There's an obvious guess for the rule about the derivative of a product—which happens to be wrong. We explain the correct product rule and why it has to be true, along with the quotient rule. We show how to compute derivatives of additional trigonometric functions and new proofs of the Power Rule.

1. Section 2.4
The Product and Quotient Rules
V63.0121, Calculus I
February 17–18, 2009
Announcements
Quiz 2 is this week, covering 1.3–1.6
Midterm I is March 4/5, covering 1.1–2.4 (today)
ALEKS is due February 27, 11:59pm

2. Outline
The Product Rule
Derivation of the product rule
Examples
The Quotient Rule
Derivation
Examples
More derivatives of trigonometric functions
Derivative of Tangent
Derivative of Cotangent
Derivative of Secant
Derivative of Cosecant
More on the Power Rule
Power Rule for Positive Integers by Induction
Power Rule for Negative Integers

3. Calculus

4. Recollection and extension
We have shown that if u and v are functions, that
(u + v ) = u + v
(u − v ) = u − v
What about uv ?

5. Is the derivative of a product the product of the
derivatives?
(uv ) = u v ?

6. Is the derivative of a product the product of the
derivatives?
(uv ) = u v !
Try this with u = x and v = x 2 .

7. Is the derivative of a product the product of the
derivatives?
(uv ) = u v !
Try this with u = x and v = x 2 .
Then uv = x 3 =⇒ (uv ) = 3x 2 .

8. Is the derivative of a product the product of the
derivatives?
(uv ) = u v !
Try this with u = x and v = x 2 .
Then uv = x 3 =⇒ (uv ) = 3x 2 .
But u v = 1 · 2x = 2x.

9. Is the derivative of a product the product of the
derivatives?
(uv ) = u v !
Try this with u = x and v = x 2 .
Then uv = x 3 =⇒ (uv ) = 3x 2 .
But u v = 1 · 2x = 2x.
So we have to be more careful.

10. Mmm...burgers
Say you work in a fast-food joint. You want to make more money.
What are your choices?

11. Mmm...burgers
Say you work in a fast-food joint. You want to make more money.
What are your choices?
Work longer hours.

12. Mmm...burgers
Say you work in a fast-food joint. You want to make more money.
What are your choices?
Work longer hours.
Get a raise.

13. Mmm...burgers
Say you work in a fast-food joint. You want to make more money.
What are your choices?
Work longer hours.
Get a raise.
Say you get a 25 cent raise in
your hourly wages and work 5
hours more per week. How
much extra money do you
make?

14. Money money money money
The answer depends on how much you work already and your
current wage. Suppose you work h hours and are paid w . You get
a time increase of ∆h and a wage increase of ∆w . Income is
wages times hours, so
∆I = (w + ∆w )(h + ∆h) − wh
FOIL
= wh + w ∆h + ∆w h + ∆w ∆h − wh
= w ∆h + ∆w h + ∆w ∆h

15. A geometric argument
Draw a box:
∆h w ∆h ∆w ∆h
h wh ∆w h
w ∆w

16. A geometric argument
Draw a box:
∆h w ∆h ∆w ∆h
h wh ∆w h
w ∆w
∆I = w ∆h + h ∆w + ∆w ∆h

17. Supose wages and hours are changing continuously over time. How
does income change?
∆I w ∆h + h ∆w + ∆w ∆h
=
∆t ∆t
∆h ∆w ∆h
=w +h + ∆w
∆t ∆t ∆t

18. Supose wages and hours are changing continuously over time. How
does income change?
∆I w ∆h + h ∆w + ∆w ∆h
=
∆t ∆t
∆h ∆w ∆h
=w +h + ∆w
∆t ∆t ∆t
So
dI ∆I dh dw
= lim =w +h +0
dt t→0 ∆t dt dt

19. Supose wages and hours are changing continuously over time. How
does income change?
∆I w ∆h + h ∆w + ∆w ∆h
=
∆t ∆t
∆h ∆w ∆h
=w +h + ∆w
∆t ∆t ∆t
So
dI ∆I dh dw
= lim =w +h +0
dt t→0 ∆t dt dt
Theorem (The Product Rule)
Let u and v be differentiable at x. Then
(uv ) (x) = u(x)v (x) + u (x)v (x)

20. Example
Apply the product rule to u = x and v = x 2 .

21. Example
Apply the product rule to u = x and v = x 2 .
Solution
(uv ) (x) = u(x)v (x) + u (x)v (x) = x · (2x) + 1 · x 2 = 3x 2
This is what we get the “normal” way.

22. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d
(3 − x 2 )(x 3 − x + 1)
dx

23. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d
(3 − x 2 )(x 3 − x + 1)
dx
Solution
by direct multiplication:
d FOIL d
(3 − x 2 )(x 3 − x + 1) = −x 5 + 4x 3 − x 2 − 3x + 3
dx dx

24. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d
(3 − x 2 )(x 3 − x + 1)
dx
Solution
by direct multiplication:
d FOIL d
(3 − x 2 )(x 3 − x + 1) = −x 5 + 4x 3 − x 2 − 3x + 3
dx dx
= −5x 4 + 12x 2 − 2x − 3

25. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d
(3 − x 2 )(x 3 − x + 1)
dx
Solution
by the product rule:
dy d d3
(3 − x 2 ) (x 3 − x + 1) + (3 − x 2 ) (x − x + 1)
=
dx dx dx

26. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d
(3 − x 2 )(x 3 − x + 1)
dx
Solution
by the product rule:
dy d d3
(3 − x 2 ) (x 3 − x + 1) + (3 − x 2 ) (x − x + 1)
=
dx dx dx
= (−2x)(x 3 − x + 1) + (3 − x 2 )(3x 2 − 1)

27. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d
(3 − x 2 )(x 3 − x + 1)
dx
Solution
by the product rule:
dy d d3
(3 − x 2 ) (x 3 − x + 1) + (3 − x 2 ) (x − x + 1)
=
dx dx dx
= (−2x)(x 3 − x + 1) + (3 − x 2 )(3x 2 − 1)

28. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d
(3 − x 2 )(x 3 − x + 1)
dx
Solution
by the product rule:
dy d d3
(3 − x 2 ) (x 3 − x + 1) + (3 − x 2 ) (x − x + 1)
=
dx dx dx
= (−2x)(x 3 − x + 1) + (3 − x 2 )(3x 2 − 1)

29. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d
(3 − x 2 )(x 3 − x + 1)
dx
Solution
by the product rule:
dy d d3
(3 − x 2 ) (x 3 − x + 1) + (3 − x 2 ) (x − x + 1)
=
dx dx dx
= (−2x)(x 3 − x + 1) + (3 − x 2 )(3x 2 − 1)

30. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d
(3 − x 2 )(x 3 − x + 1)
dx
Solution
by the product rule:
dy d d3
(3 − x 2 ) (x 3 − x + 1) + (3 − x 2 ) (x − x + 1)
=
dx dx dx
= (−2x)(x 3 − x + 1) + (3 − x 2 )(3x 2 − 1)

31. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d
(3 − x 2 )(x 3 − x + 1)
dx
Solution
by the product rule:
dy d d3
(3 − x 2 ) (x 3 − x + 1) + (3 − x 2 ) (x − x + 1)
=
dx dx dx
= (−2x)(x 3 − x + 1) + (3 − x 2 )(3x 2 − 1)
= −5x 4 + 12x 2 − 2x − 3

32. One more
Example
d
Find x sin x.
dx

33. One more
Example
d
Find x sin x.
dx
Solution
d d d
x sin x = x sin x + x sin x
dx dx dx

34. One more
Example
d
Find x sin x.
dx
Solution
d d d
x sin x = x sin x + x sin x
dx dx dx
= 1 · sin x + x · cos x

35. One more
Example
d
Find x sin x.
dx
Solution
d d d
x sin x = x sin x + x sin x
dx dx dx
= 1 · sin x + x · cos x
= sin x + x cos x

36. Mnemonic
Let u = “hi” and v = “ho”. Then
(uv ) = vu + uv = “ho dee hi plus hi dee ho”

37. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product
uvw .

38. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product
uvw .
Solution
(uvw ) = ((uv )w ) = (uv ) w + (uv )w
= (u v + uv )w + (uv )w
= u vw + uv w + uvw

39. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product
uvw .
Solution
(uvw ) = ((uv )w ) = (uv ) w + (uv )w
= (u v + uv )w + (uv )w
= u vw + uv w + uvw
So we write down the product three times, taking the derivative of
each factor once.

