The document explains the chain rule in calculus, which describes how to compute the derivative of the composition of two functions. It provides examples illustrating both linear and nonlinear cases, demonstrating that the derivative of a composition is the product of the derivatives of the individual functions evaluated at the appropriate points. The document emphasizes the importance of correctly applying derivatives when functions are combined, outlining key concepts and examples throughout.

1. Section 2.5
The Chain Rule
V63.0121.027, Calculus I
October 6, 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: ooOJasonOoo
. . . . . .

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: ooOJasonOoo
. . . . . .

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)
fractions .
.
Image credit: ooOJasonOoo
. . . . . .

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
.
2
(√ )
f′ (x) = 2 x5 − 2 + 8 1 (x5 − 2)−2/3 (5x4 )
3
3
.
Image credit: photobunny
. . . . . .

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
dA dr .
B. = 2π r +
dt dt
dA dr
C. = 2π r
dt dt
D. not enough information
.
Image credit: Jim Frazier
. . . . . .

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
dA dr .
B. = 2π r +
dt dt
dA dr
C. = 2π r
dt dt
D. not enough information
.
Image credit: Jim Frazier
. . . . . .

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!
. . . . . .