The product rule or Leibniz rule is the rule by which we can differentiate products of functions!

1. Section 2.4
The Product and Quotient Rules
V63.0121.006/016, Calculus I
February 16, 2010
Announcements
Quiz 2 is February 26, covering §§1.5–2.3
Midterm I is March 4, covering §§1.1–2.5
Office Hours W 1:30–2:30, R 9–10
do get-to-know-you survey by Thursday
. . . . . .

2. Outline
Grader’s Corner
Derivative of a Product
Derivation
Examples
The Quotient Rule
Derivation
Examples
More derivatives of trigonometric functions
Derivative of Tangent and Cotangent
Derivative of Secant and Cosecant
More on the Power Rule
Power Rule for Positive Integers by Induction
Power Rule for Negative Integers
. . . . . .

3. Problem 1.5.20
sin x
Use the theorems on continuity to show h(x) = is
x+1
continuous.
. . . . . .

4. Problem 1.5.20
sin x
Use the theorems on continuity to show h(x) = is
x+1
continuous.
Solution
By Theorem 6, f(x) = sin x and g(x) = x + 1 are continuous
because f(x) is a trigonometric function and g(x) is a polynomial.
f(x)
By Theorem 4, part 5, h(x) = is continuous wherever
g(x)
g(x) ̸= 0.
. . . . . .

5. Problem 1.5.20
sin x
Use the theorems on continuity to show h(x) = is
x+1
continuous.
Solution
By Theorem 6, f(x) = sin x and g(x) = x + 1 are continuous
because f(x) is a trigonometric function and g(x) is a polynomial.
f(x)
By Theorem 4, part 5, h(x) = is continuous wherever
g(x)
g(x) ̸= 0.
Note
The function h is not a rational function. A rational function is
the quotient of two polynomials.
. . . . . .

6. Problem 1.6.20
x3 − 2x + 3 x3 1 − 2/x2 + 3/x3
lim = lim 2 ·
x→∞ 5 − 2x2 x→∞ x 5 /x 2 − 2
1 − 2/x2 + 3/x3
= lim x · lim
x→∞ x→∞ 5 /x 2 − 2
Since the first factor tends to ∞ and the second factor tends to
1
− , the product tends to −∞.
2
Notes
Make sure the “lim” is there in each stage
Do not do arithmetic with ∞ on paper
. . . . . .

7. Explanations
Explanations are getting much better.
Please (continue to) format your papers presentably.
. . . . . .

8. Outline
Grader’s Corner
Derivative of a Product
Derivation
Examples
The Quotient Rule
Derivation
Examples
More derivatives of trigonometric functions
Derivative of Tangent and Cotangent
Derivative of Secant and Cosecant
More on the Power Rule
Power Rule for Positive Integers by Induction
Power Rule for Negative Integers
. . . . . .

9. 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?
. . . . . .

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

11. Is the derivative of a product the product of the
derivatives?
.
. uv)′ = u′ v′ !
(
Try this with u = x and v = x2 .
. . . . . .

12. Is the derivative of a product the product of the
derivatives?
.
. uv)′ = u′ v′ !
(
Try this with u = x and v = x2 .
Then uv = x3 =⇒ (uv)′ = 3x2 .
. . . . . .

13. Is the derivative of a product the product of the
derivatives?
.
. uv)′ = u′ v′ !
(
Try this with u = x and v = x2 .
Then uv = x3 =⇒ (uv)′ = 3x2 .
But u′ v′ = 1 · 2x = 2x.
. . . . . .

14. Is the derivative of a product the product of the
derivatives?
.
. uv)′ = u′ v′ !
(
Try this with u = x and v = x2 .
Then uv = x3 =⇒ (uv)′ = 3x2 .
But u′ v′ = 1 · 2x = 2x.
So we have to be more careful.
. . . . . .

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

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

17. 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.
. .
. . . . . .

18. 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?
. .
. . . . . .

19. 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?
. I = 5 × $0..25 = $1.25?
∆
. . . . . .

20. 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?
. I = 5 × $0..25 = $1.25?
∆
. . . . . .

21. 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
= w · h + w · ∆h + ∆w · h + ∆w · ∆h − wh
= w · ∆h + ∆w · h + ∆w · ∆h
. . . . . .

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

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

24. Supose wages and hours are changing continuously over time.
Over a time interval ∆t, what is the average rate of change of
income?
∆I w ∆h + h ∆w + ∆w ∆h
=
∆t ∆t
∆h ∆w ∆h
=w +h + ∆w
∆t ∆t ∆t
. . . . . .

