Lesson 11: The Chain Rule
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Transcript of Lesson 11: The Chain Rule
![Page 1: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/1.jpg)
. . . . . .
Section 2.5The Chain Rule
V63.0121, Calculus I
February 19, 2009
Announcements
I Midterm is March 4/5 (75 min., in class, covers 1.1–2.4)I ALEKS is due February 27, 11:59pm
![Page 2: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/2.jpg)
. . . . . .
Outline
Compositions
HeuristicsAnalogyThe Linear Case
The chain rule
Examples
Related rates of change
![Page 3: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/3.jpg)
. . . . . .
CompositionsSee Section 1.2 for review
DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”
.
.g .f.x .g(x) .f(g(x))
.f ◦ g
Our goal for the day is to understand how the derivative of thecomposition of two functions depends on the derivatives of theindividual functions.
![Page 4: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/4.jpg)
. . . . . .
CompositionsSee Section 1.2 for review
DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”
..g
.f
.x .g(x)
.f(g(x)).f ◦ g
Our goal for the day is to understand how the derivative of thecomposition of two functions depends on the derivatives of theindividual functions.
![Page 5: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/5.jpg)
. . . . . .
CompositionsSee Section 1.2 for review
DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”
..g .f.x .g(x)
.f(g(x)).f ◦ g
Our goal for the day is to understand how the derivative of thecomposition of two functions depends on the derivatives of theindividual functions.
![Page 6: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/6.jpg)
. . . . . .
CompositionsSee Section 1.2 for review
DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”
..g .f.x .g(x) .f(g(x))
.f ◦ g
Our goal for the day is to understand how the derivative of thecomposition of two functions depends on the derivatives of theindividual functions.
![Page 7: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/7.jpg)
. . . . . .
CompositionsSee Section 1.2 for review
DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”
..g .f.x .g(x) .f(g(x))
.f ◦ g
Our goal for the day is to understand how the derivative of thecomposition of two functions depends on the derivatives of theindividual functions.
![Page 8: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/8.jpg)
. . . . . .
CompositionsSee Section 1.2 for review
DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”
..g .f.x .g(x) .f(g(x))
.f ◦ g
Our goal for the day is to understand how the derivative of thecomposition of two functions depends on the derivatives of theindividual functions.
![Page 9: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/9.jpg)
. . . . . .
Outline
Compositions
HeuristicsAnalogyThe Linear Case
The chain rule
Examples
Related rates of change
![Page 10: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/10.jpg)
. . . . . .
Analogy
Think about riding a bike. Togo faster you can either:
I pedal fasterI change gears
.
.Image credit: SpringSun
The angular position (φ) of the back wheel depends on the positionof the front sprocket (θ):
φ(θ) =Rθ
r
And so the angular speed of the back wheel depends on thederivative of this function and the speed of the front wheel.
![Page 11: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/11.jpg)
. . . . . .
Analogy
Think about riding a bike. Togo faster you can either:
I pedal faster
I change gears
.
.Image credit: SpringSun
The angular position (φ) of the back wheel depends on the positionof the front sprocket (θ):
φ(θ) =Rθ
r
And so the angular speed of the back wheel depends on thederivative of this function and the speed of the front wheel.
![Page 12: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/12.jpg)
. . . . . .
Analogy
Think about riding a bike. Togo faster you can either:
I pedal fasterI change gears
.
.Image credit: SpringSun
The angular position (φ) of the back wheel depends on the positionof the front sprocket (θ):
φ(θ) =Rθ
r
And so the angular speed of the back wheel depends on thederivative of this function and the speed of the front wheel.
![Page 13: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/13.jpg)
. . . . . .
Analogy
Think about riding a bike. Togo faster you can either:
I pedal fasterI change gears
.
.Image credit: SpringSun
The angular position (φ) of the back wheel depends on the positionof the front sprocket (θ):
φ(θ) =Rθ
r
And so the angular speed of the back wheel depends on thederivative of this function and the speed of the front wheel.
![Page 14: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/14.jpg)
. . . . . .
The Linear Case
QuestionLet f(x) = mx + b and g(x) = m′x + b′. What can you say about thecomposition?
