lesson 10: the chain rule (section 21 slides)
DESCRIPTION
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.TRANSCRIPT
Section 2.5The Chain Rule
V63.0121.021, Calculus I
New York University
October 7, 2010
Announcements
I Quiz 2 in recitation next week (October 11-15)I Midterm in class Tuesday, october 19 on §§1.1–2.5
. . . . . .
. . . . . .
Announcements
I Quiz 2 in recitation nextweek (October 11-15)
I Midterm in class Tuesday,october 19 on §§1.1–2.5
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 2 / 36
. . . . . .
Objectives
I Given a compoundexpression, write it as acomposition of functions.
I Understand and apply theChain Rule for thederivative of a compositionof functions.
I Understand and useNewtonian and Leibniziannotations for the ChainRule.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 3 / 36
. . . . . .
CompositionsSee Section 1.2 for review
DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “dog 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.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 4 / 36
. . . . . .
CompositionsSee Section 1.2 for review
DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “dog 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.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 4 / 36
. . . . . .
CompositionsSee Section 1.2 for review
DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “dog 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.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 4 / 36
. . . . . .
CompositionsSee Section 1.2 for review
DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “dog 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.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 4 / 36
. . . . . .
CompositionsSee Section 1.2 for review
DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “dog 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.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 4 / 36
. . . . . .
CompositionsSee Section 1.2 for review
DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “dog 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.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 4 / 36
. . . . . .
Outline
HeuristicsAnalogyThe Linear Case
The chain rule
Examples
Related rates of change
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 5 / 36
. . . . . .
Analogy
Think about riding a bike. To gofaster you can either:
I pedal fasterI change gears
.
.Image credit: SpringSun
The angular position (φ) of the back wheel depends on the position ofthe front sprocket (θ):
φ(θ) =R..
.radius of front sprocket
θ
r..
.radius of back sprocket
And so the angular speed of the back wheel depends on the derivativeof this function and the speed of the front sprocket.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 6 / 36
. . . . . .
Analogy
Think about riding a bike. To gofaster you can either:
I pedal faster
I change gears
.
.Image credit: SpringSun
The angular position (φ) of the back wheel depends on the position ofthe front sprocket (θ):
φ(θ) =R..
.radius of front sprocket
θ
r..
.radius of back sprocket
And so the angular speed of the back wheel depends on the derivativeof this function and the speed of the front sprocket.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 6 / 36
. . . . . .
Analogy
Think about riding a bike. To gofaster you can either:
I pedal fasterI change gears
.
.Image credit: SpringSun
The angular position (φ) of the back wheel depends on the position ofthe front sprocket (θ):
φ(θ) =R..
.radius of front sprocket
θ
r..
.radius of back sprocket
And so the angular speed of the back wheel depends on the derivativeof this function and the speed of the front sprocket.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 6 / 36
. . . . . .
Analogy
Think about riding a bike. To gofaster you can either:
I pedal fasterI change gears
.
.Image credit: SpringSun
The angular position (φ) of the back wheel depends on the position ofthe front sprocket (θ):
φ(θ) =R..
.radius of front sprocket
θ
r..
.radius of back sprocket
And so the angular speed of the back wheel depends on the derivativeof this function and the speed of the front sprocket.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 6 / 36
. . . . . .
Analogy
Think about riding a bike. To gofaster you can either:
I pedal fasterI change gears
.
.Image credit: SpringSun
The angular position (φ) of the back wheel depends on the position ofthe front sprocket (θ):
φ(θ) =R..
.radius of front sprocket
θ
r..
.radius of back sprocket
And so the angular speed of the back wheel depends on the derivativeof this function and the speed of the front sprocket.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 6 / 36
. . . . . .
Analogy
Think about riding a bike. To gofaster you can either:
I pedal fasterI change gears
.
.Image credit: SpringSun
The angular position (φ) of the back wheel depends on the position ofthe front sprocket (θ):
φ(θ) =R..
.radius of front sprocket
θ
r..
.radius of back sprocket
And so the angular speed of the back wheel depends on the derivativeof this function and the speed of the front sprocket.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 6 / 36
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 7 / 36
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 7 / 36
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 7 / 36
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 7 / 36
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 7 / 36
. . . . . .
