math 1131q - calculus 1. · a very important theorem: differentiability is a stronger condition...
TRANSCRIPT
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Average ALEKS scores for MATH 1131Q - Section 80
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Our first Midterm Exam will be on Tuesday 9/30
Two times (check your time following these instructions)(1) 6-8 PM at AUST 108, or(2) 9-11 PM at TLS 154.
There is a practice exam and solutions in outline in the website:http://alozano.clas.uconn.edu/math1131f14
![Page 4: MATH 1131Q - Calculus 1. · A very important theorem: differentiability is a stronger condition than continuity. Theorem If f(x) is differentiable at x = a, then f(x) is continuous](https://reader033.vdocument.in/reader033/viewer/2022042103/5e813003daf38229c574b2b9/html5/thumbnails/4.jpg)
Our first Midterm Exam will be on Tuesday 9/30Two times (check your time following these instructions)(1) 6-8 PM at AUST 108, or(2) 9-11 PM at TLS 154.
There is a practice exam and solutions in outline in the website:http://alozano.clas.uconn.edu/math1131f14
![Page 5: MATH 1131Q - Calculus 1. · A very important theorem: differentiability is a stronger condition than continuity. Theorem If f(x) is differentiable at x = a, then f(x) is continuous](https://reader033.vdocument.in/reader033/viewer/2022042103/5e813003daf38229c574b2b9/html5/thumbnails/5.jpg)
Our first Midterm Exam will be on Tuesday 9/30Two times (check your time following these instructions)(1) 6-8 PM at AUST 108, or(2) 9-11 PM at TLS 154.
There is a practice exam and solutions in outline in the website:http://alozano.clas.uconn.edu/math1131f14
![Page 6: MATH 1131Q - Calculus 1. · A very important theorem: differentiability is a stronger condition than continuity. Theorem If f(x) is differentiable at x = a, then f(x) is continuous](https://reader033.vdocument.in/reader033/viewer/2022042103/5e813003daf38229c574b2b9/html5/thumbnails/6.jpg)
Our first Midterm Exam will be on Tuesday 9/30Two times (check your time following these instructions)(1) 6-8 PM at AUST 108, or(2) 9-11 PM at TLS 154.
There is a practice exam and solutions in outline in the website:http://alozano.clas.uconn.edu/math1131f14
![Page 7: MATH 1131Q - Calculus 1. · A very important theorem: differentiability is a stronger condition than continuity. Theorem If f(x) is differentiable at x = a, then f(x) is continuous](https://reader033.vdocument.in/reader033/viewer/2022042103/5e813003daf38229c574b2b9/html5/thumbnails/7.jpg)
Our first Midterm Exam will be on Tuesday 9/30Two times (check your time following these instructions)(1) 6-8 PM at AUST 108, or(2) 9-11 PM at TLS 154.
There is a practice exam and solutions in outline in the website:http://alozano.clas.uconn.edu/math1131f14
Covers Sections 1.1 - 3.3 (including 3.3, derivatives oftrigonometric functions).
There will be a review during class on Tuesday 9/30. Bring yourquestions, and/or I will go over the practice exam.
My office hours are as usual: Tuesdays 10:15-11:15 andThursdays 11-12, at MSB 312. TA’s office hours also as usual(see website).
There will be a review session on Monday 9/29 (by Amit Savkar;time and room TBA; will also be available online).
![Page 8: MATH 1131Q - Calculus 1. · A very important theorem: differentiability is a stronger condition than continuity. Theorem If f(x) is differentiable at x = a, then f(x) is continuous](https://reader033.vdocument.in/reader033/viewer/2022042103/5e813003daf38229c574b2b9/html5/thumbnails/8.jpg)
Our first Midterm Exam will be on Tuesday 9/30Two times (check your time following these instructions)(1) 6-8 PM at AUST 108, or(2) 9-11 PM at TLS 154.
There is a practice exam and solutions in outline in the website:http://alozano.clas.uconn.edu/math1131f14
Covers Sections 1.1 - 3.3 (including 3.3, derivatives oftrigonometric functions).
