The chain rule tells us how we find the derivative of the composition of two functions

1. Section 2.5
The Chain Rule
V63.0121, Calculus I
February 19, 2009
Announcements
Midterm is March 4/5 (75 min., in class, covers 1.1–2.4)
ALEKS is due February 27, 11:59pm
. . . . . .

2. Outline
Compositions
Heuristics
Analogy
The Linear Case
The chain rule
Examples
Related rates of change
. . . . . .

3. Compositions
See Section 1.2 for review
Definition
If f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means
“do g first, then f.”
.
. . . . . .

4. Compositions
See Section 1.2 for review
Definition
If f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means
“do g first, then f.”
. (x)
g
x
. .
g
.
. . . . . .

5. Compositions
See Section 1.2 for review
Definition
If f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means
“do g first, then f.”
. (x)
g
x
. .
g
. f
.
. . . . . .

6. Compositions
See Section 1.2 for review
Definition
If f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means
“do g first, then f.”
. (x) . (g(x))
g f
x
. .
g
. f
.
. . . . . .

7. Compositions
See Section 1.2 for review
Definition
If f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means
“do g first, then f.”
f .g(x) . (g(x))
f
. ◦g
x
. .
g
. f
.
. . . . . .

8. Compositions
See Section 1.2 for review
Definition
If f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means
“do g first, then f.”
f .g(x) . (g(x))
f
. ◦g
x
. .
g
. f
.
Our goal for the day is to understand how the derivative of the
composition of two functions depends on the derivatives of the
individual functions.
. . . . . .

9. Outline
Compositions
Heuristics
Analogy
The Linear Case
The chain rule
Examples
Related rates of change
. . . . . .

10. Analogy
Think about riding a bike. To
go faster you can either:
.
.
Image credit: SpringSun
. . . . . .

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

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

13. Analogy
Think about riding a bike. To
go faster you can either:
pedal faster
change gears
.
The angular position (φ) of the back wheel depends on the position
of the front sprocket (θ):
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.
.
Image credit: SpringSun
. . . . . .

14. The Linear Case
Question
Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the
composition?
. . . . . .

15. The Linear Case
Question
Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the
composition?
Answer
f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b)
. . . . . .

16. The Linear Case
Question
Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the
composition?
Answer
f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b)
The composition is also linear
. . . . . .

17. The Linear Case
Question
Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the
composition?
Answer
f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b)
The composition is also linear
The slope of the composition is the product of the slopes of the
two functions.
. . . . . .

18. The Linear Case
Question
Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the
composition?
Answer
f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b)
The composition is also linear
The slope of the composition is the product of the slopes of the
two functions.
. . . . . .

19. The Linear Case
Question
Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the
composition?
Answer
f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b)
The composition is also linear
The slope of the composition is the product of the slopes of the
two functions.
The derivative is supposed to be a local linearization of a function. So
there should be an analog of this property in derivatives.
. . . . . .

20. Outline
Compositions
Heuristics
Analogy
The Linear Case
The chain rule
Examples
Related rates of change
. . . . . .

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

22. Observations
Succinctly, the derivative of a
composition is the product of
the derivatives
.
.
Image credit: ooOJasonOoo
. . . . . .

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

24. Observations
Succinctly, the derivative of a
composition is the product of
the derivatives
The only complication is
where these derivatives are
evaluated: at the same point
the functions are
.
.
Image credit: ooOJasonOoo
. . . . . .

25. Compositions
See Section 1.2 for review
Definition
If f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means
“do g first, then f.”
f .g(x) . (g(x))
f
. ◦g
x
. .
g
. f
.
. . . . . .

26. Observations
Succinctly, the derivative of a
composition is the product of
the derivatives
The only complication is
where these derivatives are
evaluated: at the same point
the functions are
In Leibniz notation, the Chain
Rule looks like cancellation of
(fake) fractions
.
.
Image credit: ooOJasonOoo
. . . . . .

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

28. Theorem of the day: The chain rule
Theorem
Let f and g be functions, with g differentiable at x and f differentiable at
g(x). Then f ◦ g is differentiable at x and
(f ◦ g)′ (x) = f′ (g(x))g′ (x)
dy du
In Leibnizian notation, let y = f(u) and u = g(x). .Then
.
du dx
dy dy du
=
dx du dx
. . . . . .

29. Outline
Compositions
Heuristics
Analogy
The Linear Case
The chain rule
Examples
Related rates of change
. . . . . .

30. Example
Example √
3x2 + 1. Find h′ (x).
let h(x) =
. . . . . .

31. Example
Example √
3x2 + 1. Find h′ (x).
let h(x) =
Solution
First, write h as f ◦ g.
. . . . . .

32. Example
Example √
3x2 + 1. Find h′ (x).
let h(x) =
Solution √
u and g(x) = 3x2 + 1.
First, write h as f ◦ g. Let f(u) =
. . . . . .

33. Example
Example √
3x2 + 1. Find h′ (x).
let h(x) =
Solution √
First, write h as f ◦ g. Let f(u) = u and g(x) = 3x2 + 1. Then
f′ (u) = 1 u−1/2 , and g′ (x) = 6x. So
2
h′ (x) = 1 u−1/2 (6x)
2
. . . . . .

34. Example
Example √
3x2 + 1. Find h′ (x).
let h(x) =
Solution √
First, write h as f ◦ g. Let f(u) = u and g(x) = 3x2 + 1. Then
f′ (u) = 1 u−1/2 , and g′ (x) = 6x. So
2
3x
h′ (x) = 1 u−1/2 (6x) = 1 (3x2 + 1)−1/2 (6x) = √
2 2
3x2 + 1
. . . . . .

