Lesson 11 (Section 3.5)The Chain Rule

Math 1a

February 27, 2007

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Image: mklingo

Outline

Anology

The chain rule

Examples

Questions

Analogy

Think about riding a bike. Togo faster you can either:

I pedal faster

I change gears

SpringSun

The angular position of the back wheel depends on the position ofthe front wheel:

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.

Analogy

Think about riding a bike. Togo faster you can either:

I pedal faster

I change gears

SpringSun

The angular position of the back wheel depends on the position ofthe front wheel:

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.

Analogy

Think about riding a bike. Togo faster you can either:

I pedal faster

I change gears

SpringSun

The angular position of the back wheel depends on the position ofthe front wheel:

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.

Analogy

Think about riding a bike. Togo faster you can either:

I pedal faster

I change gears

SpringSun

The angular position of the back wheel depends on the position ofthe front wheel:

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.

Outline

Anology

The chain rule

Examples

Questions

Theorem of the day: The chain rule

TheoremLet f and g be functions, with g differentiable at a and fdifferentiable at g(a). Then f ◦ g is differentiable at a and

(f ◦ g)′(a) = f ′(g(a))g ′(a)

In Leibnizian notation, let y = f (u) and u = g(x). Then

dy

dx=

dy

du

du

dx

Outline

Anology

The chain rule

Examples

Questions

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) = 1

2(3x2 + 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. 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

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) = 1

2(3x2 + 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(3x2 + 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) = 1

2(3x2 + 1)−1/2(6x) =3x√

3x2 + 1

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 − 5)

= 2(

3√

x5 − 2 + 8)

13(x5 − 2)−2/3(5x4)

=10

3x4(

3√

x5 − 2 + 8)

(x5 − 2)−2/3

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 − 5)

= 2(

3√

x5 − 2 + 8)

13(x5 − 2)−2/3(5x4)

=10

3x4(

3√

x5 − 2 + 8)

(x5 − 2)−2/3

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 − 5)

= 2(

3√

x5 − 2 + 8)

13(x5 − 2)−2/3(5x4)

=10

3x4(

3√

x5 − 2 + 8)

(x5 − 2)−2/3

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 − 5)

= 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

photobunny

f ′(x) = 2(

3√

x5 − 2 + 8)

13(x5 − 2)−2/3(5x4)

Outline

Anology

The chain rule

Examples

Questions

QuestionThe area of a circle, A = πr2, changes as its radius changes. If theradius changes with respect to 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

QuestionThe area of a circle, A = πr2, changes as its radius changes. If theradius changes with respect to 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

Photo Credits

Chain Linkage by Flickr user mklingo. Usedin compliance with the Creative CommonsNo Derivative Works License