Answer
I f(g(x)) = m(m′x + b′) + b = (mm′)x + (mb′ + b)I The composition is also linearI The slope of the composition is the product of the slopes of the
two functions.
The derivative is supposed to be a local linearization of a function. Sothere should be an analog of this property in derivatives.
![Page 15: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/15.jpg)
. . . . . .
The Linear Case
QuestionLet f(x) = mx + b and g(x) = m′x + b′. What can you say about thecomposition?
Answer
I f(g(x)) = m(m′x + b′) + b = (mm′)x + (mb′ + b)
I The composition is also linearI The slope of the composition is the product of the slopes of the
two functions.
The derivative is supposed to be a local linearization of a function. Sothere should be an analog of this property in derivatives.
![Page 16: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/16.jpg)
. . . . . .
The Linear Case
QuestionLet f(x) = mx + b and g(x) = m′x + b′. What can you say about thecomposition?
Answer
I f(g(x)) = m(m′x + b′) + b = (mm′)x + (mb′ + b)I The composition is also linear
I The slope of the composition is the product of the slopes of thetwo functions.
The derivative is supposed to be a local linearization of a function. Sothere should be an analog of this property in derivatives.
![Page 17: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/17.jpg)
. . . . . .
The Linear Case
QuestionLet f(x) = mx + b and g(x) = m′x + b′. What can you say about thecomposition?
Answer
I f(g(x)) = m(m′x + b′) + b = (mm′)x + (mb′ + b)I The composition is also linearI The slope of the composition is the product of the slopes of the
two functions.
The derivative is supposed to be a local linearization of a function. Sothere should be an analog of this property in derivatives.
![Page 18: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/18.jpg)
. . . . . .
The Linear Case
QuestionLet f(x) = mx + b and g(x) = m′x + b′. What can you say about thecomposition?
Answer
I f(g(x)) = m(m′x + b′) + b = (mm′)x + (mb′ + b)I The composition is also linearI The slope of the composition is the product of the slopes of the
two functions.
The derivative is supposed to be a local linearization of a function. Sothere should be an analog of this property in derivatives.
![Page 19: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/19.jpg)
. . . . . .
The Linear Case
QuestionLet f(x) = mx + b and g(x) = m′x + b′. What can you say about thecomposition?
Answer
I f(g(x)) = m(m′x + b′) + b = (mm′)x + (mb′ + b)I The composition is also linearI The slope of the composition is the product of the slopes of the
two functions.
The derivative is supposed to be a local linearization of a function. Sothere should be an analog of this property in derivatives.
![Page 20: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/20.jpg)
. . . . . .
Outline
Compositions
HeuristicsAnalogyThe Linear Case
The chain rule
Examples
Related rates of change
![Page 21: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/21.jpg)
. . . . . .
Theorem of the day: The chain rule
TheoremLet f and g be functions, with g differentiable at x and f differentiable atg(x). Then f ◦ g is differentiable at x and
(f ◦ g)′(x) = f′(g(x))g′(x)
In Leibnizian notation, let y = f(u) and u = g(x). Then
dydx
=dydu
dudx
..dy
��du��dudx
![Page 22: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/22.jpg)
. . . . . .
Observations
I Succinctly, the derivative of acomposition is the product ofthe derivatives
I The only complication iswhere these derivatives areevaluated: at the same pointthe functions are
I In Leibniz notation, the ChainRule looks like cancellation of(fake) fractions
.
.Image credit: ooOJasonOoo
![Page 23: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/23.jpg)
. . . . . .
Theorem of the day: The chain rule
TheoremLet f and g be functions, with g differentiable at x and f differentiable atg(x). Then f ◦ g is differentiable at x and
(f ◦ g)′(x) = f′(g(x))g′(x)
In Leibnizian notation, let y = f(u) and u = g(x). Then
dydx
=dydu
dudx
..dy
��du��dudx
![Page 24: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/24.jpg)
. . . . . .
Observations
I Succinctly, the derivative of acomposition is the product ofthe derivatives
I The only complication iswhere these derivatives areevaluated: at the same pointthe functions are
I In Leibniz notation, the ChainRule looks like cancellation of(fake) fractions
.