40. Outline
The Product Rule
Derivation of the product rule
Examples
The Quotient Rule
Derivation
Examples
More derivatives of trigonometric functions
Derivative of Tangent
Derivative of Cotangent
Derivative of Secant
Derivative of Cosecant
More on the Power Rule
Power Rule for Positive Integers by Induction
Power Rule for Negative Integers

41. The Quotient Rule
What about the derivative of a quotient?

42. The Quotient Rule
What about the derivative of a quotient?
u
Let u and v be differentiable functions and let Q = . Then
v
u = Qv

43. The Quotient Rule
What about the derivative of a quotient?
u
Let u and v be differentiable functions and let Q = . Then
v
u = Qv
If Q is differentiable, we have
u = (Qv ) = Q v + Qv

44. The Quotient Rule
What about the derivative of a quotient?
u
Let u and v be differentiable functions and let Q = . Then
v
u = Qv
If Q is differentiable, we have
u = (Qv ) = Q v + Qv
u − Qv u uv
−·
=⇒ Q = =
v v vv

45. The Quotient Rule
What about the derivative of a quotient?
u
Let u and v be differentiable functions and let Q = . Then
v
u = Qv
If Q is differentiable, we have
u = (Qv ) = Q v + Qv
u − Qv u uv
−·
=⇒ Q = =
v v vv
u v − uv
u
=⇒ Q = =
v2
v

46. The Quotient Rule
What about the derivative of a quotient?
u
Let u and v be differentiable functions and let Q = . Then
v
u = Qv
If Q is differentiable, we have
u = (Qv ) = Q v + Qv
u − Qv u uv
−·
=⇒ Q = =
v v vv
u v − uv
u
=⇒ Q = =
v2
v
This is called the Quotient Rule.

47. Verifying Example
Example
x2
d
Verify the quotient rule by computing and comparing it
dx x
d
to (x).
dx

48. Verifying Example
Example
x2
d
Verify the quotient rule by computing and comparing it
dx x
d
to (x).
dx
Solution
d d
x dx x 2 − x 2 dx (x)
x2
d
=
x2
dx x
x · 2x − x 2 · 1
=
x2
2
x d
= 2 =1= (x)
x dx

49. Examples
Example
d 2x + 5
1.
dx 3x − 2
d 2x + 1
2.
dx x 2 − 1
t −1
d
3. 2+t +2
dt t

50. Solution to first example
d 2x + 5
dx 3x − 2

51. Solution to first example
d d
(3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d 2x + 5
=
(3x − 2)2
dx 3x − 2

52. Solution to first example
d d
(3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d 2x + 5
=
(3x − 2)2
dx 3x − 2

53. Solution to first example
d d
(3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d 2x + 5
=
(3x − 2)2
dx 3x − 2

54. Solution to first example
d d
(3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d 2x + 5
=
(3x − 2)2
dx 3x − 2

55. Solution to first example
d d
(3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d 2x + 5
=
(3x − 2)2
dx 3x − 2

56. Solution to first example
d d
(3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d 2x + 5
=
(3x − 2)2
dx 3x − 2

57. Solution to first example
d d
(3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d 2x + 5
=
(3x − 2)2
dx 3x − 2
(3x − 2)(2) − (2x + 5)(3)
=
(3x − 2)2

58. Solution to first example
d d
(3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d 2x + 5
=
(3x − 2)2
dx 3x − 2
(3x − 2)(2) − (2x + 5)(3)
=
(3x − 2)2

59. Solution to first example
d d
(3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d 2x + 5
=
(3x − 2)2
dx 3x − 2
(3x − 2)(2) − (2x + 5)(3)
=
(3x − 2)2

60. Solution to first example
d d
(3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d 2x + 5
=
(3x − 2)2
dx 3x − 2
(3x − 2)(2) − (2x + 5)(3)
=
(3x − 2)2

61. Solution to first example
d d
(3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d 2x + 5
=
(3x − 2)2
dx 3x − 2
(3x − 2)(2) − (2x + 5)(3)
=
(3x − 2)2

62. Solution to first example
d d
(3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d 2x + 5
=
(3x − 2)2
dx 3x − 2
(3x − 2)(2) − (2x + 5)(3)
=
(3x − 2)2
(6x − 4) − (6x + 15)
=
(3x − 2)2

63. Solution to first example
d d
(3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d 2x + 5
=
(3x − 2)2
dx 3x − 2
(3x − 2)(2) − (2x + 5)(3)
=
(3x − 2)2
(6x − 4) − (6x + 15) 19
=−
= 2 (3x − 2)2
(3x − 2)

