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. To
go faster you can either:
SpringSun

Analogy
Think about riding a bike. To
go faster you can either:
pedal faster
SpringSun

Analogy
Think about riding a bike. To
go faster you can either:
pedal faster
change gears
SpringSun

Analogy
Think about riding a bike. To
go faster you can either:
pedal faster
change gears
SpringSun
The angular position of the back wheel depends on the position of
the front wheel:
Rθ
ϕ(θ) =
r
And so the angular speed of the back wheel depends on the
derivative of this function and the speed of the front wheel.

Outline
Anology
The chain rule
Examples
Questions

Theorem of the day: The chain rule
Theorem
Let f and g be functions, with g differentiable at a and f
differentiable 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 dy du
=
dx du dx

Outline
Anology
The chain rule
Examples
Questions

Example
Example
let h(x) = 3x 2 + 1. Find h (x).

Example
Example
let h(x) = 3x 2 + 1. Find h (x).
Solution
First, write h as f ◦ g.

Example
Example
let h(x) = 3x 2 + 1. Find h (x).
Solution √
First, write h as f ◦ g . Let f (u) = u and g (x) = 3x 2 + 1.

Example
Example
let h(x) = 3x 2 + 1. Find h (x).
Solution √
First, write h as f ◦ g . Let f (u) = u and g (x) = 3x 2 + 1. Then
f (u) = 1 u −1/2 , and g (x) = 6x. So
2
h (x) = 1 u −1/2 (6x)
2

Example
Example
let h(x) = 3x 2 + 1. Find h (x).
Solution √
First, write h as f ◦ g . Let f (u) = u and g (x) = 3x 2 + 1. Then
f (u) = 1 u −1/2 , and g (x) = 6x. So
2
3x
h (x) = 1 u −1/2 (6x) = 2 (3x 2 + 1)−1/2 (6x) = √
2
1
3x 2 + 1

Example
3
2
Let f (x) = x5 − 2 + 8 . Find f (x).

Example
3
2
Let f (x) = x5 − 2 + 8 . Find f (x).
Solution
d 3
2 3 d 3
x5 − 2 + 8 =2 x5 − 2 + 8 x5 − 2 + 8
dx dx

Example
3
2
Let f (x) = x5 − 2 + 8 . Find f (x).
Solution
d 3
2 3 d 3
x5 − 2 + 8 =2 x5 − 2 + 8 x5 − 2 + 8
dx dx
3 d 3
=2 x5 − 2 + 8 x5 − 2
dx

Example
3
2
Let f (x) = x5 − 2 + 8 . Find f (x).
Solution
d 3
2 3 d 3
x5 − 2 + 8 =2 x5 − 2 + 8 x5 − 2 + 8
dx dx
3 d 3
=2 x5 − 2 + 8 x5 − 2
dx
d 5
− 2)−2/3
3 1 5
=2 x5 − 2 + 8 3 (x (x − 5)
dx
− 2)−2/3 (5x 4 )
3 1 5
=2 x5 − 2 + 8 3 (x
10 4
x 5 − 2 + 8 (x 5 − 2)−2/3
3
= x
3

A metaphor
Think about peeling an onion:
2
3
f (x) = x 5 −2 +8
5
√
3
+8
photobunny
2
− 2)−2/3 (5x 4 )
3 1 5
f (x) = 2 x5 − 2 + 8 3 (x

Outline
Anology
The chain rule
Examples
Questions

Question
The area of a circle, A = πr 2 , changes as its radius changes. If the
radius changes with respect to time, the change in area with
respect to time is
dA
A. = 2πr
dr
dA dr
B. = 2πr +
dt dt
dA dr
C. = 2πr
dt dt
D. not enough information

Question
The area of a circle, A = πr 2 , changes as its radius changes. If the
radius changes with respect to time, the change in area with
respect to time is
dA
A. = 2πr
dr
dA dr
B. = 2πr +
dt dt
dA dr
C. = 2πr
dt dt
D. not enough information

Photo Credits
Chain Linkage by Flickr user mklingo. Used
in compliance with the Creative Commons
No Derivative Works License