There will be a review during class on Tuesday 9/30. Bring yourquestions, and/or I will go over the practice exam.
My office hours are as usual: Tuesdays 10:15-11:15 andThursdays 11-12, at MSB 312. TA’s office hours also as usual(see website).
There will be a review session on Monday 9/29 (by Amit Savkar;time and room TBA; will also be available online).
![Page 9: MATH 1131Q - Calculus 1. · A very important theorem: differentiability is a stronger condition than continuity. Theorem If f(x) is differentiable at x = a, then f(x) is continuous](https://reader033.vdocument.in/reader033/viewer/2022042103/5e813003daf38229c574b2b9/html5/thumbnails/9.jpg)
Our first Midterm Exam will be on Tuesday 9/30Two times (check your time following these instructions)(1) 6-8 PM at AUST 108, or(2) 9-11 PM at TLS 154.
There is a practice exam and solutions in outline in the website:http://alozano.clas.uconn.edu/math1131f14
Covers Sections 1.1 - 3.3 (including 3.3, derivatives oftrigonometric functions).
There will be a review during class on Tuesday 9/30. Bring yourquestions, and/or I will go over the practice exam.
My office hours are as usual: Tuesdays 10:15-11:15 andThursdays 11-12, at MSB 312. TA’s office hours also as usual(see website).
There will be a review session on Monday 9/29 (by Amit Savkar;time and room TBA; will also be available online).
![Page 10: MATH 1131Q - Calculus 1. · A very important theorem: differentiability is a stronger condition than continuity. Theorem If f(x) is differentiable at x = a, then f(x) is continuous](https://reader033.vdocument.in/reader033/viewer/2022042103/5e813003daf38229c574b2b9/html5/thumbnails/10.jpg)
Our first Midterm Exam will be on Tuesday 9/30Two times (check your time following these instructions)(1) 6-8 PM at AUST 108, or(2) 9-11 PM at TLS 154.
There is a practice exam and solutions in outline in the website:http://alozano.clas.uconn.edu/math1131f14
Covers Sections 1.1 - 3.3 (including 3.3, derivatives oftrigonometric functions).
There will be a review during class on Tuesday 9/30. Bring yourquestions, and/or I will go over the practice exam.
My office hours are as usual: Tuesdays 10:15-11:15 andThursdays 11-12, at MSB 312. TA’s office hours also as usual(see website).
There will be a review session on Monday 9/29 (by Amit Savkar;time and room TBA; will also be available online).
![Page 11: MATH 1131Q - Calculus 1. · A very important theorem: differentiability is a stronger condition than continuity. Theorem If f(x) is differentiable at x = a, then f(x) is continuous](https://reader033.vdocument.in/reader033/viewer/2022042103/5e813003daf38229c574b2b9/html5/thumbnails/11.jpg)
MATH 1131Q - Calculus 1.
Álvaro Lozano-Robledo
Department of MathematicsUniversity of Connecticut
Day 9
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 5 / 30
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Derivatives(Rates of Change)
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The Derivative as a Function
DefinitionThe derivative of a function f at a number a, denoted by f ′(a), isdefined by
f ′(a) = limh→0
f (a + h)− f (a)
hif this limit exists and it is finite.
We define a new function, the derivative of f :
DefinitionLet f (x) be a function. We define the derivative of f , denoted by f ′, by
f ′(x) = limh→0
f (x + h)− f (x)
h.
The domain of f ′(x) are those values x where the limit exists and it isfinite.
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 7 / 30
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Differentiability
DefinitionA function f (x) is continuous at a number a if
1 f (x) is defined at x = a, i.e., f (a) is well-defined,
2 limx→a
f (x) exists, and
3 limx→a
f (x) = f (a).
DefinitionA function f (x) is differentiable at x = a if f ′(a) exists, i.e., if the limit
f ′(a) = limh→0
f (a + h)− f (a)
h
exists and it is finite. We say f (x) is differentiable in a set of points (e.g., anopen interval) if it is differentiable at every point in the set.
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 8 / 30
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Continuity and Differentiability
A very important theorem: differentiability is a stronger condition thancontinuity.