25. Supose wages and hours are changing continuously over time.
Over a time interval ∆t, what is the average rate of change of
income?
∆I w ∆h + h ∆w + ∆w ∆h
=
∆t ∆t
∆h ∆w ∆h
=w +h + ∆w
∆t ∆t ∆t
What is the instantaneous rate of change of income?
dI ∆I dh dw
= lim =w +h +0
dt ∆t→0 ∆t dt dt
. . . . . .

26. Eurekamen!
We have discovered
Theorem (The Product Rule)
Let u and v be differentiable at x. Then
(uv)′ (x) = u(x)v′ (x) + u′ (x)v(x)
in Leibniz notation
d du dv
(uv) = ·v+u
dx dx dx
. . . . . .

27. Example
Apply the product rule to u = x and v = x2 .
. . . . . .

28. Example
Apply the product rule to u = x and v = x2 .
Solution
(uv)′ (x) = u(x)v′ (x) + u′ (x)v(x) = x · (2x) + 1 · x2 = 3x2
This is what we get the “normal” way.
. . . . . .

29. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d [ ]
(3 − x2 )(x3 − x + 1)
dx
. . . . . .

30. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d [ ]
(3 − x2 )(x3 − x + 1)
dx
Solution
by direct multiplication:
d [ ]FOIL d [ 5 ]
(3 − x2 )(x3 − x + 1) = −x + 4x3 − x2 − 3x + 3
dx dx
. . . . . .

31. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d [ ]
(3 − x2 )(x3 − x + 1)
dx
Solution
by direct multiplication:
d [ ]FOIL d [ 5 ]
(3 − x2 )(x3 − x + 1) = −x + 4x3 − x2 − 3x + 3
dx dx
= −5x4 + 12x2 − 2x − 3
. . . . . .

32. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d [ ]
(3 − x2 )(x3 − x + 1)
dx
Solution
by the product rule:
( ) ( )
dy d 2 3 2 d 3
= (3 − x ) (x − x + 1) + (3 − x ) (x − x + 1)
dx dx dx
. . . . . .

33. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d [ ]
(3 − x2 )(x3 − x + 1)
dx
Solution
by the product rule:
( ) ( )
dy d 2 3 2 d 3
= (3 − x ) (x − x + 1) + (3 − x ) (x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
. . . . . .

34. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d [ ]
(3 − x2 )(x3 − x + 1)
dx
Solution
by the product rule:
( ) ( )
dy d 2 3 2 d 3
= (3 − x ) (x − x + 1) + (3 − x ) (x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
. . . . . .

35. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d [ ]
(3 − x2 )(x3 − x + 1)
dx
Solution
by the product rule:
( ) ( )
dy d 2 3 2 d 3
= (3 − x ) (x − x + 1) + (3 − x ) (x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
. . . . . .

36. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d [ ]
(3 − x2 )(x3 − x + 1)
dx
Solution
by the product rule:
( ) ( )
dy d 2 3 2 d 3
= (3 − x ) (x − x + 1) + (3 − x ) (x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
. . . . . .

37. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d [ ]
(3 − x2 )(x3 − x + 1)
dx
Solution
by the product rule:
( ) ( )
dy d 2 3 2 d 3
= (3 − x ) (x − x + 1) + (3 − x ) (x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
. . . . . .

38. Example
Find this derivative two ways: first by direct multiplication and
then by the product rule:
d [ ]
(3 − x2 )(x3 − x + 1)
dx
Solution
by the product rule:
( ) ( )
dy d 2 3 2 d 3
= (3 − x ) (x − x + 1) + (3 − x ) (x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
= −5x4 + 12x2 − 2x − 3
. . . . . .

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

40. 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
. . . . . .

41. 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
. . . . . .

42. 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
. . . . . .

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

44. Musical interlude
jazz bandleader and
singer
hit song “Minnie the
Moocher” featuring “hi
de ho” chorus
played Curtis in The
Blues Brothers
Cab Calloway
1907–1994
. . . . . .

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

46. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product
uvw.
Solution
(uvw)′ .
. . . . . .

47. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product
uvw.
Solution
(uvw)′ = ((uv)w)′ .
. . . . . .

48. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product
uvw. .
Apply the product rule
to uv and w
Solution
(uvw)′ = ((uv)w)′ .
. . . . . .

49. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product
uvw. .
Apply the product rule
to uv and w
Solution
(uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
. . . . . .

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

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

52. 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′
. . . . . .

53. 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.
. . . . . .

