lim

x→∞f(x) = 1, f(−x) = −f(x).

Solution: We start by drawing just a graph that is increasing and decreasing in the right regions,namely: f(x) is increasing for −2 < x < 2, and f(x) is decreasing for x < −2 and x > 2.

71

K6 K4 K2 0 2 4 6

K2

K1

1

2

Then we change this picture so that it has the right concavity, it flattens out at x = 2, it has a horizontalasymptote, and it is odd (i.e. the left side is a certain kind of mirror image of the right side).

K8 K6 K4 K2 0 2 4 6 8

K3

K2

K1

1

2

3

Example 6 (4.3#9). f(x) = 2x3 + 3x2 − 36xf ′(x) = 6x2 + 6x− 36f ′(x) = 0 at x = 2,−3f ′(x) DNE: no solutions.

−3 2

f ′(−4) = + f ′(0) = − f ′(3) = +

x = −3 is l. max, y-value is f(−3) = −44.x = 2 is l. min, y-value is f(2) = 81.f ↑: (−∞,−3, ) ∪ (2,∞)f ↓: (−3, 2)f ′′(x) = 12x+ 6f ′′(x) = 0 at x = −1/2f ′′(x) DNE: no solutions.

72

−1/2

f ′′(−1) = − f(0) = +

x = −1/2 is an inflection point.f is C.U. (−1/2,∞)f is C.D. (−∞,−1/2).

Example 7 (4.3#16). f(x) = x2 ln(x)f ′(x) = 2x ln(x) + xf ′(x) DNE at x = 0, but f(x) also DNE there.0 = 2x ln(x) + x0 = x(2 ln(x) + 1)x = 0 (which is covered above) OR 2 ln(x) + 1 = 0ln(x) = −1/2x = e−1/2 ≈ .607.f ′(x) pos/neg

0 e−1/2

f ′(0.1) = − f ′(1) = +

f ↑: (e−1/2,∞)f ↓: (0, e−1/2)x = e−1/2 is a l. min, y = f(e−1/2) ≈ −0.1839f ′′(x) = 2 ln(x) + 3f ′′(x) DNE at x = 0 (but f(x) DNE there too).0 = 2 ln(x) + 3ln(x) = −3/2x = e−3/2 ≈ .223

0 e−3/2

f ′′(0.1) = − f ′′(1) = +

Inflection points: x = e−3/2

f C.U.: (e−3/2,∞)f C.D.: (0, e−3/2)

Example 8 (4.3#43). f(θ) = 2 cos(θ) + cos2(θ), 0 ≤ θ ≤ 2πf ′(θ) = −2 sin(θ)− 2 cos(θ) sin(θ)f ′(θ) DNE: no solutions.0 = −2 sin(θ)− 2 cos(θ) sin(θ)0 = −2 sin(θ)(1 + cos(θ))sin(θ) = 0 OR 1 + cos(θ) = 0θ = 0, π, 2π.

0 π 2π

f ′(1) = − f ′(5) = +

θ = π is a l.min, y = f(π) = −1.f ′′(θ) = −2 cos(θ) + 2 sin2(θ)− 2 cos2(θ)

73

0 = −2 cos(θ) + 2 sin2(θ)− 2 cos2(θ) (use the identity sin2 + cos2 = 1)0 = −2 cos(θ)− 4 cos2(θ) + 2 (let’s divide by −2 and rearrange)0 = 2 cos2(θ) + cos(θ)− 1 (treat this as a hidden quadratic: Let X = cos(θ))0 = 2X2 +X − 10 = (2X − 1)(X + 1)2X − 1 = 0, i.e. 2 cos(x)− 1 = 0, OR X + 1 = 0, i.e. cos(θ) + 1 = 0.θ = π/3, π, 5π/3

0 π/3 π 5π/3 2π

f ′′(0.5) = − f ′′(2.5) = + f ′′(4) = + f ′′(5.5) = −Inflection points: θ = π/3, 5π/3f C.U. (π/3, π) ∪ (π, 5π/3)f C.D. (0, π/3) ∪ (5π/3, 2π)

4.4 L’Hospital’s Rule

Throughout this section, let “lim” be one of the following:

lim = limx→a

, limx→a+

, limx→a−

, limx→∞

, limx→−∞

74

Then we have seen the following limits before

limf(x)

g(x)=

lim f(x)

lim g(x)if we have

#

#(6= 0)(i.e. both limits exist and

lim g(x) 6= 0)

limf(x)

g(x)= 0 if we have

#

±∞ (i.e. lim f(x) = # 6= ±∞ and

lim g(x) = ±∞)

limf(x)

g(x)= ±∞ if we have

#(6= 0)

0(i.e. lim f(x) exists and 6= 0

and lim g(x) = 0)

limf(x)

g(x)= ±∞ if we have

±∞#

(i.e. lim f(x) = ±∞ and

lim g(x) = # 6= ±∞)

Note: we are starting to do arithmetic with “extended real numbers” here. This means we work with theusual real numbers, R, together with something we call infinity, ∞ (as well as −∞). Here, ∞ is not anumber, but we can still use some of our usual operations on it. So if r is a positive real number we have

r +∞ =∞, r · ∞ =∞, ∞− r =∞,r

∞ = 0,r

0= ±∞, etc. (See the Wikipedia article on the extended

real number line for more information about this.) However, other operations we cannot do, at least not

without some other method of calculation. For example, ∞−∞ =?,∞∞ =?, etc. These question marks

can sometimes be filled in, but not always. We will learn the technique for filling them in now.

Theorem 9 (L’Hospital’s Rule). Suppose that limf ′(x)

g′(x)exists or equals ±∞.

If limf(x)

g(x)=±∞±∞ or lim

f(x)

g(x)=

0

0then

limf(x)

g(x)= lim

f ′(x)

g′(x).

Some things to keep in mind:

• It is crucial that you check the condition limf(x)

g(x)before you use L’Hospital’s Rule, and that this

appear on your paper. It is crucial that if limf ′(x)

g′(x)does not exist, then you do not draw any

conclusions about limf(x)

g(x).

• You can use L’Hospital’s Rule more than once, so that limf(x)

g(x)= lim

f ′(x)

g′(x)= lim

f ′′(x)

g′′(x), etc.

• We often need to know a limit like limx→∞

f(x). We will say things like “plug∞ into f(x)” or “f(∞)”.

There are two ways to justify this: you can work with the extended reals, as described above, oryou should translate the phrase “plug in ∞” into “take the limit as x goes to ∞”.

75

In this manner, we can write e∞ =∞, e−∞ = 0, ln(∞) =∞, ln(0) = −∞,1

±∞ = 0,1

0= ±∞.


sin(x)

x.

Solution: Check: limx→0

sin(x)

x

=sin(0)

0=

0

0X.

limx→0

sin(x)

xLH= lim

x→0

cos(x)

1

=cos(0)

1= 1.

Example 2. Find limx→(π/2)+

cos(x)

1− sin(x)

Solution: Check: limx→(π/2)+

cos(x)

1− sin(x)

=cos(π/2)

1− sin(π/2)

=0

1− 1=

0

0X

limx→(π/2)+

cos(x)

1− sin(x)

LH= lim

x→(pi/2)+

− sin(x)

0− cos(x)

=− sin(π/2)

− cos(π/2)

=sin(π/2)

cos(π/2)

=1

0= ±∞.

It is −∞ since if x is just a little larger than π/2, then cos(x) is negative, whence cos(x) is positive.


ln(x)

x.

Solution: Check: limx→∞

ln(x)

x

=ln(∞)

∞ =∞∞X

limx→∞

ln(x)

xLH= lim

x→∞

1x

1

=1∞1

=0

1= 0

76


cos(x)

x.

Solution: WRONG SOLUTION: limx→0

− sin(x)

1

=− sin(0)

1= 0.

There are four mistakes here. First: we need to check for∞∞ or

0

0. Secondly, in this case we don’t

have one of these limits, so it’s wrong to use L’Hospital’s Rule. Third, the original limit can be figuredout without L’Hospital’s Rule. Fourth, the answer we just got was wrong!

