Lesson 10: The Chain Rule (Section 21 handout)

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 the derivativeof a composition offunctions.

I Understand and useNewtonian and Leibniziannotations for the Chain Rule.


Notes

Notes

Notes

1

Section 2.5 : The Chain RuleV63.0121.021, Calculus I October 7, 2010

CompositionsSee Section 1.2 for review

Definition

If f and g are functions, the composition (f ◦ g)(x) = f (g(x)) means “dog first, then f .”

g fx g(x) f (g(x))

f ◦ g

Our goal for the day is to understand how the derivative of the compositionof two functions depends on the derivatives of the individual functions.


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 faster

I change gears

Image credit: SpringSun

The angular position (ϕ) of the back wheel depends on the position of thefront 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 derivative ofthis function and the speed of the front sprocket.


Notes

Notes

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2


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

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

The Linear Case

Question

Let 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 the twofunctions.

The derivative is supposed to be a local linearization of a function. Sothere should be an analog of this property in derivatives.


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



Notes

Notes

Notes

3


Theorem of the day: The chain rule

Theorem

Let 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

dy

dx=

dy

du

du

dx

dy

��du��du

dx


Observations

I Succinctly, the derivative of acomposition is the product ofthe derivatives

I The only complication is wherethese derivatives are evaluated:at the same point the functionsare

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

Outline


The chain rule

Examples



Notes

Notes

Notes

4


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

Example

Example

let h(x) =√

3x2 + 1. Find h′(x).

Solution

First, 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


Corollary

Corollary (The Power Rule Combined with the Chain Rule)

If n is any real number and u = g(x) is differentiable, then

d

dx(un) = nun−1 du

dx.


Order matters!

Example

Findd

dx(sin 4x) and compare it to

d

dx(4 sin x).

Solution

I For the first, let u = 4x and y = sin(u). Then

dy

dx=

dy

du· du

dx= cos(u) · 4 = 4 cos 4x .

I For the second, let u = sin x and y = 4u. Then

dy

dx=

dy

du· du

dx= 4 · cos x


Notes

Notes

Notes

5


Example

Let f (x) =(

3√

x5 − 2 + 8)2

. Find f ′(x).

Solution

d

dx

(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)

=10

3x4(

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(x5 − 2)−2/3(5x4)


Combining techniques

Example

Findd

dx

((x3 + 1)10 sin(4x2 − 7)

)Solution

The “last” part of the function is the product, so we apply the productrule. Each factor’s derivative requires the chain rule:

d

dx

((x3 + 1)10 · sin(4x2 − 7)

)=

(d

dx(x3 + 1)10

)· sin(4x2 − 7) + (x3 + 1)10 ·

(d

dxsin(4x2 − 7)

)= 10(x3 + 1)9(3x2) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x)


Notes

Notes

Notes

6


http://flickr.com/photos/71502646@N00/993311118/

http://flickr.com/photos/71502646@N00/

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 − 1

z + 16. 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.

Solution

y ′ = 10(1− x2)9(−2x) = −20x(1− x2)9


Solution to #2

Example

Find the derivative of y =√

sin x .

Solution

Writing√

sin x as (sin x)1/2, we have

y ′ = 12 (sin x)−1/2 (cos x) =

cos x

2√

sin x


Notes

Notes

Notes

7


Solution to #3

Example

Find the derivative of y = sin√

x .

Solution

y ′ =d

dxsin(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

Solution

We 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

)


Solution to #5

Example

Find the derivative of F (z) =

√z − 1

z + 1.

Solution

y ′ =1

2

(z − 1

z + 1

)−1/2((z + 1)(1)− (z − 1)(1)

(z + 1)2

)=

1

2

(z + 1

z − 1

)1/2( 2

(z + 1)2

)=

1

(z + 1)3/2(z − 1)1/2


Notes

Notes

Notes

8


Solution to #6

Example

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

Solution

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


Solution to #7

Example

Find the derivative of y = csc2(sin θ).

Solution

Remember 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)))))).

Solution

Relax! 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))


Notes

Notes

Notes

9


Outline


The chain rule

Examples



Related rates of change at the Deli

Question

Suppose a deli clerk can slice a stick of pepperoni (assume the taperedends have been removed) by hand at the rate of 2 inches per minute, whilea machine can slice pepperoni at the rate of 10 inches per minute. ThendV

dtfor the machine is 5 times greater than

dV

dtfor the deli clerk. This is

explained by the

A. chain rule

B. product rule

C. quotient Rule

D. addition rule


Related rates of change in the ocean

Question

The area of a circle, A = πr2,changes as its radius changes. Ifthe radius changes with respectto time, the change in area withrespect to time is

A.dA

dr= 2πr

B.dA

dt= 2πr +

dr

dt

C.dA

dt= 2πr

dr

dtD. not enough information

Image credit: Jim FrazierV63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 35 / 36

Notes

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10


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!


Notes

Notes

Notes

11


Documents

Lesson 10: The Chain Rule (Section 21 handout)