TheoremIf f (x) is differentiable at x = a, then f (x) is continuous at x = a.
Warning! The converse is not true. The function f (x) = |x | iscontinuous at x = 0, but not differentiable at x = 0.
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 9 / 30
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Differentiation Rules
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The Derivative of Elementary Functions
TheoremLet f (x) be a function.
If f (x) = c is constant for all x, then f ′(x) = 0.
If f (x) = x for all x, then f ′(x) = 1.
If f (x) = xn where n is a real number, then f ′(x) = nxn−1.
If f (x) = ax where a is a positive real number, then
f ′(x) = f ′(0)f (x) = f ′(0)ax where f ′(0) = limh→0
ah − 1
h.
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 11 / 30
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DefinitionWe define the number e as the unique real number a with the property
limh→0
ah − 1
h= 1.
Corollaryd
dx(ex) = ex .
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 12 / 30
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DefinitionWe define the number e as the unique real number a with the property
limh→0
ah − 1
h= 1.
Corollaryd
dx(ex) = ex .
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 12 / 30
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New Derivatives from Old
Some helpful rules of differentiation:
TheoremLet c be a constant, and let f and g be differentiable functions. Then:
1 ddx (cf (x)) = c d
dx f (x), i.e., (cf (x))′ = cf ′(x).
2 ddx (f (x) + g(x)) = d
dx f (x) + ddx g(x), i.e., (f + g)′ = f ′ + g′.
3 ddx (f (x)− g(x)) = d
dx f (x)− ddx g(x), i.e., (f − g)′ = f ′ − g′.
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 13 / 30
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Example
What is the derivative of f (x) = x2ex?
We need a “product rule” for differentiation.
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 14 / 30
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Example
What is the derivative of f (x) = x2ex?
We need a “product rule” for differentiation.
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 14 / 30
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The Product Rule
Let f (x) and g(x) be two differentiable functions, and letp(x) = f (x)g(x). Then:
p′(x) = limh→0
p(x + h)− p(x)
h
= limh→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)− f (x)
hg(x + h) + f (x)
g(x + h)− g(x)
h
= limh→0
f (x + h)− f (x)
hg(x + h) + lim
h→0f (x)
g(x + h)− g(x)
h= f ′(x)g(x) + f (x)g′(x).
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 15 / 30
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The Product Rule
Let f (x) and g(x) be two differentiable functions, and letp(x) = f (x)g(x). Then:
p′(x) = limh→0
p(x + h)− p(x)
h
= limh→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)− f (x)
hg(x + h) + f (x)
g(x + h)− g(x)
h
= limh→0
f (x + h)− f (x)
hg(x + h) + lim
h→0f (x)
g(x + h)− g(x)
h= f ′(x)g(x) + f (x)g′(x).
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 15 / 30
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The Product Rule
Let f (x) and g(x) be two differentiable functions, and letp(x) = f (x)g(x). Then:
p′(x) = limh→0
p(x + h)− p(x)
h
= limh→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)− f (x)
hg(x + h) + f (x)
g(x + h)− g(x)
h
= limh→0
f (x + h)− f (x)
hg(x + h) + lim
h→0f (x)
g(x + h)− g(x)
h= f ′(x)g(x) + f (x)g′(x).
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 15 / 30
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The Product Rule
Let f (x) and g(x) be two differentiable functions, and letp(x) = f (x)g(x). Then:
p′(x) = limh→0
p(x + h)− p(x)
h
= limh→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)− f (x)
hg(x + h) + f (x)
g(x + h)− g(x)
h
= limh→0
f (x + h)− f (x)
hg(x + h) + lim
h→0f (x)
g(x + h)− g(x)
h= f ′(x)g(x) + f (x)g′(x).
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 15 / 30
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The Product Rule
Let f (x) and g(x) be two differentiable functions, and letp(x) = f (x)g(x). Then:
p′(x) = limh→0
p(x + h)− p(x)
h
= limh→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)− f (x)
hg(x + h) + f (x)
g(x + h)− g(x)
h
= limh→0
f (x + h)− f (x)
hg(x + h) + lim
h→0f (x)
g(x + h)− g(x)
h= f ′(x)g(x) + f (x)g′(x).