RIGHT SOLUTION: limx→0

cos(x)

x=

cos(0)

0=

1

0= ±∞.


x2

e0.01x

Solution: Check: limx→∞

x2

e0.01x

=∞2

e0.01∞

=∞e∞

=∞∞X.

limx→∞

x2

e0.01xLH= lim

x→∞

2x

0.01e0.01x

=2∞

0.01e0.01∞

=∞∞X.

So we should use L’Hospital’s Rule again.LH= lim

x→∞

2

(0.01)2e0.01x

=2

(0.01)2e∞=

2

∞ = 0


1− cos(x10)

x20. Hint: Do LH once, simplify x’s, and then do LH again. Compare

your answer to the graph of1− cosx10

x20.

Solution: This is one of my favorite examples, for reasons I’ll give below.

Check: limx→0

1− cos(x10)

x20

=1− cos(0)

0

=0

0X

limx→0

1− cos(x10)

x20

77

LH= lim

x→0

sin(x10) · 10x9

20x19SIMPLIFY!

= limx→0

sin(x10)

2x10

= limx→0

sin(0)

0=

0

0X

LH= lim

x→0

cos(x10) · 10x9

2 · 10x9SIMPLIFY!

= limx→0

cos(x10)

2=

cos(0)

2=

1

2.

One of the reasons I like this example so much is that involved both simplification and doing L’Hospitaltwice. But the other reason I like it is that your calculator cannot do this problem. In the picture belowwe show a simulation of a calculator graph of this function.

K2 K1 0 1 2

K1

1

2

According to the graph, the limit looks like it will be 0 at x = 0. Do you know what is going on? Is thecalculator right and Calculus wrong? No way! The calculator is running into round error. The numberson top of the original fraction are the same out to the 14th decimal place (or so), but these differencesare what’s keeping the limit from really being equal to 0. But the 14th decimal place is the limit of thecalculator’s accuracy, so the calculations are not quite able to give the right values.

Variations on L’Hospital’s Rule.

• lim f(x)g(x) = 0 · ∞: rewrite as a fraction, and use LH.

(Also we could have 0 · (−∞), and ±∞ · 0, etc. To rewrite f(x)g(x) as a fraction, note that

f(x)g(x) =f(x)

1/g(x)=

g(x)

1/f(x).)

• lim f(x)− g(x) =∞−∞: rewrite as a fraction, and use LH.

(To rewrite f(x) − g(x) as a fraction may require creativity: maybe get a common denominator,

maybe write it asf(x)

1− g(x)

1and multiply by conjugate square roots, etc.).

• lim(f(x))g(x) = 00, ∞0, 1±∞: Let y = f(x)g(x), take natural log, rewrite as a product, rewrite asa fraction, apply L’Hospital, get back to original limit using e: lim y = elim g(x) ln(f(x)).

78

Example 7. Find limx→0+

x ln(x)

Solution: Check: limx→0+

x ln(x)

= 0 · ln(0+)= 0 · (−∞)X

Use L’Hospital: limx→0+

x ln(x)

= limx→0+

ln(x)

1/x

LH= lim

x→0+

1x

−1/x2

SIMPLIFY!

= limx→0+

1

x· −x

2

1= lim

x→0+−x = 0


ln(x) tan(πx/2).


ln(x) tan(πx/2)

= ln(1) tan(π/2+)= 0 · (−∞)X

Use L’Hospital: limx→1+

ln(x) tan(πx/2)

= limx→1+

ln(x)

1/ tan(πx/2)

= limx→1+

ln(x)

cot(πx/2)

LH= lim

x→1+

1x

− csc2(πx/2) · π/2

=11

− csc2(π/2) · π/2=

1

−1 · π/2= − 2

π

Example 9. Find limx→π/2−

(sec(x)− tan(x))

Solution: Check: limx→(π/2)−

(sec(x)− tan(x))

= sec(π/2−)− tan(π/2−)=∞−∞X(note: you need to know what the vertical asymptotes of sec(x) and tan(x) look like, see the graphs

in front cover of the book.)Use L’Hospital: lim

x→(π/2)−(sec(x)− tan(x))

WRONG:LH= lim

x→(π/2)−sec(x) tan(x)− sec2(x).

79

This is wrong because we first need to put this function into a fraction, then take the derivative ofthe top and bottom.

= limx→(π/2)−

(1

cos(x)− sin(x)

cos(x)

)

= limx→(π/2)−

1− sin(x)

cos(x)

LH= lim

x→(π/2)−

0− cos(x)

− sin(x)

= limx→(π/2)−

cos(x)

sin(x)

=0

1= 0.


(csc(x)− cot(x)).


(csc(x)− cot(x))

=∞−∞X (Again, check the graphs to see that each function gives positive infinity.)

Rewrite as fraction, us LH: limx→0+

(csc(x)− cot(x))

= limx→0+

(1

sin(x)− cos(x)

sin(x)

)

= limx→0+

1− cos(x)

sin(x)LH= lim

x→0+

0 + sin(x)

cos(x)

=0 + sin(0)

cos(0)= 0

Example 11. If you have a debt of P (in dollars), compounded n times per year, at an annual interestrate of r, then the amount you will owe after t years is

A = P(

1 +r

n

)nt.

(a) Suppose you have $100,000 of debt, at 7% interest, for 5 years. How much will you owe if it’scompounded monthly? (b) Daily? (c) Compounded every instant?

Solution: (a) 100000(1 + .07

1

)1·5= $140, 255.17.

(b) 100000(1 + .07

12

)12·5= $141, 762.52.

(c) We want to find limn→∞

100000(1 + .07

n

)5n. This will be 10000 times lim

n→∞

(1 + .07

n

)5n. We’ll leave

the 10000 alone for now, and just worry about this limit.Check: lim

n→∞

(1 + .07

n

)5n

=(1 + .07

∞)5∞

= (1 + 0)∞

= 1∞X.We need to use L’Hospital to figure this out.

80

y =

(1 +

.07

x

)5x

.

ln(y) = ln

((1 +

.07

x

)5x)

.

ln(y) = 5x ln

(1 +

.07

x

)

limx→∞

5x ln

(1 +

.07

x

)

limx→∞

ln(1 + .07/x)

1/(5x)

L.H. limx→∞

11+.07/x · (0− .07/x2)

−15x−2

limx→∞

.071+.07/x

15

=.07

1 + 0· 5

1= .07 · 5.

limx→∞

y = elim

x→∞ln(y)

= e5(.07)

= e5(.07)

Total debt: 10000e5(.07) = 141906.75Note: the final formula was 105e5(.07). Thus, we have seen/proved that

P(

1 +r

n

)nt−→ Pert

as n→∞.


(1 + sin(4x))cot(x)


(1 + sin(4x))cot(x)

= (1 + sin(0))cot(0)

= 1∞X

Take ln( ), rewrite as a product, rewrite as a fraction, use LH.y = (1 + sin(4x))cot(x)

ln(y) = ln(

(1 + sin(4x))cot(x))

ln(y) = cot(x) ln(1 + sin(4x))limx→0+

cot(x) ln(1 + sin(4x))

= limx→0+

ln(1 + sin(4x))

tan(x)

LH= lim

x→0+

11+sin(4x) cos(4x)4

sec2(x)

=1

1+0 · 1 · 4sec2(0)

= 4

limx→0+

y = elim

x→0+g(x) ln(f(x))

= e4.

81

Some Homework problems solutions

Example 13 (4.4#39). limx→∞

x sin(π/x)

limx→∞

x sin(π/x) =∞ sin(π/∞) =∞ · sin(0) =∞ · 0X

limx→∞

sin(π/x)

1/x

LH= lim

x→∞

cos(π/x)(−πx2

)−1x2

(Simplify!)

limx→∞

cos(π/x)π

1cos(π/∞)π = cos(0) = πNote: you could have tried to turn this product into a fraction the other way, using the fact that

1

sin= csc and gotten this

limx→∞

x

csc(π/x)

but then this wouldn’t have helped, because the next step would have been

LH= lim

x→∞

1

− csc(π/x) cot(π/x)(−πx2

)=

1

− csc(0) cot(0)0

But csc(0) =∞, and so this didn’t work, at least not without additional steps.