The Nonlinear Case
Let u = g(x) and y = f(u). Suppose x is changed by a small amount∆x. Then
∆y ≈ f′(y)∆u
and∆u ≈ g′(u)∆x.
So∆y ≈ f′(y)g′(u)∆x =⇒ ∆y
∆x≈ f′(y)g′(u)
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 8 / 36
. . . . . .
Outline
HeuristicsAnalogyThe Linear Case
The chain rule
Examples
Related rates of change
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 9 / 36
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 10 / 36
. . . . . .
Observations
I Succinctly, the derivative of acomposition is the product ofthe derivatives
I The only complication is wherethese derivatives areevaluated: at the same pointthe functions are
I In Leibniz notation, the ChainRule looks like cancellation of(fake) fractions
.
.Image credit: ooOJasonOooV63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 11 / 36
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 12 / 36
. . . . . .
Observations
I Succinctly, the derivative of acomposition is the product ofthe derivatives
I The only complication is wherethese derivatives areevaluated: at the same pointthe functions are
I In Leibniz notation, the ChainRule looks like cancellation of(fake) fractions
.
.Image credit: ooOJasonOooV63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 13 / 36
. . . . . .
CompositionsSee Section 1.2 for review
DefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “dog 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.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 14 / 36
. . . . . .
Observations
I Succinctly, the derivative of acomposition is the product ofthe derivatives
I The only complication is wherethese derivatives areevaluated: at the same pointthe functions are
I In Leibniz notation, the ChainRule looks like cancellation of(fake) fractions .
.Image credit: ooOJasonOooV63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 15 / 36
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 16 / 36
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 16 / 36
. . . . . .
Outline
HeuristicsAnalogyThe Linear Case
The chain rule
Examples
Related rates of change
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 17 / 36
. . . . . .
Example
Example
let 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(3x
2 + 1)−1/2(6x) =3x√
3x2 + 1
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 18 / 36
. . . . . .
Example
Example
let 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(3x
2 + 1)−1/2(6x) =3x√
3x2 + 1
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 18 / 36
. . . . . .
Example
Example
let 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) = 12(3x
2 + 1)−1/2(6x) =3x√
3x2 + 1
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 18 / 36
. . . . . .
Example
Example
let 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(3x
2 + 1)−1/2(6x) =3x√
3x2 + 1
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 18 / 36
. . . . . .
Example
Example
let 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(3x
2 + 1)−1/2(6x) =3x√
3x2 + 1
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 18 / 36
. . . . . .
Corollary
Corollary (The Power Rule Combined with the Chain Rule)
If n is any real number and u = g(x) is differentiable, then
ddx
(un) = nun−1dudx
.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 19 / 36
. . . . . .
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 · cos x
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 20 / 36
. . . . . .
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 · cos x
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 20 / 36
. . . . . .
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 · cos x
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 20 / 36
. . . . . .
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 · cos x
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 20 / 36
. . . . . .
Example
Let f(x) =(
3√
x5 − 2+ 8)2
. Find f′(x).
Solution
ddx
(3√
x5 − 2+ 8)2
= 2(
3√
x5 − 2+ 8) ddx
(3√
x5 − 2+ 8)
= 2(
3√
x5 − 2+ 8) ddx
3√
x5 − 2
= 2(
3√
x5 − 2+ 8)
13(x
5 − 2)−2/3 ddx
(x5 − 2)
= 2(
3√
x5 − 2+ 8)
13(x
5 − 2)−2/3(5x4)
=103x4
(3√
x5 − 2+ 8)(x5 − 2)−2/3
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 21 / 36
. . . . . .
Example
Let f(x) =(
3√
x5 − 2+ 8)2
. Find f′(x).
Solution
ddx
(3√
x5 − 2+ 8)2
= 2(
3√
x5 − 2+ 8) ddx
(3√
x5 − 2+ 8)
= 2(
3√
x5 − 2+ 8) ddx
3√
x5 − 2
= 2(
3√
x5 − 2+ 8)
13(x
5 − 2)−2/3 ddx
(x5 − 2)
= 2(
3√
x5 − 2+ 8)
13(x
5 − 2)−2/3(5x4)
=103x4
(3√
x5 − 2+ 8)(x5 − 2)−2/3
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 21 / 36
. . . . . .