64. Examples
Example
d 2x + 5
1.
dx 3x − 2
d 2x + 1
2.
dx x 2 − 1
t −1
d
3. 2+t +2
dt t
Answers
19
1. −
(3x − 2)2

65. Solution to second example
d 2x + 1
dx x 2 − 1

66. Solution to second example
(x 2 − 1)(2) − (2x + 1)(2x)
d 2x + 1
=
dx x 2 − 1 (x 2 − 1)2

67. Solution to second example
(x 2 − 1)(2) − (2x + 1)(2x)
d 2x + 1
=
dx x 2 − 1 (x 2 − 1)2
2 − 2) − (4x 2 + 2x)
(2x
=
(x 2 − 1)2
2 x2 + x + 1
=−
(x 2 − 1)2

68. Examples
Example
d 2x + 5
1.
dx 3x − 2
d 2x + 1
2.
dx x 2 − 1
t −1
d
3. 2+t +2
dt t
Answers
19
1. −
(3x − 2)2
2 x2 + x + 1
2. −
(x 2 − 1)2

69. Solution to third example
t −1
d
2+t +2
dt t

70. Solution to third example
(t 2 + t + 2)(1) − (t − 1)(2t + 1)
t −1
d
=
dt t 2 + t + 2 (t 2 + t + 2)2

71. Solution to third example
(t 2 + t + 2)(1) − (t − 1)(2t + 1)
t −1
d
=
dt t 2 + t + 2 (t 2 + t + 2)2
(t 2 + t + 2) − (2t 2 − t − 1)
=
(t 2 + t + 2)2
−t 2 + 2t + 3
=2
(t + t + 2)2

72. Examples
Example
d 2x + 5
1.
dx 3x − 2
d 2x + 1
2.
dx x 2 − 1
t −1
d
3. 2+t +2
dt t
Answers
19
1. −
(3x − 2)2
2 x2 + x + 1
2. −
(x 2 − 1)2
−t 2 + 2t + 3
3.
(t 2 + t + 2)2

73. Mnemonic
Let u = “hi” and v = “lo”. Then
vu − uv
u
= = “lo dee hi minus hi dee lo over lo lo”
v2
v

74. Outline
The Product Rule
Derivation of the product rule
Examples
The Quotient Rule
Derivation
Examples
More derivatives of trigonometric functions
Derivative of Tangent
Derivative of Cotangent
Derivative of Secant
Derivative of Cosecant
More on the Power Rule
Power Rule for Positive Integers by Induction
Power Rule for Negative Integers

75. Derivative of Tangent
Example
d
Find tan x
dx

76. Derivative of Tangent
Example
d
Find tan x
dx
Solution
d d sin x
tan x =
dx dx cos x

77. Derivative of Tangent
Example
d
Find tan x
dx
Solution
cos x · sin x − sin x · (− sin x)
d d sin x
tan x = =
cos2 x
dx dx cos x

78. Derivative of Tangent
Example
d
Find tan x
dx
Solution
cos x · sin x − sin x · (− sin x)
d d sin x
tan x = =
cos2 x
dx dx cos x
cos2 x + sin2 x
=
cos2 x

79. Derivative of Tangent
Example
d
Find tan x
dx
Solution
cos x · sin x − sin x · (− sin x)
d d sin x
tan x = =
cos2 x
dx dx cos x
cos2 x + sin2 x 1
= =
2x cos2 x
cos

80. Derivative of Tangent
Example
d
Find tan x
dx
Solution
cos x · sin x − sin x · (− sin x)
d d sin x
tan x = =
cos2 x
dx dx cos x
cos2 x + sin2 x 1
= sec2 x
= =
2x cos2 x
cos

81. Derivative of Cotangent
Example
d
Find cot x
dx

82. Derivative of Cotangent
Example
d
Find cot x
dx
Answer
d 1
cot x = − 2 = − csc2 x
dx sin x

83. Derivative of Secant
Example
d
Find sec x
dx

84. Derivative of Secant
Example
d
Find sec x
dx
Solution
d d 1
sec x =
dx dx cos x

85. Derivative of Secant
Example
d
Find sec x
dx
Solution
cos x · 0 − 1 · (− sin x)
d d 1
sec x = =
cos2 x
dx dx cos x

86. Derivative of Secant
Example
d
Find sec x
dx
Solution
cos x · 0 − 1 · (− sin x)
d d 1
sec x = =
cos2 x
dx dx cos x
sin x
=
cos2 x