Example 14 (4.4#40). limx→−∞

x2ex = (−∞)2e−∞ =∞ · 1

∞ =∞ · 0XRewrite as a fraction, put ex on the bottom as 1/ex = e−x.

limx→−∞

x2

e−xLH= = lim

x→−∞

2x

−e−x =∞

−e−(−∞)=∞∞X

Do L’Hospital’s Rule again:LH= lim

x→−∞

2

e−x=

2

e−(−∞)=

2

∞ = 0

Note: you could have tried putting x2 on the bottom of the fraction. You’d getex

x−2, and after

applying L’Hospital’s Rule, you’d getex

−2x−3. This is still indeterminate, but note that if you apply

L’Hospital’s Rule again, it doesn’t help. You started with x−2, and then get x−3, and then x−4, etc. It’sbetter to leave x2 on top, so you can get rid of it by taking the derivative.

Example 15 (4.4#47). limx→1

(x

x− 1− 1

ln(x)

)=

1

1− 1− 1

ln(1)=

1

0− 1

0=∞−∞X

Rewrite as a fraction:

limx→1

x ln(x)

(x− 1) ln(x)− (x− 1)

(x− 1) ln(x)= lim

x→1

x ln(x)− (x− 1)

(x− 1) ln(x)

LH= lim

x→1

1 ln(x) + x · 1x − 1

1 ln(x) + (x− 1) · 1x= lim

x→1

ln(x)

ln(x) + 1− 1x

=ln(1)

ln(1) + 1− 11

=0

0X

LH= lim

x→1

1x

1x + 0 + 1

x2

=1

1 + 1=

1

2

82

Example 16 (4.4 #49). limx→∞

(√x2 + x− x) =

√∞−∞ =∞−∞X

= limx→∞

(√x2 + x− x) ·

√x2 + x+ x√x2 + x+ x

= limx→∞

(x2 + x)− x2√x2 + x+ x

= limx→∞

x√x2 + x+ x

Divide the top and bottom by the largest simplified power of x from the bottom. This is x since wehave x =

√x2.

limx→∞

1x · x

1x(√x2 + x+ x)

= limx→∞

1√1x2

(x2 + x) + 1x · x

limx→∞

1√1 + 1

x + 1=

1√1 + 0 + 1

=1

2

Note: We didn’t do L’Hospital’s rule here for two reasons. First of all, we know how to do thisproblem without L’Hospital’s rule: this is straight from section 2.6. Secondly, if we did L’Hospital’s rule,it doesn’t work. When you take the derivative of the bottom, part of what you will get is the derivative

of the square root, which isx√

x2 + x. This part alone is of the form

∞∞ , and so you would need to do

L’Hospital’s rule again. Then you will get1x√x2+x

=

√x2 + x

x. This is again of the form

∞∞ , and so

you would need to do L’Hospital’s rule again. But doing it again just starts the cycle over again withthe square root.

Example 17 (4.4 #55). limx→0

(1− 2x)1/x = (1− 0)1/0 = 1∞X.

y = (1− 2x)1/x

ln(y) =1

xln(1− 2x)

limx→0

1

xln(1− 2x) = lim

x→0

ln(1− 2x)

x

LH= lim

x→0

11−2x(−2)

1=−21−01

= −2

e−2

Example 18 (4.4, #64). limx→∞

(2x− 3

2x+ 5

)2x+1

=

(2

2

)∞= 1∞X

y =

(2x− 3

2x+ 5

)2x+1

ln(y) = (2x+ 1) ln

(2x− 3

2x+ 5

)= (2x+ 1)(ln(2x− 3)− ln(2x+ 5))

limx→∞

(2x+ 1)(ln(2x− 3)− ln(2x+ 5)) = limx→∞

ln(2x− 3)− ln(2x+ 5)1

2x+1

LH= lim

x→∞

12x−3(2)− 1

2x+5(2)

−(2x+ 1)−2(2)Cancel the three (2)’s, and rewrite the big fraction, so that instead of dividing by −(2x + 1)−2,

multiply by its reciprocal.

= limx→∞

(1

2x− 3− 1

2x+ 5

)(−(2x+ 1)2)

= limx→∞

((2x+ 5)− (2x− 3)

(2x− 3)(2x+ 5)

) −(4x2 + 4x+ 1)

1= lim

x→∞

8

4x2 + 4x− 15· −(4x2 + 4x+ 1)

1

83

= limx→∞

−8 · 4x2 + 4x+ 1

4x2 + 4x− 15= −8 · 4

4= −8

e−8

Note: I used twice the short-cut rule for horizontal asymptotes of rational functions (see page 26).Namely: given the same powers of x on the top and bottom of the fraction, the limit as x goes to infinity,i.e. the horizontal asymptote, is the ratio of the leading coefficients.

4.5 (and some of 4.6) Function analysis

In a sense, this section has nothing new in it, it’s a combination of our previous work. The point is toanalyze a function using calculus, and then to make a sketch of its graph.

To sketch the graph of a function f(x), find (some of) the following:

f(x) parts

• Symmetry:

∗ if f(−x) = f(x) then f(x) is even. This means that the graph to the left of the y-axis isthe mirror image of the right side. Note: x = 0 will be a local max or min. Examples ofeven functions: cos(x) or any even power of x.

∗ if f(−x) = −f(x) then f(x) is odd. This means that the graph to the left of the y-axisis the negative mirror image of the right side. Note: the graph will go through the origin(unless x = 0 is a vertical asymptote). Examples of odd function: sin(x) or any odd powerof x.

∗ Note: even×even=even, even×odd=odd, odd×odd = even (so multiplying functions islike adding even/odd numbers). Thus, x2 sin(x) will be odd, and x4 cos(x) will be even.Also, if g = even is even then f(g(x)) = even for any function f .

• if f(x) is periodic (like sin(x), cos(x), etc.), then we only need to analyze it for one full period(like [0, 2π]).

• x and y intercepts.

• Vertical and horizontal asymptotes.

• intervals of positivity/negativity.

f ′(x) parts Find the critical points, local max/mins (including y-values), intervals of increase/decrease.

f ′′(x) parts Find when f ′′(x) equals 0 or is undefined, find the intervals of concavity and inflectionpoints.

Note: Not all of the above will be needed in every problem. Finding symmetry is not absolutelynecessary, but can save time.

Example 1. Analyze the function xe−x2

and graph the result by hand.Solution:

• f(x) parts.

Symmetry: f(−x) = (−x)e−(−x)2

= −xe−x2 = −f(x), so f(x) is odd.

x-int: 0 = xe−x2

=⇒ x = 0. y-int: f(0) = 0

84

Pos/neg: test f(−1) = − and f(1) = +. So f(x) = −, (−∞, 0) and f(x) = +, (0,∞).

Asymptotes: no vertical asymptotes.

Horizontal: limx→∞

xe−x2.

Rewrite as fraction: limx→∞

x

ex2. L.H. lim

x→∞

1

2xex2=

1

∞ = 0 So the Horizontal Asymptote is y = 0.

• First derivative stuff. f ′(x) = 1e−x2

+ xe−x2(−2x) = e−x

2(1− 2x2)

Critical points: 0 = e−x2 − 2x2e−x

2 → 0 = e−x2(1− 2x2) → 0 = 1− 2x2 → x2 =

1

2→ x = ±

√1

2Increase/decrease: test f ′(−1) = −, f ′(0) = +, f ′(1) = −Increase:

(− 1√

2, 1√

2

)

Decrease: (−∞, 1√2) ∪ ( 1√

2,∞)

local max: x =1√2

, y = f(1/√

2) ≈ .42

local min: x = − 1√2

, y ≈ −.42

• Second derivative stuff.

f ′′(x) = e−x2(−2x)(1− 2x2) + e−x

2(0− 4x) = e−x

2(4x3 − 6x)

Solve f ′′(x) = 0 for 0 = −6x+ 4x3 = 2x(−3 + 2x2) get x = 0,±√

3/2.

Test: f ′′(−2) = −, f ′′(−1) = +, f ′′(1) = −, f ′′(2) = +.

Concave down: (−∞,−√

3/2) ∪ (0,√

3/2)

Concave up: (−√

3/2, 0) ∪ (√

3/2,∞)

Inflection points: 0, ±√

3/2.

Putting it all together, we see the following graph:

K3 K2 K1 0 1 2 3

K1.0

K0.5

0.5

1.0

85

Example 2. Analyze f(x) =1

x8− 2× 108

x4, use your calculator where necessary, but graph the result

by hand.Solution:

• f stuff without derivative.