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 15 / 30
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The Product Rule
Let f (x) and g(x) be two differentiable functions, and letp(x) = f (x)g(x). Then:
p′(x) = limh→0
p(x + h)− p(x)
h
= limh→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)− f (x)
hg(x + h) + f (x)
g(x + h)− g(x)
h
= limh→0
f (x + h)− f (x)
hg(x + h) + lim
h→0f (x)
g(x + h)− g(x)
h
= f ′(x)g(x) + f (x)g′(x).
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 15 / 30
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The Product Rule
Let f (x) and g(x) be two differentiable functions, and letp(x) = f (x)g(x). Then:
p′(x) = limh→0
p(x + h)− p(x)
h
= limh→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)− f (x)
hg(x + h) + f (x)
g(x + h)− g(x)
h
= limh→0
f (x + h)− f (x)
hg(x + h) + lim
h→0f (x)
g(x + h)− g(x)
h= f ′(x)g(x) + f (x)g′(x).
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 15 / 30
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The Product Rule
TheoremIf f (x) and g(x) are both differentiable, then
d
dx(f (x)g(x)) =
df
dxg(x) + f (x)
dg
dx.
Example
What is the derivative of f (x) = x2ex?
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 16 / 30
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The Product Rule
TheoremIf f (x) and g(x) are both differentiable, then
d
dx(f (x)g(x)) =
df
dxg(x) + f (x)
dg
dx.
Example
What is the derivative of f (x) = x2ex?
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 16 / 30
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The Product Rule
TheoremIf f (x) and g(x) are both differentiable, then
d
dx(f (x)g(x)) =
df
dxg(x) + f (x)
dg
dx.
Example
If R(t) = v(t)p
t , where v(4) = 2 and v ′(4) = 3, find f ′(4).
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 17 / 30
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The Product Rule
TheoremIf f (x) and g(x) are both differentiable, then
d
dx(f (x)g(x)) =
df
dxg(x) + f (x)
dg
dx.
Example
Find the first derivative of f (x) =x2
ex.
We need a “quotient rule” of differentiation.
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 18 / 30
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The Product Rule
TheoremIf f (x) and g(x) are both differentiable, then
d
dx(f (x)g(x)) =
df
dxg(x) + f (x)
dg
dx.
Example
Find the first derivative of f (x) =x2
ex.
We need a “quotient rule” of differentiation.
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 18 / 30
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The Quotient Rule
Let g(x) be a differentiable function, and let Q(x) = 1g(x) . Then:
Q′(x) = limh→0
Q(x + h)−Q(x)
h
= limh→0
1g(x+h) −
1g(x)
h
= limh→0
g(x)−g(x+h)g(x)g(x+h)
h
= limh→0
g(x)−g(x+h)h
g(x)g(x + h)
= limh→0
− g(x+h)−g(x)h
g(x)g(x + h)
=−g′(x)
(g(x))2
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 19 / 30