Example
Let f(x) =(
3√
x5 − 2+ 8)2
. Find f′(x).
Solution
ddx
(3√
x5 − 2+ 8)2
= 2(
3√
x5 − 2+ 8) ddx
(3√
x5 − 2+ 8)
= 2(
3√
x5 − 2+ 8) ddx
3√
x5 − 2
= 2(
3√
x5 − 2+ 8)
13(x
5 − 2)−2/3 ddx
(x5 − 2)
= 2(
3√
x5 − 2+ 8)
13(x
5 − 2)−2/3(5x4)
=103x4
(3√
x5 − 2+ 8)(x5 − 2)−2/3
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 21 / 36
. . . . . .
Example
Let f(x) =(
3√
x5 − 2+ 8)2
. Find f′(x).
Solution
ddx
(3√
x5 − 2+ 8)2
= 2(
3√
x5 − 2+ 8) ddx
(3√
x5 − 2+ 8)
= 2(
3√
x5 − 2+ 8) ddx
3√
x5 − 2
= 2(
3√
x5 − 2+ 8)
13(x
5 − 2)−2/3 ddx
(x5 − 2)
= 2(
3√
x5 − 2+ 8)
13(x
5 − 2)−2/3(5x4)
=103x4
(3√
x5 − 2+ 8)(x5 − 2)−2/3
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 21 / 36
. . . . . .
Example
Let f(x) =(
3√
x5 − 2+ 8)2
. Find f′(x).
Solution
ddx
(3√
x5 − 2+ 8)2
= 2(
3√
x5 − 2+ 8) ddx
(3√
x5 − 2+ 8)
= 2(
3√
x5 − 2+ 8) ddx
3√
x5 − 2
= 2(
3√
x5 − 2+ 8)
13(x
5 − 2)−2/3 ddx
(x5 − 2)
= 2(
3√
x5 − 2+ 8)
13(x
5 − 2)−2/3(5x4)
=103x4
(3√
x5 − 2+ 8)(x5 − 2)−2/3
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 21 / 36
. . . . . .
Example
Let f(x) =(
3√
x5 − 2+ 8)2
. Find f′(x).
Solution
ddx
(3√
x5 − 2+ 8)2
= 2(
3√
x5 − 2+ 8) ddx
(3√
x5 − 2+ 8)
= 2(
3√
x5 − 2+ 8) ddx
3√
x5 − 2
= 2(
3√
x5 − 2+ 8)
13(x
5 − 2)−2/3 ddx
(x5 − 2)
= 2(
3√
x5 − 2+ 8)
13(x
5 − 2)−2/3(5x4)
=103x4
(3√
x5 − 2+ 8)(x5 − 2)−2/3
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 21 / 36
. . . . . .
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(x
5 − 2)−2/3(5x4)
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 22 / 36
. . . . . .
Combining techniques
Example
Findddx
((x3 + 1)10 sin(4x2 − 7)
)
SolutionThe “last” part of the function is the product, so we apply the productrule. 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)
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 23 / 36
. . . . . .
Combining techniques
Example
Findddx
((x3 + 1)10 sin(4x2 − 7)
)SolutionThe “last” part of the function is the product, so we apply the productrule. 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)
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 23 / 36
. . . . . .
Combining techniques
Example
Findddx
((x3 + 1)10 sin(4x2 − 7)
)SolutionThe “last” part of the function is the product, so we apply the productrule. 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)
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 23 / 36
. . . . . .
Combining techniques
Example
Findddx
((x3 + 1)10 sin(4x2 − 7)
)SolutionThe “last” part of the function is the product, so we apply the productrule. 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)
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 23 / 36
. . . . . .
Your Turn
Find derivatives of these functions:1. y = (1− x2)10
2. y =√sin x
3. y = sin√x
4. y = (2x− 5)4(8x2 − 5)−3
5. F(z) =√
z− 1z+ 1
6. y = tan(cos x)7. y = csc2(sin θ)8. y = sin(sin(sin(sin(sin(sin(x))))))
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 24 / 36
. . . . . .
Solution to #1
Example
Find the derivative of y = (1− x2)10.
Solutiony′ = 10(1− x2)9(−2x) = −20x(1− x2)9
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 25 / 36
. . . . . .