Symmetry:

f(−x) =1

(−x)8− 2× 108

(−x)4=

1

x8− 2× 108

x4.

Since this equals f(x) we see that f(x) is even.

Horizontal Asymptote: limx→∞

1

x8− 2× 108

x4= lim

x→0

1− 2× 108x4

x8

Since the degree on the bottom is larger than the degree on top, this will have a limit of 0, i.e. thex-axis is the horizontal asymptote (if you don’t recall this fact, then apply L’Hospital).

x-int:

0 =1

x8− 2× 108

x4

multiply by x8

0 = 1− 2× 108x4

whence

x = ± 4

√1

2× 108= ± 1

100 4√

2≈ ±0.008408964155

y-int: f(0) is not defined, so there is no y-intercept.

Vertical Asymptote: Probably there is one at x = 0, but trying to plug it in gives an indeterminate

form ∞−∞. So, we use L’Hospital. limx→0

1

x8− 2× 108

x4= lim

x→0

1

x8− 2× 108x4

x8

= limx→0

1− 2× 108x4

x8=

1− 0

0= ±∞

So there is a vertical asymptote at x = 0.

f(x) pos/neg. There are two ways to do this. One, write f(x) like this

f(x) =1− 2× 108x4

x8=

1− 2× 108x4

pos #

to see that the sign of f(x) equals the sign of 1 − 2 × 108x4, which looks like an upside downparabola. Or you can test at f(−1), f(−0.00001), f(0.00001) and f(1).

f(x) = ± − 1

100 4√

20

1

100 4√

2| | |− + + −

86

• First derivative stuff

f ′(x) = − 8

x9+

8× 108

x5.

f ′(x) DNE @x = 0 (but f(x) DNE there too).

0 = − 8

x9+

8× 108

x5

multiply by x9

0 = −8 + 8× 108x4

x = ± 4

√8

8× 108= ± 1

100= ±0.01.

Increasing/decreasing: There are two ways to find the sign of f ′(x). You can write f ′(x) like this

f ′(x) =−8 + 8× 108x4

x9

Now x9 is positive for x > 0, and the top of f ′(x) looks like a parabola opening upwards. Thus, forx > 0 we have that f ′(x) is negative, then positive. Or, you can test f ′(x) at appropriate values,for instance f ′(−1), f ′(−0.001), f ′(0.001), f ′(1).

f ′(x) = ± − 1

1000

1

100| | |− + − +

f(x) is increasing:(− 1

100 , 0)∪(

1100 , ∞

)

f(x) is decreasing:(−∞, − 1

100

)∪(0, 1

100

)

local min: x = ± 1

100, y = −1016

• Second derivative stuff

f ′′(x) =72

x10− 40× 108

x6.

This is undefined at x = 0, and solving

0 =72

x10− 40× 108

x6

gives

x = ±√

3

100 4√

5≈ ±0.01158292186.

Thus, there are three places where f ′′(x) can change sign, and we get

f ′′(x) = ± −√

3

100 4√

50

√3

100 4√

5| | |− + + −

87

Now we can state the intervals of concavity, and the inflection points:

f(x) is concave down:(−∞,−

√3

100 4√5

)∪( √

3100 4√5

,∞)

f(x) is concave up:(−√3

100 4√5, 0)∪(

0,√3

100 4√5

)

Inflection points: x = ±√

3

100 4√

5

Now, we turn to graphing this function. From the above work, we know that it should look aboutlike this

However, when we plot f(x), with −1 ≤ x ≤ 1 and −10 ≤ y ≤ 10, we get this,

K1.0 K0.5 0 0.5 1.0

K10

K5

5

10

Obviously, we need to change the window size. The x-values of the critical points are very small,

and the y-values are very large. Thus, a better window is given by − 2

100≤ x ≤ 2

100and −1016 ≤

y ≤ 1016. This graph looks like this

88

K0.02 K0.01 0 0.01 0.02

K1#1016

K5#1015

5#1015

1#1016

That doesn’t look quite like the first graph we got. The function is going up so quickly near 0 thatwe can’t even see enough of the graph for it to get close to the y-axis. Let’s change the windowsize again, and see more of the graph get close to the y-axis.

K0.0010 K0.0005 0 0.0005 0.0010

2#1031

4#1031

6#1031

8#1031

1#1032

Now we have a different problem, we can’t see the local mins! Well, it’s not possible to get everythingwe want in one graph. That’s part of the reason we use Calculus, and not calculators!

Example 3 (Extra example: not covered in class). Analyze the functionx

x3 − 1.

Solution:

• f(x) parts.

Symmetry. f(−x) =−x

(−x)3 − 1=

−x−x3 − 1

=x

x3 + 1. This last formula does not equal f(x) or

−f(x), so this function is not even or odd.

x-int: 0 =x

x3 − 1⇒ x = 0.

y-int: f(0) = 0, i.e. f(x) goes through the origin.

89

f(x) undefined: There is division by 0 at x = 1. Note that the top is not also zero there, so wehave a vertical asymptote at x = 1.

Pos/neg: Test f(x) between 0 and 1, and to the left of 0 and to the right of 1.

x = 0 x = 1

f(−1) = + f(.1) = − f(1) = +

Thus, f(x) is positive for (−∞, 0) ∪ (1,∞), and negative for (0, 1).

Horizontal Asymptote: y = 0 because the degree on the bottom is larger than on top.

• First derivative stuff.

f ′(x) =−2x3 − 1

(x3 − 1)2.

f ′(x) DNE at x = 1.

f ′(x) = 0

0 =−2x3 − 1

(x3 − 1)2

0 = −2x3 − 1

x = − 13√

2.

Test values

x = − 13√

2x = 1

f ′(−1) = + f ′(0) = − f ′(2) = −

f(x) increasing (−∞,− 13√

2)

f(x) decreasing (− 13√

2, 1) ∪ (1,∞).

There is a local max at x = − 13√

2.

• Second derivative stuff.

f ′′(x) =6x2(x3 + 2)(x3 − 1)

(x3 − 1)4

f ′′(x) =6x2(x3 + 2)

(x3 − 1)3

f ′′(x) DNE at x = 1.

f ′′(x) = 0

0 =6x2(x3 + 2)

(x3 − 1)3

0 = 6x2(x3 + 2)x = 0, − 3

√2

Test valuesx = − 3

√2 x = 0

f ′′(−2) = + f ′′(−1) = − f ′′(1) = +

90

f(x) is concave up: (−∞,− 3√

2) ∪ (1,∞)

f(x) is concave down: (− 3√

2, 0) ∪ (0, 1).

The point x = − 3√

2 is an inflection point.

Putting it all together we have the following graph

K4 K3 K2 K1 0 1 2 3 4

K1.0

K0.5

0.5

1.0

Example 4 (4.5,#11). f(x) =1

x2 − 9.

f(x)-stuff.

x-int: 0 =1

x2 − 9, no solution.

y-int: f(0) =1

0− 9= −1

9Symmetry: even.HA: y = 0 since x2 is on the bottom and there are no x’s on top.VA: x = ±3 since this gives division by 0.f pos/neg:

−3 3

f(−4) = + f(0) = − f(4) = +

f ′(x)-stuff.

f ′(x) = − 2x

(x2 − 9)2

f ′(x) = 0 at x = 0.f ′(x) DNE at x = ±3.f ′ pos/neg

−3 0 3

f ′(−4) = + f ′(−1) = + f ′(1) = − f ′(4) = −

x = 0 is l.max, y = −1

9.

f ↑: (−∞,−3) ∪ (3, 0)f ↓: (0, 3) ∪ (3,∞)f ′′(x)-stuff

91

f ′′(x) =6(x2 + 3)

(x2 − 9)3

f ′′(x) = 0: no solution.f ′′(x) DNE at x = ±3.f ′′(x) pos/neg

−3 3

f ′′(−4) = + f ′′(0) = − f ′′(4) = +

f is C.U. (−∞, 3) ∪ (3,∞)f is C.D. (−3, 3)Putting it all together:

Example 5 (4.5 #31). f(x) = 3 sin(x)− sin3(x)y-int: f(0) = 0.x-int: f(x) = 0, 0 = sin(x)

(3− sin2(x)

), sin(x) = 0, x = 0, π, 2π, 3π, etc.