54. Outline
Grader’s Corner
Derivative of a Product
Derivation
Examples
The Quotient Rule
Derivation
Examples
More derivatives of trigonometric functions
Derivative of Tangent and Cotangent
Derivative of Secant and Cosecant
More on the Power Rule
Power Rule for Positive Integers by Induction
Power Rule for Negative Integers
. . . . . .

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

56. 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
. . . . . .

57. 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′
. . . . . .

58. 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′ u v′
=⇒ Q′ = = − ·
v v v v
. . . . . .

59. 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′ u v′
=⇒ Q′ = = − ·
v v v v
( u )′ u′ v − uv′
=⇒ Q′ = =
v v2
. . . . . .

60. 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′ u v′
=⇒ Q′ = = − ·
v v v v
( u )′ u′ v − uv′
=⇒ Q′ = =
v v2
This is called the Quotient Rule.
. . . . . .

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

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

63. Examples
Example
d 2x + 5
1.
dx 3x − 2
d 2x + 1
2.
dx x2 − 1
d t−1
3.
dt t2 + t + 2
. . . . . .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

78. Examples
Example
d 2x + 5
1.
dx 3x − 2
d 2x + 1
2.
dx x2 − 1
d t−1
3.
dt t2 + t + 2
Answers
19
1. −
(3x − 2)2
. . . . . .

79. Solution to second example
d 2x + 1
dx x2 − 1
. . . . . .

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

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

82. Examples
Example
d 2x + 5
1.
dx 3x − 2
d 2x + 1
2.
dx x2 − 1
d t−1
3.
dt t2 + t + 2
Answers
19
1. −
(3x − 2)2
( )
2 x2 + x + 1
2. −
(x2 − 1)2
. . . . . .

83. Solution to third example
d t−1
dt t2 + t + 2
. . . . . .

84. Solution to third example
d t−1 (t2 + t + 2)(1) − (t − 1)(2t + 1)
=
dt t2 + t + 2 (t2 + t + 2)2
. . . . . .

85. Solution to third example
d t−1 (t2 + t + 2)(1) − (t − 1)(2t + 1)
=
dt t2 + t + 2 (t2 + t + 2)2
(t2 + t + 2) − (2t2 − t − 1)
=
(t2 + t + 2)2
−t2 + 2t + 3
= 2
(t + t + 2)2
. . . . . .

86. Examples
Example
d 2x + 5
1.
dx 3x − 2
d 2x + 1
2.
dx x2 − 1
d t−1
3.
dt t2 + t + 2
Answers
19
1. −
(3x − 2)2
( )
2 x2 + x + 1
2. −
(x2 − 1)2
−t2 + 2t + 3
3. 2
(t2 + t + 2)
. . . . . .

87. Mnemonic
Let u = “hi” and v = “lo”. Then
( u )′ vu′ − uv′
= = “lo dee hi minus hi dee lo over lo lo”
v v2
. . . . . .

88. Outline
Grader’s Corner
Derivative of a Product
Derivation
Examples
The Quotient Rule
Derivation
Examples
More derivatives of trigonometric functions
Derivative of Tangent and Cotangent
Derivative of Secant and Cosecant
More on the Power Rule
Power Rule for Positive Integers by Induction
Power Rule for Negative Integers
. . . . . .

89. Derivative of Tangent
Example
d
Find tan x
dx
. . . . . .

90. Derivative of Tangent
Example
d
Find tan x
dx
Solution
( )
d d sin x
tan x =
dx dx cos x
. . . . . .

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

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

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

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

95. Derivative of Cotangent
Example
d
Find cot x
dx
. . . . . .

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

97. Derivative of Cotangent
Example
d
Find cot x
dx
Answer
d 1
cot x = − 2 = − csc2 x
dx sin x
Solution
d d ( cos x ) sin x · (− sin x) − cos x · cos x
cot x = =
dx dx sin x sin2 x
. . . . . .

98. Derivative of Cotangent
Example
d
Find cot x
dx
Answer
d 1
cot x = − 2 = − csc2 x
dx sin x
Solution
d d ( cos x ) sin x · (− sin x) − cos x · cos x
cot x = =
dx dx sin x sin2 x
− sin2 x − cos2 x
=
sin2 x
. . . . . .

99. Derivative of Cotangent
Example
d
Find cot x
dx
Answer
d 1
cot x = − 2 = − csc2 x
dx sin x
Solution
d d ( cos x ) sin x · (− sin x) − cos x · cos x
cot x = =
dx dx sin x sin2 x
− sin2 x − cos2 x 1
= 2
=− 2
sin x sin x
. . . . . .