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The Quotient Rule
Let g(x) be a differentiable function, and let Q(x) = 1g(x) . Then:
Q′(x) = limh→0
Q(x + h)−Q(x)
h
= limh→0
1g(x+h) −
1g(x)
h
= limh→0
g(x)−g(x+h)g(x)g(x+h)
h
= limh→0
g(x)−g(x+h)h
g(x)g(x + h)
= limh→0
− g(x+h)−g(x)h
g(x)g(x + h)
=−g′(x)
(g(x))2
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 19 / 30
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The Quotient Rule
Let g(x) be a differentiable function, and let Q(x) = 1g(x) . Then:
Q′(x) = limh→0
Q(x + h)−Q(x)
h
= limh→0
1g(x+h) −
1g(x)
h
= limh→0
g(x)−g(x+h)g(x)g(x+h)
h
= limh→0
g(x)−g(x+h)h
g(x)g(x + h)
= limh→0
− g(x+h)−g(x)h
g(x)g(x + h)
=−g′(x)
(g(x))2
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 19 / 30
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The Quotient Rule
Let g(x) be a differentiable function, and let Q(x) = 1g(x) . Then:
Q′(x) = limh→0
Q(x + h)−Q(x)
h
= limh→0
1g(x+h) −
1g(x)
h
= limh→0
g(x)−g(x+h)g(x)g(x+h)
h
= limh→0
g(x)−g(x+h)h
g(x)g(x + h)
= limh→0
− g(x+h)−g(x)h
g(x)g(x + h)
=−g′(x)
(g(x))2
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 19 / 30
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The Quotient Rule
Let g(x) be a differentiable function, and let Q(x) = 1g(x) . Then:
Q′(x) = limh→0
Q(x + h)−Q(x)
h
= limh→0
1g(x+h) −
1g(x)
h
= limh→0
g(x)−g(x+h)g(x)g(x+h)
h
= limh→0
g(x)−g(x+h)h
g(x)g(x + h)
= limh→0
− g(x+h)−g(x)h
g(x)g(x + h)
=−g′(x)
(g(x))2
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 19 / 30
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The Quotient Rule
Let g(x) be a differentiable function, and let Q(x) = 1g(x) . Then:
Q′(x) = limh→0
Q(x + h)−Q(x)
h
= limh→0
1g(x+h) −
1g(x)
h
= limh→0
g(x)−g(x+h)g(x)g(x+h)
h
= limh→0
g(x)−g(x+h)h
g(x)g(x + h)
= limh→0
− g(x+h)−g(x)h
g(x)g(x + h)
=−g′(x)
(g(x))2
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 19 / 30
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The Quotient Rule
Let g(x) be a differentiable function, and let Q(x) = 1g(x) . Then:
Q′(x) = limh→0
Q(x + h)−Q(x)
h
= limh→0
1g(x+h) −
1g(x)
h
= limh→0
g(x)−g(x+h)g(x)g(x+h)
h
= limh→0
g(x)−g(x+h)h
g(x)g(x + h)
= limh→0
− g(x+h)−g(x)h
g(x)g(x + h)
=−g′(x)
(g(x))2
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 19 / 30
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The Quotient Rule
Let g(x) be a differentiable function. Then:
�
1
g(x)
�′
=−g′(x)
(g(x))2
Now, let f (x) be a diff. function, and consider f (x)g(x) = f (x) · 1
g(x) . Then:
�
f (x) ·1
g(x)
�′
=
f ′(x)1
g(x)+ f (x)
�
1
g(x)
�′
= f ′(x)1
g(x)+ f (x)
−g′(x)
(g(x))2
=f ′(x)g(x)− f (x)g′(x)
(g(x))2
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 20 / 30
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The Quotient Rule
Let g(x) be a differentiable function. Then:
�
1
g(x)
�′
=−g′(x)
(g(x))2
Now, let f (x) be a diff. function, and consider f (x)g(x) = f (x) · 1
g(x) .