Symmetry: f(x) is odd.Periodic: yes. So, if we want, we can do only work from 0 to 2π, and then extend our results.VA: none.HA: none.f pos/neg (between 0 and 2π)

0 π 2π

f(1) = + f(4) = −f ′(x) stuff.f ′(x) = 3 cos(x)− 3 sin2(x) cos(x)0 = 3 cos(x)− 3 sin2(x) cos(x)

92

0 = 3 cos(x)(1− sin2(x))cos(x) = 0 OR 1− sin2(x) = 0x = π/2, 3π/2, 5π/2, 7π/2,. . .f ′(x) pos/neg (between 0 and 2π)

0 π/2 3π/2 2π

f ′(0.5) = + f ′(3) = − f ′(5.5) = +

x = π/2 is l. max, y = f(π/2) = 2x = 3π/2 is l.min, y = f(3π/2) = −2f ↑: (0, π/2) ∪ (3π/2, 2π)f ↓: (π/2, 3π/2)f ′′(x)-stufff ′′(x) = −3 sin(x)− 6 sin(x) cos2(x) + 3 sin3(x)f ′′(x) = 3 sin(x)

(1− 2 cos2(x) + sin2(x)

)

f ′′(x) = −9 sin(x) cos2(x) (using the identity sin2(x) + cos2(x) = 1)0 = −9 sin(x) cos2(x) means sin(x) = 0 OR cos2(x) = 0.x = 0, π/2, π, 3π/2, 2π, 5π/2,. . .

0 π/2 π 3π/2 2π

f ′′(0.5) = − f ′′(2.5) = − f ′′(3.5) = + f ′′(5.5) = +

Inflection points: x = πf C.U. (π, 3π/2) ∪ (3π/2, 2π)f C.D. (0, π/2) ∪ (π/2, π)Putting it all together:

93

4.7 Optimization

The basic idea of this section is to find absolute max/mins of real-world problems. Sometimes you aregiven a single function to optimize, but most often you are asked to combine two functions.

Example 1. (a). Find formulas for the following the area A and the perimeter P of a rectangle.

x

y

(b). Set A = 100 and solve for y. Plug this into the formula for P and simplify.

(c). Find the minimum of P (include x and y values).Solution: (a) A = xy

P = 2x+ 2y

(b) 100 = xy

y =100

x

P = 2x+ 2 · 100

x

(c) P ′ = 2− 2001

x2

0 = 2− 2001

x2

0 = 2x2 − 200x = ±

√100 = ±10

x = 10

y =100

10= 10

P = 2 · 10 + 2 · 10 = 40

Example 2. (a). Find formulas for the area A and the cost C of a rectangular fence which costs $30per meter on the north and south sides, and $50 per meter on the east and west sides.

(b). Set C = $5000 and solve for y. Plug this into the formula for A and simplify.

(c). Find the maximum of A (include x and y values).Solution: (a) A = xy

C = 2 · 30x+ 2 · 50y

(b) 5000 = 60x+ 100y

y =5000− 60x

100= 50− .6x

A = x(50− .6x)A = 50x− .6x2

(c) A′ = 50− 1.2x0 = 50− 1.2x

94

x =50

1.2≈ 41.66

y = 50− .6(41.66) = 25A = 41.666(25) = 1041.66

Example 3. (a). Find formulas for the area A and the perimeter P of the shape below (which is arectangle with a half-circle on top. The perimeter does not include the dashed line between thecircle and the rectangle).

x

y

(b). Set P = 10 and solve for y. Plug this into the formula for A and simplify.

(c). Maximize A (include x and y values)

Solution: (a) A = xy +1

2π(x/2)2

P = x+ 2y +1

2πx

(b) 10 = x+ 2y +1

2πx

y =10− x− π

2x

2

y = 5− 2 + π

4x

A = x

(5− 2 + π

4x

)+

1

2π(x/2)2

A = 5x− 2 + π

4x2 +

π

8x2

A = 5x+−4− 2π + π

8x2

A = 5x+−4− π

8x2

(c) A′ = 5 +−4− π

4x

0 = 5 +−4− π

4x

x = −5 · 4

−4− πx =

20

4 + π≈ 2.800

95

y = 5− 2 + π

4x =≈ 1.400

A = 7.00

Example 4. (a). Find formulas for the area A and the volume V of the cylinder (including the topand bottom) below.

h

r

(b). Set A = 1 and solve for h. Plug this into the formula for V and simplify.

(c). Maximize V (include h and r values).Solution: (a) V = area of end× length = πr2 · h

A = area of two ends + area of sides

= 2πr2 + area of rectangle “wrapped around side”

= 2πr2 + 2πrh

(b) 1 = 2πr2 + 2πrh1− 2πr2 = 2πrh

h =1− 2πr2

2πr

V = πr2(

1− 2πr2

2πr

)

V = r

(1− 2πr2

2

)

V =1

2r − πr3

(c) V ′ =1

2− 3πr2

0 =1

2− 3πr2

r2 =1

6π

r =1√6π≈ 0.23

96

h =1− 2πr2

2πr≈ 0.46

V = πr2h ≈ .076

4.9 Anti-derivatives

Definition. An anti-derivative of f(x) is a function F (x) such that F ′(x) = f(x). As an abbreviation

for “the anti-derivative of f(x)” we write

∫f(x) dx. Note that both “

∫” and “ dx” are part of this

notation, they are somewhat like parentheses ( and ), the important stuff goes between them. Also, dx

plays the role of telling us what variable we are using for the anti-derivative, just liked

dxdoes for the

derivative.

Example 1. Find an anti-derivative of 3x2 and verify this.Solution: Guess F (x) = x3

Check F ′(x) =d

dxx3 = 3x2.

Theorem 10. If F (x) is an anti-derivative of f(x), then F (x) + C is also an anti-derivative for eachconstant C.

Proof.d

dxF (x) + C = F ′(x) + 0 = f(x)

Example 2. Find three anti-derivatives of 2 sin(x).Solution: We guess F (x) = −2 cos(x).

Check: F ′(x) =d

dx− 2 cos(x) = 2 sin(x).

Now we need some other anti-derivatives: F (x) + 8 = −2 cos(x) + 8 and F (x) + 5 = −2 cos(x) + π.

Checkd

dx(−2 cos(x) + 1) = 2 sin(x) + 0, etc.

As a result of the previous theorem, we always write anti-derivatives using +C at the end of theformula.

Each derivative formula that we know can be turned around to make an anti-derivative

d

dxxn = nxn−1 =⇒

∫nxn−1 dx = xn + C

d

dxex = ex =⇒

∫ex dx = ex + C

d

dxsin(x) = cos(x) =⇒

∫cos(x) dx = sin(x) + C

d

dxcos(x) = − sin(x) =⇒

∫− sin(x) dx = cos(x) + C

d

dxtan(x) = sec2(x) =⇒

∫sec2(x) dx = tan(x) + C

d

dxln(x) =

1

x=⇒

∫1

xdx = ln(x) + C

97

But sometimes we want to write the anti-derivative formula differently. For instance, we want to knowthe anti-derivative of sin(x), not − sin(x), and similarly we want to know the anti-derivative of xn, notnxn−1. This is easy to do once we use the constant multiple rule:

d

dxCf(x) = Cf ′(x)⇒

∫Cf(x) dx = C

∫f(x) dx

Now it’s easy to rewrite the formulas for sin(x) and xn:

∫sin(x) dx = − cos(x)

∫xn dx =

xn+1

n+ 1if n 6= −1

Example 3. Find the anti-derivative F (x) of f(x) = x7 − 5 sin(x) such that F (0) = 7.

Solution: We have F (x) =

∫x7 − 5 sin(x) dx =

x8

8+ 5 cos(x) + C.

F (0) = 7⇒ 0

8+ 5 cos(0) + C = 7

C = 2

F (x) =x8

8+ 5 cos(x) + 2.