100. Derivative of Cotangent
Example
d
Find cot x
dx
Answer
d 1
cot x = − 2 = − csc2 x
dx sin x
Solution
d d ( cos x ) sin x · (− sin x) − cos x · cos x
cot x = =
dx dx sin x sin2 x
− sin2 x − cos2 x 1
= 2
= − 2 = − csc2 x
sin x sin x
. . . . . .

101. Derivative of Secant
Example
d
Find sec x
dx
. . . . . .

102. Derivative of Secant
Example
d
Find sec x
dx
Solution
( )
d d 1
sec x =
dx dx cos x
. . . . . .

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

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

105. Derivative of Secant
Example
d
Find sec x
dx
Solution
( )
d d 1 cos x · 0 − 1 · (− sin x)
sec x = =
dx dx cos x cos2 x
sin x 1 sin x
= 2x
= ·
cos cos x cos x
. . . . . .

106. Derivative of Secant
Example
d
Find sec x
dx
Solution
( )
d d 1 cos x · 0 − 1 · (− sin x)
sec x = =
dx dx cos x cos2 x
sin x 1 sin x
= 2x
= · = sec x tan x
cos cos x cos x
. . . . . .

107. Derivative of Cosecant
Example
d
Find csc x
dx
. . . . . .

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

109. Derivative of Cosecant
Example
d
Find csc x
dx
Answer
d
csc x = − csc x cot x
dx
Solution
( )
d d 1 sin x · 0 − 1 · (cos x)
csc x = =
dx dx sin x sin2 x
. . . . . .

110. Derivative of Cosecant
Example
d
Find csc x
dx
Answer
d
csc x = − csc x cot x
dx
Solution
( )
d d 1 sin x · 0 − 1 · (cos x)
csc x = =
dx dx sin x sin2 x
cos x
=− 2
sin x
. . . . . .

111. Derivative of Cosecant
Example
d
Find csc x
dx
Answer
d
csc x = − csc x cot x
dx
Solution
( )
d d 1 sin x · 0 − 1 · (cos x)
csc x = =
dx dx sin x sin2 x
cos x 1 cos x
=− 2 =− ·
sin x sin x sin x
. . . . . .

112. Derivative of Cosecant
Example
d
Find csc x
dx
Answer
d
csc x = − csc x cot x
dx
Solution
( )
d d 1 sin x · 0 − 1 · (cos x)
csc x = =
dx dx sin x sin2 x
cos x 1 cos x
=− 2 =− · = − csc x cot x
sin x sin x sin x
. . . . . .

113. Recap: Derivatives of trigonometric functions
y y′
Functions come in pairs
sin x cos x
(sin/cos, tan/cot, sec/csc)
cos x − sin x Derivatives of pairs
tan x sec x 2 follow similar patterns,
with functions and
cot x − csc2 x co-functions switched
sec x sec x tan x and an extra sign.
csc x − csc x cot x
. . . . . .

114. Outline
Grader’s Corner
Derivative of a Product
Derivation
Examples
The Quotient Rule
Derivation
Examples
More derivatives of trigonometric functions
Derivative of Tangent and Cotangent
Derivative of Secant and Cosecant
More on the Power Rule
Power Rule for Positive Integers by Induction
Power Rule for Negative Integers
. . . . . .

115. Power Rule for Positive Integers by Induction
Theorem
Let n be a positive integer. Then
d n
x = nxn−1
dx
. . . . . .

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

117. Principle of Mathematical Induction
.
Suppose S(1) is
true and S(n + 1)
is true whenever
.
S(n) is true. Then
S(n) is true for all
n.
.
.
Image credit: Kool Skatkat
. . . . . .

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

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

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

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

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

123. 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 d 1
x =
dx dx xn
. . . . . .

124. 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 d 1
x =
dx dx xn
d d n
xn · dx 1 − 1 · dx x
=
x2n
. . . . . .

125. 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 d 1
x =
dx dx xn
d d n
xn · dx 1 − 1 · dx x
=
x2n
0 − nx n −1
=
x2n
. . . . . .

126. 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 d 1
x =
dx dx xn
d d
xn · dx 1 − 1 · dx xn
=
x2n
0 − nx n −1
= = −nx−n−1
x2n
. . . . . .

127. What have we learned today?
The Product Rule: (uv)′ = u′ v + uv′
( u )′ vu′ − uv′
The Quotient Rule: =
v v2
Derivatives of tangent/cotangent, secant/cosecant
d d
tan x = sec2 x sec x = sec x tan x
dx dx
d d
cot x = − csc2 x csc x = − csc x cot x
dx dx
The Power Rule is true for all whole number powers,
including negative powers:
d n
x = nxn−1
dx
. . . . . .