Then:
�
f (x) ·1
g(x)
�′
=
f ′(x)1
g(x)+ f (x)
�
1
g(x)
�′
= f ′(x)1
g(x)+ f (x)
−g′(x)
(g(x))2
=f ′(x)g(x)− f (x)g′(x)
(g(x))2
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 20 / 30
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The Quotient Rule
Let g(x) be a differentiable function. Then:
�
1
g(x)
�′
=−g′(x)
(g(x))2
Now, let f (x) be a diff. function, and consider f (x)g(x) = f (x) · 1
g(x) . Then:
�
f (x) ·1
g(x)
�′
=
f ′(x)1
g(x)+ f (x)
�
1
g(x)
�′
= f ′(x)1
g(x)+ f (x)
−g′(x)
(g(x))2
=f ′(x)g(x)− f (x)g′(x)
(g(x))2
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 20 / 30
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The Quotient Rule
Let g(x) be a differentiable function. Then:
�
1
g(x)
�′
=−g′(x)
(g(x))2
Now, let f (x) be a diff. function, and consider f (x)g(x) = f (x) · 1
g(x) . Then:
�
f (x) ·1
g(x)
�′
= f ′(x)1
g(x)+ f (x)
�
1
g(x)
�′
= f ′(x)1
g(x)+ f (x)
−g′(x)
(g(x))2
=f ′(x)g(x)− f (x)g′(x)
(g(x))2
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 20 / 30
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The Quotient Rule
Let g(x) be a differentiable function. Then:
�
1
g(x)
�′
=−g′(x)
(g(x))2
Now, let f (x) be a diff. function, and consider f (x)g(x) = f (x) · 1
g(x) . Then:
�
f (x) ·1
g(x)
�′
= f ′(x)1
g(x)+ f (x)
�
1
g(x)
�′
= f ′(x)1
g(x)+ f (x)
−g′(x)
(g(x))2
=f ′(x)g(x)− f (x)g′(x)
(g(x))2
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 20 / 30
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The Quotient Rule
Let g(x) be a differentiable function. Then:
�
1
g(x)
�′
=−g′(x)
(g(x))2
Now, let f (x) be a diff. function, and consider f (x)g(x) = f (x) · 1
g(x) . Then:
�
f (x) ·1
g(x)
�′
= f ′(x)1
g(x)+ f (x)
�
1
g(x)
�′
= f ′(x)1
g(x)+ f (x)
−g′(x)
(g(x))2
=f ′(x)g(x)− f (x)g′(x)
(g(x))2
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 20 / 30
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The Quotient Rule
TheoremLet f (x) and g(x) be two differentiable functions. Then:
d
dx
�
f (x)
g(x)
�
=
dfdx · g(x)− f (x) · dg
dx
(g(x))2.
Example
Find the first derivative of f (x) =x2
ex.
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 21 / 30
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The Quotient Rule
TheoremLet f (x) and g(x) be two differentiable functions. Then:
d
dx
�
f (x)
g(x)
�
=
dfdx · g(x)− f (x) · dg
dx
(g(x))2.
Example
Find the first derivative of f (x) =x2
ex.
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 21 / 30
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The Quotient Rule
How to remember the formula?�
f (x)
g(x)
�′
=g(x)f ′(x)− f (x)g′(x)
g(x)2
The Quotient Rule Song
If the quotient rule you wish to know,it’s low d-high less high d-low.Draw the line and down belowdenominator squared will go!
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 22 / 30
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Example
Find the equation of the tangent line to the curve y =ex
1 + x2at x = 0.
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 23 / 30
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Example
Find the derivative of f (x) = x3 sin(x).
We need the derivative of sin(x) first!
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 24 / 30
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Example
Find the derivative of f (x) = x3 sin(x).
We need the derivative of sin(x) first!
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 24 / 30
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Derivatives of trigonometric functions
f (x) = sin(x)
−3 −2 −1 1 2 3 4 5 6 7
−1
1
0
f
f ′(x) = (sin(x))′
−3 −2 −1 1 2 3 4 5 6 7
−1
1
0
f
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Derivatives of trigonometric functions
f (x) = cos(x)
−3 −2 −1 1 2 3 4 5 6 7
−1
1
0
f
f ′(x) = (cos(x))′
−3 −2 −1 1 2 3 4 5 6 7
−1
1
0
f
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Derivatives of trigonometric functions
Theorem
(sin(x))′ = cos(x), and (cos(x))′ = − sin(x).
Example
Find the derivative of f (x) = x3 sin(x).
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 27 / 30
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Derivatives of trigonometric functions
Theorem
(sin(x))′ = cos(x), and (cos(x))′ = − sin(x).
Example
Find the derivative of f (x) = x3 sin(x).
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 27 / 30
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ExampleFind the derivative of f (x) = tan x .
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 28 / 30
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Two important limits
Recall:
limx→0
sin x
x= 1.
The proof is in the book.
Let us calculate: limx→0
cos x − 1
x
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 29 / 30
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Two important limits
Recall:
limx→0
sin x
x= 1.
The proof is in the book. Let us calculate: limx→0
cos x − 1
x
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 29 / 30
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This slide left intentionally blank
Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 30 / 30