The chain rule helps us find a derivative of a composition of functions. It turns out that it's the product of the derivatives of the composed functions.

1. Section 2.5
The Chain Rule
V63.0121.034, Calculus I
October 5, 2009
Announcements
Quiz 2 this week
Midterm in class on §§1.1–1.4
. . . . . .

2. 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.”
.
. . . . . .

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.”
x
. g
. (x)
g
. .
. . . . . .

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
. . f
.
. . . . . .

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) f
.(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.”
f .
. ◦ g
x
. g
. (x) f
.(g(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
.
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.
. . . . . .

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

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

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

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

12. 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
. . . . . .

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

14. 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)
. . . . . .

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)
The composition is also linear
. . . . . .

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
The slope of the composition is the product of the slopes of
the two functions.
. . . . . .

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.
The derivative is supposed to be a local linearization of a
function. So there should be an analog of this property in
derivatives.
. . . . . .

18. The Nonlinear Case
See the Mathematica applet
Let u = g(x) and y = f(u). Suppose x is changed by a small
amount ∆x. Then
∆y ≈ f′ (y)∆u
and
∆u ≈ g′ (u)∆x.
So
∆y
∆y ≈ f′ (y)g′ (u)∆x =⇒ ≈ f′ (y)g′ (u)
∆x
. . . . . .

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

20. 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
. . . . . .

21. Observations
Succinctly, the derivative of
a composition is the
product of the derivatives
.
Image credit: o
.
. . . . . .

22. 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
. . . . . .

23. 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: o
.
. . . . . .

24. 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
.
. . . . . .

25. 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) .
Image credit: o
fractions .
. . . . . .

26. 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
. . . . . .

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

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

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

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

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

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

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

34. Corollary
Corollary (The Power Rule Combined with the Chain Rule)
If n is any real number and u = g(x) is differentiable, then
d n du
(u ) = nun−1 .
dx dx
. . . . . .

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

36. Does order matter?
Example
d d
Find (sin 4x) and compare it to (4 sin x).
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
Find (sin 4x) and compare it to (4 sin x).
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 · cos x
dx du dx
. . . . . .

38. Order matters!
Example
d d
Find (sin 4x) and compare it to (4 sin x).
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 · cos x
dx du dx
. . . . . .

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

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

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

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

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

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

45. A metaphor
Think about peeling an onion:
(√ )2
3 5
f(x) = x −2 +8
5
√
3
+8 .
Image credit: p
.
2
(√ )
f′ (x) = 2 x5 − 2 + 8 1 (x5 − 2)−2/3 (5x4 )
3
3
. . . . . .

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

47. Combining techniques
Example
d ( 3 )
Find (x + 1)10 sin(4x2 − 7)
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 )
Find (x + 1)10 sin(4x2 − 7)
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
( ) ( )
d 3 d
= (x + 1)10 · sin(4x2 − 7) + (x3 + 1)10 · sin(4x2 − 7)
dx dx
. . . . . .

49. Combining techniques
Example
d ( 3 )
Find (x + 1)10 sin(4x2 − 7)
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
( ) ( )
d 3 d
= (x + 1)10 · sin(4x2 − 7) + (x3 + 1)10 · sin(4x2 − 7)
dx dx
= 10(x3 + 1)9 (3x2 ) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x)
. . . . . .

50. Your Turn
Find derivatives of these functions:
1. y = (1 − x2 )10
√
2. y = sin x
√
3. y = sin x
4. y = (2x − 5)4 (8x2 − 5)−3
√
z−1
5. F(z) =
z+1
6. y = tan(cos x)
7. y = csc2 (sin θ)
8. y = sin(sin(sin(sin(sin(sin(x))))))
. . . . . .

51. Solution to #1
Example
Find the derivative of y = (1 − x2 )10 .
Solution
y′ = 10(1 − x2 )9 (−2x) = −20x(1 − x2 )9
. . . . . .

52. Solution to #2
Example
√
Find the derivative of y = sin x.
Solution
√
Writing sin x as (sin x)1/2 , we have
cos x
y′ = 1
2 (sin x)−1/2 (cos x) = √
2 sin x
. . . . . .

53. Solution to #3
Example
√
Find the derivative of y = sin x.
Solution
(√ )
′d 1 /2 1/2 1 −1/2 cos x
y = sin(x ) = cos(x ) 2 x = √
dx 2 x
. . . . . .

54. Solution to #4
Example
Find the derivative of y = (2x − 5)4 (8x2 − 5)−3
Solution
We need to use the product rule and the chain rule:
y′ = 4(2x − 5)3 (2)(8x2 − 5)−3 + (2x − 5)4 (−3)(8x2 − 5)−4 (16x)
The rest is a bit of algebra, useful if you wanted to solve the
equation y′ = 0:
[ ]
y′ = 8(2x − 5)3 (8x2 − 5)−4 (8x2 − 5) − 6x(2x − 5)
( )
= 8(2x − 5)3 (8x2 − 5)−4 −4x2 + 30x − 5
( )
= −8(2x − 5)3 (8x2 − 5)−4 4x2 − 30x + 5
. . . . . .

55. Solution to #5
Example √
z−1
Find the derivative of F(z) = .
z+1
Solution
( )−1/2 ( )
′ 1 z−1 (z + 1)(1) − (z − 1)(1)
y =
2 z+1 (z + 1)2
( )1/2 ( )
1 z+1 2 1
= 2
=
2 z−1 (z + 1) (z + 1)3/2 (z − 1)1/2
. . . . . .

56. Solution to #6
Example
Find the derivative of y = tan(cos x).
Solution
y′ = sec2 (cos x) · (− sin x) = − sec2 (cos x) sin x
. . . . . .

57. Solution to #7
Example
Find the derivative of y = csc2 (sin θ).
Solution
Remember the notation:
y = csc2 (sin θ) = [csc(sin θ)]2
So
y′ = 2 csc(sin θ) · [− csc(sin θ) cot(sin θ)] · cos(θ)
= −2 csc2 (sin θ) cot(sin θ) cos θ
. . . . . .

58. Solution to #8
Example
Find the derivative of y = sin(sin(sin(sin(sin(sin(x)))))).
Solution
Relax! It’s just a bunch of chain rules. All of these lines are
multiplied together.
y′ = cos(sin(sin(sin(sin(sin(x))))))
· cos(sin(sin(sin(sin(x)))))
· cos(sin(sin(sin(x))))
· cos(sin(sin(x)))
· cos(sin(x))
· cos(x))
. . . . . .

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

60. Related rates of change
Question
The area of a circle, A = π r2 ,
changes as its radius
changes. If the radius
changes with respect to time,
the change in area with
respect to time is
dA
A. = 2π r
dr
.
Image
dA dr .
B. = 2π r +
dt dt
dA dr
C. = 2π r
dt dt
D. not enough information
. . . . . .

61. Related rates of change
Question
The area of a circle, A = π r2 ,
changes as its radius
changes. If the radius
changes with respect to time,
the change in area with
respect to time is
dA
A. = 2π r
dr
.
Image
dA dr .
B. = 2π r +
dt dt
dA dr
C. = 2π r
dt dt
D. not enough information
. . . . . .

62. What have we learned today?
The derivative of a
composition is the
product of derivatives
In symbols:
(f ◦ g)′ (x) = f′ (g(x))g′ (x)
Calculus is like an
onion, and not because
it makes you cry!
. . . . . .