One of the most important types of derivatives and anti-derivatives involve position, velocity andacceleration. Since we know

velocity =d

dtposition and acceleration =

d

dtvelocity

this means, by definition, that

velocity =

∫acceleration dt and position =

∫velocity dt

Example 4. On the moon, acceleration due to gravity is given by −1.6 m/s2. A ball is thrown straightup (at time t = 0) with a velocity of 0.5 m/s from a height of 2 m. (a) Find the velocity function v(t).(b) Find the height function h(t).

Solution: (a) v(t) =

∫−1.6 dt

v(t) = −1.6t+ CSolve for C: v(0) = 0.5 so −1.6(0) + C = 0.5 and C = 0.5.v(t) = −1.6t+ 0.5

(b) h(t) =

∫v(t) dt

h(t) =

∫−1.6t+ 0.5 dt

h(t) = −1.6

2t2 + 0.5t+ C

Solve for C: h(0) = 2 so 02 + 0 + C = 2 and C = 2h(t) = −1.6

2 t2 + 0.5t+ 2

98

Chapter 5

The Definite Integral (CompressedVersion)

5.2 The Definite Integral

Definition. Let f(x) be a function defined on the interval [a, b]. We define the following symbol

∫ b

af(x) dx

to mean the signed area between the x-axis and f(x), and between x = a and x = b, where “signed”means that if part of the graph is above the x-axis and part below, we calculate the area in each part,and subtract the part below from the part on top.

We have three ways we will find

∫ b

af(x) dx.

•∫ b

af(x) dx can be found using basic formulas for area, if f(x) is made of simple shapes such as

rectangles, straight lines, parts circles, etc.

•∫ b

af(x) dx can be approximated by adding the areas of rectangles whose tops are on f(x): this is

called the Riemann Sum.

•∫ b

af(x) dx = F (b)− F (a) where F (x) is an anti-derivative of f(x): this is called the Fundamental

Theorem of Calculus.

Example 1. Let f(x) be defined by the curve below, and find

∫ 10

0f(x) dx exactly.

99

0 2 4 6 8 100

1

2

3

4

5

6

Solution: Since the word “exact” appears in the problem we cannot use Riemann sums. Instead, webreak the area into a rectangle, half of a circle, and a triangle.

Area = rect + half circle + triangle

= 2 · 10 +1

2π22 +

1

24 · 4

Example 2. Let f(x) be defined by the curve below, and find

∫ 10

0f(x) dx

Solution: We break the shape into simple geometric parts as shown.

100

We calculate the area in each part, and subtract the part below the x-axis from the part on top.

1

2π(1.5)2 +

1

2· 1 · 4 + 3 · 4− (

1

2· 2 · 4 + 3 · 4 +

1

2· 1 · 4)

Finishing the above calculation we have

∫ 10

0f(x) dx =

9

8π − 4

Example 3. Approximate the area between the x-axis and the curve y =√

1− x2 from x = −1/2 tox = 1/2 using n = 5 (i.e. use 5, equally spaced, rectangles) and the Left Hand Rule (i.e. each rectangleshould have it’s top intersect f(x) at its left edge). Write your answer using integral notation and as aRiemann Sum, and draw a picture illustrating f(x) and the Riemann Sum.Solution: The first left edge comes at x − 1/2. Since they are equally spaced, the distance betweenedges is 1/5th of the total distance from x = −1/2 to x = 1/2, i.e. the distance is 0.2.

Here are where we should have all the left edges:

x = −0.5, −0.3, −0.1, 0.1, 0.3

We are supposed to have 5 rectangles, where the top of the rectangle is where the left edge intersectsf(x). Thus, the y-value should be f(x) evaluated at each of the above left edges:

f(−0.5), f(−0.3), f(−0.1), f(0.1), f(0.3)

Putting these numbers inside of f , we have our solution

∫ 1/2

−1/2

√1− x2 dx

≈√

1− (−0.5)2 ·(0.2)+√

1− (−0.3)2 ·(0.2)+√

1− (−0.1)2 ·(0.2)+√

1− (0.1)2 ·(0.2)+√

1− (0.3)2 ·(0.2).

101

In general, we can give a formula for the values needed for a Riemann sum, but I think it’s usuallyeasier to figure them out as you go, as we did in the previous problem. But, if you are interested, hereare the formulas:

We divide the interval [a, b] into n equal pieces, each of width ∆x =b− an

a b

x1 x2 x3 xn+1· · · · · ·

∆x

Then the Left Hand Riemann Sum is

∫ b

af(x) dx ≈ f(x1)∆x+ f(x2)∆x+ · · ·+ f(xn)∆x

where

x1 = a,

x2 = x1 + ∆x = a+ ∆x,

x3 = x2 + ∆x = a+ 2∆x,

. . . ,

xn = xn−1 + ∆x = a+ (n− 1)∆x = b−∆x.

There are other ways of forming a Riemann sum: the simplest ones are called the Right Hand Sumand the Midpoint sum. The difference between these is which values of x we plug into f(x). For the right

102

hand rule we use the right edge of each rectangle: x1 = a+ ∆x, x2 = x1 + ∆x,. . . , xn = xn−1 + ∆x = b.

For the midpoint rule we plug the middle x-value into f(x): x1 = a+∆x

2, x2 = x1 + ∆x, x3 = x2 + ∆x,

. . . , xn = xn−1 + ∆x = b− ∆x

2.

Example 4. Find an approximation of

∫ π/2

0sin(x2) dx using n = 5, and a Left Hand Riemann Sum,

and interpret this on the graph of sin(x2).

Solution: We have π/2 ≈ 1.571. Thus ∆x =1.571− 0

5= 0.314. Thus the x-values are

x-values 0, 0.314, 0.628, 0.942, 1.257.

Putting it all together we have

∫ π/2

0sin(x2) dx

≈ 0.314 sin(0.3142) + 0.314 sin(0.6282) + 0.314 sin(0.9422) + 0.314 sin(1.2572) + 0.314 sin(1.5712)

= 0.906

To interpret this number in terms of the graph of sin(x2), think about what f(0.314) is: it’s a y-value, aheight. Think about what 0.314 is, it’s a width along the x-axis. Therefore, 0.314f(0.314) is the productof a width and a height, it’s the area of a rectangle. Therefore, the sum we have the sum of areas ofrectangles. The rectangles have their heights on the curve of sin(x2). We show this is the graph below

103

5.3 The Fundamental Theorem of Calculus

Theorem 11. If f is continuous on the interval [a, b], then

∫ b

af(x) dx = F (b)− F (a)

where F (x) is any anti-derivative of f(x).

Example 1. Find

∫ π

0sin(x) dx.

Solution: FTC says we should start by finding an anti-derivative. F (x) = − cos(x) so

∫ π

0sin(x) dx = − cos(x)

∣∣∣∣π

0

= − cos(π)− (− cos(0)) = −(−1)− (−1) = 2

Example 2. Find

∫ 2

1

2

x− 7

5√x3 dx.

Solution: FTC says we should start by finding an anti-derivative. Here’s the thoughts that should go

through your mind when you find F (x). You should think of2

xas 2 · 1

x, and ask yourself “what would I

take the derivative of to get1

x?” Hopefully you said ln(x). Then you should think of

5√x3 as x3/5 and

add one to the exponent, and divide by this result. Putting this together we get

∫ 2

1

2

x− 7

5√x3 dx = (2 ln(x)− 7x8/5 · 5

8)

∣∣∣∣2

1

= 2 ln(2)− 35

828/5 −

(2 ln(1)− 35

818/5

)

= 2 ln(2)− 35

828/5 +

35

8

5.4 The Net Change Theorem

Recall that the FTC had two functions f(x) and F (x) where F (x) is an anti-derivative of f(x). Thismeans that f(x) = F ′(x). If we write the theorem this way, it tells us about the net change of F (x).

Theorem 12. The net change of a function F (x) on an interval [a, b] is defined as F (b) − F (a). If weknow the derivative F ′(x), then the net change can be calculated as

F (b)− F (a) =

∫ b

aF ′(x) dx

Example 1. 5.4#61. Recall our linear density example from section 3.7. We have a rod with mass, andm(x) is the amount of mass from position 0 to position x in the rod. Then ρ(x) is the derivative of m(x),and we call ρ(x) the linear density.

Suppose that ρ(x) = 9 + 2√x kg/m for a rod of length 4 m. Find the total mass of the rod.