87. Derivative of Secant
Example
d
Find sec x
dx
Solution
cos x · 0 − 1 · (− sin x)
d d 1
sec x = =
cos2 x
dx dx cos x
sin x 1 sin x
·
= =
cos2 x cos x cos x

88. Derivative of Secant
Example
d
Find sec x
dx
Solution
cos x · 0 − 1 · (− sin x)
d d 1
sec x = =
cos2 x
dx dx cos x
sin x 1 sin x
·
= = = sec x tan x
cos2 x cos x cos x

89. Derivative of Cosecant
Example
d
Find csc x
dx

90. Derivative of Cosecant
Example
d
Find csc x
dx
Answer
d
csc x = − csc x cot x
dx

91. Recap: Derivatives of trigonometric functions
y y
Functions come in pairs
sin x cos x (sin/cos, tan/cot,
sec/csc)
− sin x
cos x
Derivatives of pairs
sec2 x
tan x follow similar patterns,
− csc2 x with functions and
cot x
co-functions switched
sec x sec x tan x and an extra sign.
− csc x cot x
csc x

92. Outline
The Product Rule
Derivation of the product rule
Examples
The Quotient Rule
Derivation
Examples
More derivatives of trigonometric functions
Derivative of Tangent
Derivative of Cotangent
Derivative of Secant
Derivative of Cosecant
More on the Power Rule
Power Rule for Positive Integers by Induction
Power Rule for Negative Integers

93. Power Rule for Positive Integers by Induction
Theorem
Let n be a positive integer. Then
dn
x = nx n−1
dx

94. Power Rule for Positive Integers by Induction
Theorem
Let n be a positive integer. Then
dn
x = nx n−1
dx
Proof.
By induction on n.

95. Principle of Mathematical Induction
Suppose S(1) is
true and S(n + 1)
is true when-
ever S(n) is true.
Then S(n) is true
for all n.
Image credit: Kool Skatkat

96. Power Rule for Positive Integers by Induction
Theorem
Let n be a positive integer. Then
dn
x = nx n−1
dx
Proof.
By induction on n. We can show it to be true for n = 1 directly.

97. Power Rule for Positive Integers by Induction
Theorem
Let n be a positive integer. Then
dn
x = nx n−1
dx
Proof.
By induction on n. We can show it to be true for n = 1 directly.
dn
x = nx n−1 . Then
Suppose for some n that
dx
d n+1 d
(x · x n )
x =
dx dx

98. Power Rule for Positive Integers by Induction
Theorem
Let n be a positive integer. Then
dn
x = nx n−1
dx
Proof.
By induction on n. We can show it to be true for n = 1 directly.
dn
x = nx n−1 . Then
Suppose for some n that
dx
d n+1 d
(x · x n )
x =
dx dx
d dn
x xn + x
= x
dx dx

99. Power Rule for Positive Integers by Induction
Theorem
Let n be a positive integer. Then
dn
x = nx n−1
dx
Proof.
By induction on n. We can show it to be true for n = 1 directly.
dn
x = nx n−1 . Then
Suppose for some n that
dx
d n+1 d
(x · x n )
x =
dx dx
d dn
x xn + x
= x
dx dx
= 1 · x n + x · nx n−1 = (n + 1)x n

100. Power Rule for Negative Integers
Use the quotient rule to prove
Theorem
d −n
x = (−n)x −n−1
dx
for positive integers n.

101. Power Rule for Negative Integers
Use the quotient rule to prove
Theorem
d −n
x = (−n)x −n−1
dx
for positive integers n.
Proof.
d −n d1
x=
dx x n
dx

102. Power Rule for Negative Integers
Use the quotient rule to prove
Theorem
d −n
x = (−n)x −n−1
dx
for positive integers n.
Proof.
d −n d1
x=
dx x n
dx
d dn
x n · dx 1 − 1 · dx x
=
x 2n

103. Power Rule for Negative Integers
Use the quotient rule to prove
Theorem
d −n
x = (−n)x −n−1
dx
for positive integers n.
Proof.
d −n d1
x=
dx x n
dx
d dn
x n · dx 1 − 1 · dx x
=
x 2n
n−1
0 − nx
=
x 2n

104. Power Rule for Negative Integers
Use the quotient rule to prove
Theorem
d −n
x = (−n)x −n−1
dx
for positive integers n.
Proof.
d −n d1
x=
dx x n
dx
d d
x n · dx 1 − 1 · dx x n
=
x 2n
n−1
0 − nx
= −nx −n−1
=
x 2n