1. Lesson 15 (Section 3.5)
The Chain Rule
Math 1a
October 29, 2007
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2. Analogy
Think about riding a bike. To
go faster you can either:

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

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

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

7. 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

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

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

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

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

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

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

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

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

17. Example
2
3
x5 − 2 + 8
Let f (x) = . Find f (x).
Solution
d d
2
3 3 3
x5 − 2 + 8 x5 − 2 + 8 x5 − 2 + 8
=2
dx dx
d
3 3
x5 − 2 + 8 x5 − 2
=2
dx
d5
− 2)−2/3
3 15
x5 − 2 + 8 (x − 5)
=2 3 (x dx
− 2)−2/3 (5x 4 )
3 15
x5 − 2 + 8
=2 3 (x
10 4
x 5 − 2 + 8 (x 5 − 2)−2/3
3
= x
3

19. A metaphor
Think about peeling an onion:
2
3
x 5 −2 +8
f (x) =
5
√
3
+8
2
− 2)−2/3 (5x 4 )
3 15
x5 − 2 + 8
f (x) = 2 3 (x

20. 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

21. 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