Solution to #2
Example
Find the derivative of y =√sin x.
SolutionWriting
√sin x as (sin x)1/2, we have
y′ = 12 (sin x)−1/2 (cos x) =
cos x2√sin x
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 26 / 36
. . . . . .
Solution to #3
Example
Find the derivative of y = sin√x.
Solution
y′ =ddx
sin(x1/2) = cos(x1/2)12x−1/2 =
cos(√
x)
2√x
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 27 / 36
. . . . . .
Solution to #4
Example
Find the derivative of y = (2x− 5)4(8x2 − 5)−3
SolutionWe need to use the product rule and the chain rule:
y′ = 4(2x− 5)3(2)(8x2 − 5)−3 + (2x− 5)4(−3)(8x2 − 5)−4(16x)
The rest is a bit of algebra, useful if you wanted to solve the equationy′ = 0:
y′ = 8(2x− 5)3(8x2 − 5)−4[(8x2 − 5)− 6x(2x− 5)
]= 8(2x− 5)3(8x2 − 5)−4
(−4x2 + 30x− 5
)= −8(2x− 5)3(8x2 − 5)−4
(4x2 − 30x+ 5
)V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 28 / 36
. . . . . .
Solution to #5
Example
Find the derivative of F(z) =√
z− 1z+ 1
.
Solution
y′ =12
(z− 1z+ 1
)−1/2((z+ 1)(1)− (z− 1)(1)(z+ 1)2
)=
12
(z+ 1z− 1
)1/2( 2(z+ 1)2
)=
1(z+ 1)3/2(z− 1)1/2
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 29 / 36
. . . . . .
Solution to #6
Example
Find the derivative of y = tan(cos x).
Solutiony′ = sec2(cos x) · (− sin x) = − sec2(cos x) sin x
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 30 / 36
. . . . . .
Solution to #7
Example
Find the derivative of y = csc2(sin θ).
SolutionRemember the notation:
y = csc2(sin θ) = [csc(sin θ)]2
So
y′ = 2 csc(sin θ) · [− csc(sin θ) cot(sin θ)] · cos(θ)= −2 csc2(sin θ) cot(sin θ) cos θ
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 31 / 36
. . . . . .
Solution to #8
Example
Find the derivative of y = sin(sin(sin(sin(sin(sin(x)))))).
SolutionRelax! It’s just a bunch of chain rules. All of these lines are multipliedtogether.
y′ = cos(sin(sin(sin(sin(sin(x))))))· cos(sin(sin(sin(sin(x)))))
· cos(sin(sin(sin(x))))· cos(sin(sin(x)))
· cos(sin(x))· cos(x))
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 32 / 36
. . . . . .
Outline
HeuristicsAnalogyThe Linear Case
The chain rule
Examples
Related rates of change
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 33 / 36
. . . . . .
Related rates of change at the Deli
QuestionSuppose a deli clerk can slice a stick of pepperoni (assume thetapered ends have been removed) by hand at the rate of 2 inches perminute, while a machine can slice pepperoni at the rate of 10 inches
per minute. ThendVdt
for the machine is 5 times greater thandVdt
forthe deli clerk. This is explained by theA. chain ruleB. product ruleC. quotient RuleD. addition rule
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 34 / 36
. . . . . .
Related rates of change at the Deli
QuestionSuppose a deli clerk can slice a stick of pepperoni (assume thetapered ends have been removed) by hand at the rate of 2 inches perminute, while a machine can slice pepperoni at the rate of 10 inches
per minute. ThendVdt
for the machine is 5 times greater thandVdt
forthe deli clerk. This is explained by theA. chain ruleB. product ruleC. quotient RuleD. addition rule
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 34 / 36
. . . . . .
Related rates of change in the ocean
QuestionThe area of a circle, A = πr2,changes as its radius changes.If the 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 FrazierV63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 35 / 36
. . . . . .
Related rates of change in the ocean
QuestionThe area of a circle, A = πr2,changes as its radius changes.If the 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 FrazierV63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 35 / 36
. . . . . .
Summary
I The derivative of acomposition is the productof derivatives
I In symbols:(f ◦ g)′(x) = f′(g(x))g′(x)
I Calculus is like an onion,and not because it makesyou cry!
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 36 / 36