.Image credit: ooOJasonOoo
![Page 25: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/25.jpg)
. . . . . .
CompositionsSee Section 1.2 for review
DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”
..g .f.x .g(x) .f(g(x))
.f ◦ g
Our goal for the day is to understand how the derivative of thecomposition of two functions depends on the derivatives of theindividual functions.
![Page 26: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/26.jpg)
. . . . . .
Observations
I Succinctly, the derivative of acomposition is the product ofthe derivatives
I The only complication iswhere these derivatives areevaluated: at the same pointthe functions are
I In Leibniz notation, the ChainRule looks like cancellation of(fake) fractions .
.Image credit: ooOJasonOoo
![Page 27: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/27.jpg)
. . . . . .
Theorem of the day: The chain rule
TheoremLet f and g be functions, with g differentiable at x and f differentiable atg(x). Then f ◦ g is differentiable at x and
(f ◦ g)′(x) = f′(g(x))g′(x)
In Leibnizian notation, let y = f(u) and u = g(x). Then
dydx
=dydu
dudx
..dy
��du��dudx
![Page 28: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/28.jpg)
. . . . . .
Theorem of the day: The chain rule
TheoremLet f and g be functions, with g differentiable at x and f differentiable atg(x). Then f ◦ g is differentiable at x and
(f ◦ g)′(x) = f′(g(x))g′(x)
In Leibnizian notation, let y = f(u) and u = g(x). Then
dydx
=dydu
dudx
..dy
��du��dudx
![Page 29: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/29.jpg)
. . . . . .
Outline
Compositions
HeuristicsAnalogyThe Linear Case
The chain rule
Examples
Related rates of change
![Page 30: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/30.jpg)
. . . . . .
Example
Examplelet h(x) =
√3x2 + 1. Find h′(x).
SolutionFirst, write h as f ◦ g. Let f(u) =
√u and g(x) = 3x2 + 1. Then
f′(u) = 12u−1/2, and g′(x) = 6x. So
h′(x) = 12u−1/2(6x) = 1
2(3x2 + 1)−1/2(6x) =3x√
3x2 + 1
![Page 31: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/31.jpg)
. . . . . .
Example
Examplelet h(x) =
√3x2 + 1. Find h′(x).
SolutionFirst, write h as f ◦ g.
Let f(u) =√
u and g(x) = 3x2 + 1. Thenf′(u) = 1
2u−1/2, and g′(x) = 6x. So
h′(x) = 12u−1/2(6x) = 1
2(3x2 + 1)−1/2(6x) =3x√
3x2 + 1
![Page 32: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/32.jpg)
. . . . . .
Example
Examplelet h(x) =
√3x2 + 1. Find h′(x).
SolutionFirst, write h as f ◦ g. Let f(u) =
√u and g(x) = 3x2 + 1.
Thenf′(u) = 1
2u−1/2, and g′(x) = 6x. So
h′(x) = 12u−1/2(6x) = 1
2(3x2 + 1)−1/2(6x) =3x√
3x2 + 1
![Page 33: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/33.jpg)
. . . . . .
Example
Examplelet h(x) =
√3x2 + 1. Find h′(x).
SolutionFirst, write h as f ◦ g. Let f(u) =
√u and g(x) = 3x2 + 1. Then
f′(u) = 12u−1/2, and g′(x) = 6x. So
h′(x) = 12u−1/2(6x)
= 12(3x2 + 1)−1/2(6x) =
3x√3x2 + 1
![Page 34: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/34.jpg)
. . . . . .
Example
Examplelet h(x) =
√3x2 + 1. Find h′(x).
SolutionFirst, write h as f ◦ g. Let f(u) =
√u and g(x) = 3x2 + 1. Then
f′(u) = 12u−1/2, and g′(x) = 6x. So
h′(x) = 12u−1/2(6x) = 1
2(3x2 + 1)−1/2(6x) =3x√
3x2 + 1
![Page 35: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/35.jpg)
. . . . . .
Does order matter?
Example
Findddx
(sin 4x) and compare it toddx
(4 sin x).
Solution
I For the first, let u = 4x and y = sin(u). Then
dydx
=dydu
· dudx
= cos(u) · 4 = 4 cos 4x.