Solution: The total mass is m(4)−m(0). We find this using the net change theorem

m(4)−m(0) =

∫ 4

0m′(x) dx

104

=

∫ 4

0ρ(x) dx =

∫ 4

09 + 2

√x dx

= 9x+ 2 · x3/2

3/2

∣∣∣∣4

0

= 9x+4

3x3/2

∣∣∣∣4

0

= (9 · 4 +4

3· 43/2)− (0 + 0)

= 36 +4

3· 8 =

140

3kg

Example 2. Let v(t) =

√t

16be a velocity function. Find the total distance travelled from t = 0 to t = 60.

Solution: Let P (t) be the position function. Then P ′(t) = v(t) and the net distance travelled isP (60)− P (0). Since the velocity is positive, the total distance travelled is the same as the net distancetravelled. Thus, our answer is given by

P (60)− P (0) =

∫ 60

0

√t

60dt

=1

60

t3/2

3/2

∣∣∣∣60

0

=1

60

(603/2

3/2− 03/2

3/2

)

=4

3

√15

5.5 U-substitution

Example 1. (a) Find the derivative of sin(x2).

(b) Find the anti-derivative

∫3x cos(x2) dx.

Solution: (a) Using the chain rule we getd

dxsin(x2) = cos(x2)2x.

(b) What we are starting with is so close to cos(x2)2x, so we should be able to guess the answer.

What’s the difference between

∫3x cos(x2) dx and

∫cos(xx)2x dx? A constant multiple. Thus, our

answer should be different by a constant multiple than sin(x2). Constant multiples should be prettyeasy, so we should be able to guess the following:

d

dxC sin(x2) = 3x cos(x2)

What should C be? It should cancel the 2 that comes fromd

dxx2 = 2x, and it should include a multiple

of 3. Thus, we should have C =3

2. Let’s double check our answer

d

dx

3

2sin(x2) =

3

22x cos(x2) = 3 cos(x2)

105

This shows that ∫3x cos(x2) dx =

3

2sin(x2) + C

In the previous example, part (a) was the chain rule, and then part (b) was essentially doing thechain rule backwards. The main idea of the technique we learn next is to make a system for doing thechain rule backwards.

The anti-derivative technique of u-substitution is the chain rule backwards. It’s very hard to put intowords or formulas, but here goes anyway.

106

Procedure for basic u-substitution

0. You are given an integral

∫ somethinginvolving x dx where

somethinginvolving x is some complicated

function of x.

1. Try view to the integrand (the thing you are integrating) this way

somethinginvolving x = f(g(x))g′(x)C

where f and g are two functions you know and C is a constant left over fromsomethinginvolving x

that’s not part of f(g(x))g′(x) or C is a constant you needed to make g′(x) equal to the derivative ofthe inside (for more complicated u-substitution, C might need to be replaced by something besidesa constant). If you can’t see how to view the integrand this way, then you’ll need to try to guesswhat u is at the next step.

2. Choose what u equals: either you take a guess, or it’s the inside function from the previous step.Either way, let’s call it g(x), then take the derivative.

let u = g(x)

then du = g′(x) dx (a rewritten form ofdu

dx= g′(x) )

3. Now, even if you didn’t see it before, you need to be able to finish step 1, using the equations fromstep 2

somethinginvolving x dx = f(g(x))g′(x)C dx = f(u)C du

Notice that we included dx and du in the step too. This means you can translate the originalintegral as shown

∫ somethinginvolving x dx =

∫Cf(u) du

Make sure that all the x’s (including dx) cancel by the last step.

4. Find the anti-derivative F (u) of what you had in step 3,∫Cf(u) du = CF (u)

then replace u with g(x) again to get CF (g(x)).

Comments

1. The crucial steps are 1 and 3. Step 1 is the hard one because you must figure out what to call u,without a formula to guide you, and do it in such a way that step 3 works out correctly.

107

2. In step 1, the main way to find u is through trial and error, practice, and experience (includingtrying things that don’t work.)

3. There are two common ways to do step 3. Sometimes people find du, and then look at what’soutside of f , and figure out how that could be part of of du. Sometimes people solve for dx likethis

dx =1

g′(x)du

and then substitute the right hand side for dx in the original integral, and try to cancel all the x’sthat are outside of f . We will do the examples below both ways. Note: if you are uncomortablewith u-substitution, then solving for dx and cancelling is probably a little safer. But, you shouldpractice seeing how the other approach works at least a little for the following reason: you needto learn to look ahead a little, and see if the stuff outside of f equals the derivative of the inside(give or take the multiple of C). Only by looking ahead a little can you quickly figure out the rightu-substitution.

Example 2. (a) Find the anti-derivative

∫sin(3x+ 7) dx.

(b) Find the anti-derivative

∫x7 sin(x8) dx.

(c) Find the anti-derivative

∫ex sin(ex) dx.

Solution: (a) Let u = 3x+ 7.du = (3x+ 7)′ dxdu = 3 dx

dx =1

3du

So we can translate everything in the original integral to one involving u’s:

∫sin(u)

1

3du =

1

3

∫sin(u) du =

1

3(− cos(u)).

We finish by putting x’s back in:

∫sin(3x+ 7) dx = −1

3cos(3x+ 7) + C

(b) We do this one two ways: solving for dx, and seeing du in there.

Solving for dx

Let u = x8

du = 8x7 dx

dx =1

8x7du

∫x7 sin(u)

1

8x7du

=

∫1

8sin(u) du

= −1

8cos(u) + C

= −1

8cos(x8) + C

108

Seeing du in there (give or take the constant)Let u = x8

du = 8x7 dx1

8du = x7 dx

1

8du

∫x7 sin(x8) dx

=

∫1

8sin(u) du

= − 1

8cos(u) + C

= − 1

8cos(x8)

(c) Let u = ex

du = ex dx

dx =1

exdu

∫ex sin(ex) dx =

∫ex sin(u)

1

exdu =

∫sin(u) du = − cos(u) = − cos(ex) + C

One special case of u-substitution, that’s very easy, is an integral of the form

∫f(ax+ b) dx.

This always equals1

aF (ax+ b). Eventually, you should become very fast at this type of u-substitution,

just like you are for doing the chain rule for a function of the form f(ax+ b).

Example 3. Find

∫(3x+ 10)7.5 dx.

Solution: u = 3x+ 10du = 3 dx

dx =1

3du

∫u7.5

1

3du

1

3

u8.5

8.51

3

(3x+ 10)8.5

8.5+ C

Example 4. Find

∫sec2(1/x)

x2dx using u = 1/x.

Solution: We do this problem 4 times, in ever decreasing detail. Pick the column you’re mostcomfortable with, and practice that.

109

Solving

∫sec2(1/x

)

x2

dxusingfourapproaches

tou-substitution.

Solu

tion

1:M

axim

um

det

ail

,so

lvin

gfo

rdx

u=

1 x

du

=

(1 x

) ′dx

du

=−

1 x2dx

dx

=−x2du

Now

we

tran

slate

the

inte

gra

lan

dso

lve

it∫

sec2

(1/x

)

x2

dx

=

∫se

c2(u

)

x2

(−x2)du

=

∫−

sec2

(u)du

=−

tan

(u)

+C

=−

tan

(1/x

)+C.

Th

em

ain

diff

eren

ceb

etw

een

this

and

the

foll

owin

gap

pro

ach

esis

that

her

ew

eso

lve

the

der

iva-

tive

equ

ati

on

fordx

.A

sa

con

-se

qu

ence

,w

hen

we

plu

gth

isin

,w

eh

ave

toca

nce

lx2

onth

eto

pan

db

ott

om

.In

this

app

roach

,yo

um

ust

alw

ays

mak

esu

reth

atall

thex

’sca

nce

l.

Solu

tion

2:Ju

sta

litt

lele

ssd

e-ta

il,

isola

tin

gth

eou

tsid

e

u=

1 x

du

=

(1 x

) ′dx

du

=−

1 x2dx

(−1)du

=1 x2dx

Now

we

tran

slat

eth

ein

tegr

alan

dso

lve

it∫

sec2

(1/x

)

x2

dx

=

∫se

c2(u

)1 x2dx

=

∫−

sec2

(u)du

=−

tan

(u)

+C

=−

tan

(1/x

)+C.