35. Does order matter?
Example
d d
(sin 4x) and compare it to (4 sin x).
Find
dx dx
. . . . . .

36. Does order matter?
Example
d d
(sin 4x) and compare it to (4 sin x).
Find
dx dx
Solution
For the first, let u = 4x and y = sin(u). Then
dy dy du
· = cos(u) · 4 = 4 cos 4x.
=
dx du dx
. . . . . .

37. Does order matter?
Example
d d
(sin 4x) and compare it to (4 sin x).
Find
dx dx
Solution
For the first, let u = 4x and y = sin(u). Then
dy dy du
· = cos(u) · 4 = 4 cos 4x.
=
dx du dx
For the second, let u = sin x and y = 4u. Then
dy dy du
· = 4 · sin x
=
dx du dx
. . . . . .

38. Order matters!
Example
d d
(sin 4x) and compare it to (4 sin x).
Find
dx dx
Solution
For the first, let u = 4x and y = sin(u). Then
dy dy du
· = cos(u) · 4 = 4 cos 4x.
=
dx du dx
For the second, let u = sin x and y = 4u. Then
dy dy du
· = 4 · sin x
=
dx du dx
. . . . . .

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

40. Example
(√ )2
. Find f′ (x).
x5 − 2 + 8
3
Let f(x) =
Solution
d (√ 5 )2 (√ ) d (√ )
x −2+8 =2 x5 − 2 + 8 x5 − 2 + 8
3 3 3
dx dx
. . . . . .

41. Example
(√ )2
. Find f′ (x).
x5 − 2 + 8
3
Let f(x) =
Solution
d (√ 5 )2 (√ ) d (√ )
x −2+8 =2 x5 − 2 + 8 x5 − 2 + 8
3 3 3
dx dx
(√ ) d√
x5 − 2 + 8 x5 − 2
3 3
=2
dx
. . . . . .

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

43. Example
(√ )2
. Find f′ (x).
x5 − 2 + 8
3
Let f(x) =
Solution
d (√ 5 )2 (√ ) d (√ )
x −2+8 =2 x5 − 2 + 8 x5 − 2 + 8
3 3 3
dx dx
(√ ) d√
x5 − 2 + 8 x5 − 2
3 3
=2
dx
(√ ) d
x5 − 2 + 8 1 (x5 − 2)−2/3 (x5 − 2)
3
=2 3 dx
(√ )
x5 − 2 + 8 1 (x5 − 2)−2/3 (5x4 )
3
=2 3
. . . . . .

44. Example
(√ )2
. Find f′ (x).
x5 − 2 + 8
3
Let f(x) =
Solution
d (√ 5 )2 (√ ) d (√ )
x −2+8 =2 x5 − 2 + 8 x5 − 2 + 8
3 3 3
dx dx
(√ ) d√
x5 − 2 + 8 x5 − 2
3 3
=2
dx
(√ ) d
x5 − 2 + 8 1 (x5 − 2)−2/3 (x5 − 2)
3
=2 3 dx
(√ )
x5 − 2 + 8 1 (x5 − 2)−2/3 (5x4 )
3
=2 3
10 4 (√ 5 )
x − 2 + 8 (x5 − 2)−2/3
3
=x
3
. . . . . .

45. A metaphor
Think about peeling an onion:
(√ )2
3
−2 +8
5
f(x) = x
5
√
3
+8
.
(√ )
2
f′ (x) = 2 − 2)−2/3 (5x4 )
x5 − 2 + 8
3 15
3 (x
.
Image credit: photobunny
. . . . . .

46. Combining techniques
Example
d( 3 )
(x + 1)10 sin(4x2 − 7)
Find
dx
. . . . . .

47. Combining techniques
Example
d( 3 )
(x + 1)10 sin(4x2 − 7)
Find
dx
Solution
The “last” part of the function is the product, so we apply the product rule.
Each factor’s derivative requires the chain rule:
. . . . . .

48. Combining techniques
Example
d( 3 )
(x + 1)10 sin(4x2 − 7)
Find
dx
Solution
The “last” part of the function is the product, so we apply the product rule.
Each factor’s derivative requires the chain rule:
d( 3 )
(x + 1)10 · sin(4x2 − 7)
dx
( ) ( )
d3 d
· sin(4x − 7) + (x + 1) · sin(4x − 7)
10 2 3 10 2
= (x + 1)
dx dx
. . . . . .

49. Combining techniques
Example
d( 3 )
(x + 1)10 sin(4x2 − 7)
Find
dx
Solution
The “last” part of the function is the product, so we apply the product rule.
Each factor’s derivative requires the chain rule:
d( 3 )
(x + 1)10 · sin(4x2 − 7)
dx
( ) ( )
d3 d
· sin(4x − 7) + (x + 1) · sin(4x − 7)
10 2 3 10 2
= (x + 1)
dx dx
= 10(x3 + 1)9 (3x2 ) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x)
. . . . . .

50. Outline
Compositions
Heuristics
Analogy
The Linear Case
The chain rule
Examples
Related rates of change
. . . . . .

51. Related rates of change
Question
The area of a circle, A = πr2 ,
changes as its radius changes. If
the radius changes with
respect to time, the change in
area with respect to time is
dA
= 2πr
A.
dr
dA dr
= 2πr +
B. .
dt dt
dA dr
= 2πr
C.
dt dt
D. not enough information
.
Image credit: Jim Frazier
. . . . . .

52. Related rates of change
Question
The area of a circle, A = πr2 ,
changes as its radius changes. If
the radius changes with
respect to time, the change in
area with respect to time is
dA
= 2πr
A.
dr
dA dr
= 2πr +
B. .
dt dt
dA dr
= 2πr
C.
dt dt
D. not enough information
.
Image credit: Jim Frazier
. . . . . .