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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.


. . . . . .

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.


. . . . . .



..g

.f

.x .g(x)

.f(g(x)).f ◦ g



. . . . . .



..g .f.x .g(x)

.f(g(x)).f ◦ g



. . . . . .



..g .f.x .g(x) .f(g(x))

.f ◦ g



. . . . . .

Outline

HeuristicsAnalogyThe Linear Case

The chain rule

Examples

Related rates of change


. . . . . .

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.


http://flickr.com/photos/63543004@N00/208409989/

http://flickr.com/photos/63543004@N00/

. . . . . .

Analogy


I pedal faster

I change gears

.



φ(θ) =R..


θ

r..






. . . . . .

Analogy


I pedal fasterI change gears

.



φ(θ) =R..


θ

r..






. . . . . .

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.


. . . . . .

The Linear Case


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 Linear Case


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 Linear Case


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 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)


. . . . . .

Outline


The chain rule

Examples



. . . . . .

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


. . . . . .

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

http://flickr.com/photos/restlessglobetrotter/

. . . . . .



(f ◦ g)′(x) = f′(g(x))g′(x)


dydx

=dydu

dudx

..dy��du

��dudx


. . . . . .

Observations



I In Leibniz notation, the ChainRule looks like cancellation of(fake) fractions

.



. . . . . .



..g .f.x .g(x) .f(g(x))

.f ◦ g



. . . . . .

Observations



I In Leibniz notation, the ChainRule looks like cancellation of(fake) fractions .



. . . . . .



(f ◦ g)′(x) = f′(g(x))g′(x)


dydx

=dydu

dudx

..dy��du

��dudx


. . . . . .

Outline


The chain rule

Examples



. . . . . .

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


. . . . . .

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


. . . . . .

Example

Example

let h(x) =√3x2 + 1. Find h′(x).


√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


. . . . . .

Example

Example

let h(x) =√3x2 + 1. Find h′(x).



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


. . . . . .

Example

Example

let h(x) =√3x2 + 1. Find h′(x).



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


. . . . . .

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

.


. . . . . .

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


. . . . . .

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


. . . . . .

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


. . . . . .

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)




. . . . . .

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)


. . . . . .

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)


. . . . . .

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))))))


. . . . . .

Solution to #1

Example

Find the derivative of y = (1− x2)10.

Solutiony′ = 10(1− x2)9(−2x) = −20x(1− x2)9


. . . . . .

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


. . . . . .

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


. . . . . .

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


. . . . . .

Solution to #6

Example

Find the derivative of y = tan(cos x).

Solutiony′ = sec2(cos x) · (− sin x) = − sec2(cos x) sin x


. . . . . .

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


. . . . . .

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))


. . . . . .

Outline


The chain rule

Examples



. . . . . .

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


. . . . . .

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

http://flickr.com/photos/jimfrazier/

. . . . . .

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!