Th

em

ain

diff

eren

ceb

etw

een

this

an

dth

efo

llow

ing

app

roac

hes

isth

atw

efi

rst

plu

gged

inju

stu

,an

dis

ola

ted

the

rest

ofth

ep

art

on

the

outs

ide,

gett

ing

1 x2dx

.

Th

en,

we

sub

stit

ute

din

(−1)du

for

this

par

t.

Sol

uti

on3:

Med

ium

amou

nt

ofd

etai

l,se

ein

gdu

inth

ere

u=

1 x

du

=−

1 x2dx

(−1)du

=1 x2dx

Now

we

tran

slat

eth

ein

tegr

al

and

solv

eit

∫se

c2(1/x

)

x2

dx(−

1)du

=

∫−

sec2

(u)du

=−

tan

(u)

+C

=−

tan

(1/x

)+C.

Inm

yop

inio

n,

this

isa

good

amou

nt

ofd

etai

lto

use

.A

lso,

Ip

refe

rit

toth

efi

rst

app

roac

hb

e-ca

use

Ith

ink

that

atso

me

poi

nt

you

nee

dto

be

able

tolo

okah

ead

ali

ttle

,an

dse

eif

the

der

ivat

ive

ofu

ison

the

outs

ide,

rath

erth

an

wai

tin

gu

nti

lyo

uplu

gindx

tose

eif

all

thex

’sca

nce

l.W

ith

ali

ttle

pra

ctic

e,you

can

see

inyo

ur

hea

dth

atm

akin

gu

=1/x

wil

lgi

veyo

uth

est

uff

circ

led

inre

dab

ove.

Solu

tion

4:M

inim

um

det

ail

:∫

sec2

(1/x

)

x2

dx

=−

tan

(1/x

)+C

Iam

not

atall

advo

cati

ng

that

you

try

tod

osu

cha

pro

ble

min

you

rh

ead

.B

ut,

itis

just

abou

tp

ossi

ble

.Y

oulo

ok

atth

eori

gin

alin

tegr

al,

see

sec2

an

dre

ali

zeth

atth

efi

nal

answ

erw

ill

be

tan

sin

ceth

isis

the

anti

-der

ivati

ve.

Th

enyo

uch

eck

that

the

der

ivat

ive

of

wh

at’s

insi

de

sec2

,eq

ual

sw

hat’

son

the

outs

ide,

giv

eor

take

am

ult

iple

of−

1.

Th

isex

pla

ins

the

neg

ati

veth

atap

pea

rsin

the

fin

al

an

swer

.

110

Example 5. Find

∫sin(x)

1 + cos2(x)dx

Solution:

Solving for dxu = cos(x)du = − sin(x) dx

dx = − 1

sin(x)du

∫sin(x)

1 + cos2(x)dx

=

∫sin(x)

1 + u2· −1

sin(x)du

=

∫ −1

1 + u2du

= − tan−1(u) = − tan−1(cos(x)) + C.

Seeing du in thereu = cos(x)du = − sin(x) dx

∫sin(x)

1 + cos2(x)dx

(−1) du

=

∫1

1 + u2· (−1) du

= − tan−1(u) = − tan−1(cos(x)) + C.

Example 6. Find

∫2 sin(x) cos(x)

1 + cos2(x)dx

Solution:

Solving for dxu = 1 + cos2(x)du = −2 sin(x) cos(x) dx

dx =−1

2 sin(x) cos(x)du

∫2 sin(x) cos(x)

1 + cos2(x)dx

=

∫2 sin(x) cos(x)

u· −1

2 sin(x) cos(x)du

=

∫ −1

udu = − ln |u| = − ln(1 + cos2(x)) + C

Seeing du in there

u = 1 + cos2(x)

du = −2 sin(x) cos(x) dx

∫2 sin(x) cos(x)

1 + cos2(x)dx

(−1) du

=

∫1

u· (−1) du

= − ln |u| = − ln(1 + cos2(x)) + C

Definite Integrals with U-substitution

Suppose you are given an integral

∫ b

a, dx, and that you use a u-substitution of the form u = g(x) to

get an integral

∫f(u) du = F (u). Then there are two ways to finish the integral:

• Get F (u), plug x’s back in to get F (g(x)), and plug in x = a and x = b,

• Get F (u), plug x = a and x = b into u = g(x) to get u = A and u = B, then plug in u = A andu = B into F (u).

Both of these ways are correct and equally good.

Note, when we change the endpoints, this sometimes produces an integral

∫ B

Awhere B < A (as

opposed to usual where we have a < b). We finish this integral as usual: F (B) − F (A). Equivalently,

111

you can rewrite it first with the endpoints being in the usual order, but with an extra negative sign∫ B

A= −

∫ A

B.

Example 7. Find

∫ 2

0x(3x2 + 1)25 dx

Solution: Let me start by setting up the indefinite integral, i.e. setting up the integral without a andb.

Solving for dx u = 3x2 + 1

du = 6x dx

dx =1

6xdu.

∫x(3x2 + 1)25 dx =

∫xu25

1

6xdu

=

∫u25

1

6du

Seeing du inside.

u = 3x2 + 1

du = 6x dx1

6du = x dx

∫(3x2 + 1)25x dx

1

6du

=

∫u25

1

6du

Now, there are two ways to finish, and both are equally good.If we do not change the endpoints, then we should be very explicit about which variable goes with

the numbers, like this:

∫ x=2

x=0

1

6u25 du =

1

6

u26

26

∣∣∣∣x=2

x=0

=1

6

(3x2 + 1)26

26

∣∣∣∣2

0

=1

6

[1326

26− 126

26

]=

1

6· 1

26[1326 − 1]

If we do change the endpoints, then we might as well do this right away: x = 0→ u = 3 · 02 + 1 = 1and x = 2→ u = 3 · 22 + 1 = 13. Then the original integral becomes

∫ 13

1

1

6u25 du =

1

6

u26

26

∣∣∣∣13

1

=1

6

[1326

26− 126

26

]=

1

6· 1

26[1326 − 1]

Backwards substitution

Sometimes we are given

∫, dx and it’s easy to see , = f(g(x))?, but ? is not equal to a constant

multiple of du = g′(x) dx. In these cases, we need to use backwards substitution to get rid of all the x’sin ?. To do this, we solve u = g(x) for x = g−1(u), and then replace each x in ? with g−1(u).

Example 8. Find

∫x(3x+ 10)99 dx

Solution: u = 3x+ 10du = 3 dx

dx =1

3du

∫xu99

1

3du

112

The problem is, there appears to be no way to cancel that other x. To get rid of it we need “backwards

substitution”. We have an equation u = 3x+10 and we can solve this equation for x! We get x =1

3(u−10)

∫1

3(u− 10)u99

1

3du =

1

9

∫(u− 10)u99 du =

1

9

∫u100 − 10u99 du =

1

9

(u101

101− 10

u100

100

)+ C

Example 9. Find

∫x2√1− x dx (hint: you will need to solve your substitution for x, then square what

you get to replace x2, then split things up into powers of u).Solution: u = 1− x

du = −dx∫x2√u

(−1) du so now we need to get rid of x2.

go back to u = 1− x, solve this for x = u− 1 and square to get x2 = (u− 1)2∫(u− 1)2√

udu =

∫u2 − 2u+ 1√

udu =

∫u2√u− 2

u√u

+1√udu =

∫u3/2− 2u1/2 + u−1/2 du =

2

5u5/2−

2 · 2

3u3/2 + 2u1/2 =

2

5(1− x)5/2 − 4

3(1− x)3/2 + 2(1− x)1/2 + C

Example 10. Find

∫x3√x2 + 1 dx.

Solution: u = x2 + 1du = 2x dx

dx =1

2xdu

∫x3√u

1

2xdu

=1

2

∫x2√u du

We need to work harder to get rid of all the x’s. Go back to u = x2 + 1, solve this for x2 = u− 11

2

∫(u− 1)

√u du

1

2

∫u3/2 − u1/2 du

1

2

(2

5u5/2 − 2

3u3/2

)

1

2

(2

5(x2 + 1)5/2 − 2

3(x2 + 1)3/2

)+C.

113

ma 251, calculus i, fall 2010 lecture...

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