CHAIN RULE

1. The Chain Rule

2. Compositions
Definition
If f and g are functions, the composition (f ◦ g)(x) = f(g(x))
means “do g first, then f.”

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

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.”
g f
x g(x)

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.”
g f
x g(x) f(g(x))

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.”
g f
x f(g(x))
f◦
g(x
g
)

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.”
g f
x f(g(x))
f◦
g(x
g
)
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):
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
φ(θ) =
Rθ
r

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 ≈ f′
(y)g′(u)∆x =⇒
∆y
∆x
≈ f′
(y)g′(u)

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:

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:

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.”
g f
x f(g(x))
f◦
g(x
g
)

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)
fractions
Image credit:

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
dy dy du
dx
=
du dx
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 =
dy
g(
/
d
x
/
u
). Then
/
d
/
u 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 √ 2
First, write h as f◦ g. Let f(u) = u and g(x) = 3x + 1.

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

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

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

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

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

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

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

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

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

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

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

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

45. A metaphor
5
`˛¸x
5
f(x) = x −
` √
3
˛
¸ x
2 +8
Think about peeling an onion:
(√3
) 2
` x
˛
+
¸
8
` ˛¸
2
( √
x
Image credit: p
)
3
f′
(x) = 2 3
x5 − 2 +8 (
1 5 −2/3 4
x − 2) (5x )

46. Combining techniques
Example
d
dx
(
Find ( 3 10 2
x +1) sin(4x − 7)
)

47. Combining techniques
Example
d
dx
(
Find ( 3 10 2
x +1) sin(4x − 7)
)
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
dx
(
Find ( 3 10 2
x +1) sin(4x − 7)
)
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
dx
3 10
(x +1) · sin( 2
4x − 7)
( )
(
d
dx
= ( 3 10
)
x +1) ·sin( 2 3 10
4x −7)+(x +1) ·
(
d
dx
sin( 2
4x − 7)
)

49. Combining techniques
Example
d
dx
(
Find ( 3 10 2
x +1) sin(4x − 7)
)
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
dx
3 10
(x +1) · sin( 2
4x − 7)
( )
(
d
dx
= (
)
3 10 2 3 10
x +1) ·sin(4x −7)+(x +1) ·
(
d
dx
2
sin(4x − 7)
)
= 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
5. F(z) =
√
z − 1
z + 1
6. y = tan(cosx)
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
√ 1/2
Writing sin x as (sinx) , we have
′ 1 −1/2 cos x
y = 2 (sin x) (cosx) =
2
√
sin x

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

56. Solution to #6
Example
Find the derivative of y = tan(cosx).
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 = πr2,
changes as its radius
changes. If the radius
changes with respect to time,
the change in area with
respect to time is
dr
A.
dA
= 2πr
dA dr
B.
dt
= 2πr +
dt
C.
dA
= 2πr
dr
dt dt
D. not enough information
Image

61. 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
dr
A.
dA
= 2πr
dA dr
B.
dt
= 2πr +
dt
C.
dA
= 2πr
dr
dt dt
D. not enough information
Image

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!