I For the second, let u = sin x and y = 4u. Then
dydx
=dydu
· dudx
= 4 · sin x
![Page 36: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/36.jpg)
. . . . . .
Does order matter?
Example
Findddx
(sin 4x) and compare it toddx
(4 sin x).
Solution
I For the first, let u = 4x and y = sin(u). Then
dydx
=dydu
· dudx
= cos(u) · 4 = 4 cos 4x.
I For the second, let u = sin x and y = 4u. Then
dydx
=dydu
· dudx
= 4 · sin x
![Page 37: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/37.jpg)
. . . . . .
Does order matter?
Example
Findddx
(sin 4x) and compare it toddx
(4 sin x).
Solution
I For the first, let u = 4x and y = sin(u). Then
dydx
=dydu
· dudx
= cos(u) · 4 = 4 cos 4x.
I For the second, let u = sin x and y = 4u. Then
dydx
=dydu
· dudx
= 4 · sin x
![Page 38: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/38.jpg)
. . . . . .
Order matters!
Example
Findddx
(sin 4x) and compare it toddx
(4 sin x).
Solution
I For the first, let u = 4x and y = sin(u). Then
dydx
=dydu
· dudx
= cos(u) · 4 = 4 cos 4x.
I For the second, let u = sin x and y = 4u. Then
dydx
=dydu
· dudx
= 4 · sin x
![Page 39: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/39.jpg)
. . . . . .
Example
Let f(x) =(
3√
x5 − 2 + 8)2
. Find f′(x).
Solution
ddx
(3√
x5 − 2 + 8)2
= 2(
3√
x5 − 2 + 8) d
dx
(3√
x5 − 2 + 8)
= 2(
3√
x5 − 2 + 8) d
dx3√
x5 − 2
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3 d
dx(x5 − 2)
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3(5x4)
=103
x4(
3√
x5 − 2 + 8)
(x5 − 2)−2/3
![Page 40: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/40.jpg)
. . . . . .
Example
Let f(x) =(
3√
x5 − 2 + 8)2
. Find f′(x).
Solution
ddx
(3√
x5 − 2 + 8)2
= 2(
3√
x5 − 2 + 8) d
dx
(3√
x5 − 2 + 8)
= 2(
3√
x5 − 2 + 8) d
dx3√
x5 − 2
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3 d
dx(x5 − 2)
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3(5x4)
=103
x4(
3√
x5 − 2 + 8)
(x5 − 2)−2/3
![Page 41: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/41.jpg)
. . . . . .
Example
Let f(x) =(
3√
x5 − 2 + 8)2
. Find f′(x).
Solution
ddx
(3√
x5 − 2 + 8)2
= 2(
3√
x5 − 2 + 8) d
dx
(3√
x5 − 2 + 8)
= 2(
3√
x5 − 2 + 8) d
dx3√
x5 − 2
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3 d
dx(x5 − 2)
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3(5x4)
=103
x4(
3√
x5 − 2 + 8)
(x5 − 2)−2/3
![Page 42: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/42.jpg)
. . . . . .
Example
Let f(x) =(
3√
x5 − 2 + 8)2
. Find f′(x).
Solution
ddx
(3√
x5 − 2 + 8)2
= 2(
3√
x5 − 2 + 8) d
dx
(3√
x5 − 2 + 8)
= 2(
3√
x5 − 2 + 8) d
dx3√
x5 − 2
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3 d
dx(x5 − 2)
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3(5x4)
=103
x4(
3√
x5 − 2 + 8)
(x5 − 2)−2/3
![Page 43: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/43.jpg)
. . . . . .
Example
Let f(x) =(
3√
x5 − 2 + 8)2
. Find f′(x).
Solution
ddx
(3√
x5 − 2 + 8)2
= 2(
3√
x5 − 2 + 8) d
dx
(3√
x5 − 2 + 8)
= 2(
3√
x5 − 2 + 8) d
dx3√
x5 − 2
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3 d
dx(x5 − 2)
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3(5x4)
=103
x4(
3√
x5 − 2 + 8)
(x5 − 2)−2/3
![Page 44: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/44.jpg)
. . . . . .
