Lesson 15: The Chain Rule

Lesson 15 (Section 3.5)The Chain Rule

Math 1a

October 29, 2007

Announcements

I Come to office hours if you don’t have your midterm yet

Analogy

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

I pedal faster

I change gears

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

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

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

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.

Math 1a - October 29, 2007.GWBMonday, Oct 29, 2007

Page1of6

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

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

Math 1a - October 29, 2007.GWBMonday, Oct 29, 2007

Page4of6

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

Math 1a - October 29, 2007.GWBMonday, Oct 29, 2007

Page6of6

A metaphor

Think about peeling an onion:

f (x) =

(3√

x5︸︷︷︸�5

−2

︸ ︷︷ ︸3√�

+8

︸ ︷︷ ︸�+8

)2

︸ ︷︷ ︸�2

f ′(x) = 2(

3√

x5 − 2 + 8)

13(x5 − 2)−2/3(5x4)

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