Example
Let f(x) =(
3√
x5 − 2 + 8)2
. Find f′(x).
Solution
ddx
(3√
x5 − 2 + 8)2
= 2(
3√
x5 − 2 + 8) d
dx
(3√
x5 − 2 + 8)
= 2(
3√
x5 − 2 + 8) d
dx3√
x5 − 2
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3 d
dx(x5 − 2)
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3(5x4)
=103
x4(
3√
x5 − 2 + 8)
(x5 − 2)−2/3
![Page 45: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/45.jpg)
. . . . . .
A metaphor
Think about peeling an onion:
f(x) =
(3√
x5︸︷︷︸�5
−2
︸ ︷︷ ︸3√�
+8
︸ ︷︷ ︸�+8
)2
︸ ︷︷ ︸�2
.
.Image credit: photobunny
f′(x) = 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3(5x4)
![Page 46: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/46.jpg)
. . . . . .
Combining techniques
Example
Findddx
((x3 + 1)10 sin(4x2 − 7)
)
SolutionThe “last” part of the function is the product, so we apply the product rule.Each factor’s derivative requires the chain rule:
ddx
((x3 + 1)10 · sin(4x2 − 7)
)=
(ddx
(x3 + 1)10)· sin(4x2 − 7) + (x3 + 1)10 ·
(ddx
sin(4x2 − 7)
)= 10(x3 + 1)9(3x2) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x)
![Page 47: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/47.jpg)
. . . . . .
Combining techniques
Example
Findddx
((x3 + 1)10 sin(4x2 − 7)
)SolutionThe “last” part of the function is the product, so we apply the product rule.Each factor’s derivative requires the chain rule:
ddx
((x3 + 1)10 · sin(4x2 − 7)
)=
(ddx
(x3 + 1)10)· sin(4x2 − 7) + (x3 + 1)10 ·
(ddx
sin(4x2 − 7)
)= 10(x3 + 1)9(3x2) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x)
![Page 48: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/48.jpg)
. . . . . .
Combining techniques
Example
Findddx
((x3 + 1)10 sin(4x2 − 7)
)SolutionThe “last” part of the function is the product, so we apply the product rule.Each factor’s derivative requires the chain rule:
ddx
((x3 + 1)10 · sin(4x2 − 7)
)=
(ddx
(x3 + 1)10)· sin(4x2 − 7) + (x3 + 1)10 ·
(ddx
sin(4x2 − 7)
)
= 10(x3 + 1)9(3x2) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x)
![Page 49: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/49.jpg)
. . . . . .
Combining techniques
Example
Findddx
((x3 + 1)10 sin(4x2 − 7)
)SolutionThe “last” part of the function is the product, so we apply the product rule.Each factor’s derivative requires the chain rule:
ddx
((x3 + 1)10 · sin(4x2 − 7)
)=
(ddx
(x3 + 1)10)· sin(4x2 − 7) + (x3 + 1)10 ·
(ddx
sin(4x2 − 7)
)= 10(x3 + 1)9(3x2) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x)
![Page 50: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/50.jpg)
. . . . . .
Outline
Compositions
HeuristicsAnalogyThe Linear Case
The chain rule
Examples
Related rates of change
![Page 51: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/51.jpg)
. . . . . .
Related rates of change
QuestionThe area of a circle, A = πr2,changes as its radius changes. Ifthe radius changes withrespect to time, the change inarea with respect to time is
A.dAdr
= 2πr
B.dAdt
= 2πr +drdt
C.dAdt
= 2πrdrdt
D. not enough information
.
.Image credit: Jim Frazier
![Page 52: Lesson 11: The Chain Rule](https://reader034.fdocuments.net/reader034/viewer/2022042814/5559e5ecd8b42a34498b4eac/html5/thumbnails/52.jpg)
. . . . . .
Related rates of change
QuestionThe area of a circle, A = πr2,changes as its radius changes. Ifthe radius changes withrespect to time, the change inarea with respect to time is
A.dAdr
= 2πr
B.dAdt
= 2πr +drdt
C.dAdt
= 2πrdrdt
D. not enough information
.
.Image credit: Jim Frazier