The derivative of a sum of functions is the sum of the derivatives of those functions, but the derivative of a product or a quotient of functions is not so simple. We'll derive and use the product and quotient rule for these purposes. It will allow us to find the derivatives of other trigonometric functions, and derivatives of power functions with negative whole number exponents.

1. Section 2.4
The Product and Quotient Rules
V63.0121.021, Calculus I
New York University
October 5, 2010
Announcements
Quiz 2 next week on §§1.5, 1.6, 2.1, 2.2
Midterm in class (covers all sections up to 2.5)
. . . . . . .

2. Announcements
Quiz 2 next week on §§1.5,
1.6, 2.1, 2.2
Midterm in class (covers all
sections up to 2.5)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 2 / 41

3. Help!
Free resources:
Math Tutoring Center
(CIWW 524)
College Learning Center
(schedule on Blackboard)
TAs’ office hours
my office hours
each other!
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 3 / 41

4. Objectives
Understand and be able to
use the Product Rule for
the derivative of the
product of two functions.
Understand and be able to
use the Quotient Rule for
the derivative of the
quotient of two functions.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 4 / 41

5. Outline
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 Negative Integers
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 5 / 41

6. 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?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 6 / 41

7. Is the derivative of a product the product of the
derivatives?
. uv)′ = u′ v′ ?
( .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 7 / 41

8. Is the derivative of a product the product of the
derivatives?
. uv)′ = u′ v′ !
( .
Try this with u = x and v = x2 .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 7 / 41

9. 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 .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 7 / 41

10. 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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 7 / 41

11. 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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 7 / 41

12. Mmm...burgers
Say you work in a fast-food joint. You want to make more money.
What are your choices?
. .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 8 / 41

13. Mmm...burgers
Say you work in a fast-food joint. You want to make more money.
What are your choices?
Work longer hours.
. .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 8 / 41

14. 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.
. .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 8 / 41

15. 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?
. .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 8 / 41

16. 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?
∆
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 8 / 41

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.
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?
∆
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 8 / 41

18. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 9 / 41

19. A geometric argument
Draw a box:
. h
∆ w
. ∆h . w ∆h
∆
h
. w
. h . wh
∆
.
w
. . w
∆
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 10 / 41

20. A geometric argument
Draw a box:
. h
∆ w
. ∆h . w ∆h
∆
h
. w
. h . wh
∆
.
w
. . w
∆
∆I = w ∆h + h ∆w + ∆w ∆h
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 10 / 41

21. Cash flow
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 11 / 41

22. Cash flow
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 11 / 41

23. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 12 / 41

24. Sanity Check
Example
Apply the product rule to u = x and v = x2 .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 13 / 41

25. Sanity Check
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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 13 / 41

26. Which is better?
Example
Find this derivative two ways: first by direct multiplication and then by
the product rule:
d [ ]
(3 − x2 )(x3 − x + 1)
dx
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41

27. Which is better?
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
[ ]
(3 − x2 )(x3 − x + 1) = −x5 + 4x3 − x2 − 3x + 3
dx dx
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41

28. Which is better?
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
[ ]
(3 − x2 )(x3 − x + 1) = −x5 + 4x3 − x2 − 3x + 3
dx dx
= −5x4 + 12x2 − 2x − 3
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41

29. Which is better?
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 d 3
= (3 − x ) (x − x + 1) + (3 − x )
2 3 2
(x − x + 1)
dx dx dx
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41

30. Which is better?
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 d 3
= (3 − x ) (x − x + 1) + (3 − x )
2 3 2
(x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41

31. Which is better?
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 d 3
= (3 − x ) (x − x + 1) + (3 − x )
2 3 2
(x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41

32. Which is better?
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 d 3
= (3 − x ) (x − x + 1) + (3 − x )
2 3 2
(x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41

33. Which is better?
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 d 3
= (3 − x ) (x − x + 1) + (3 − x )
2 3 2
(x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41

34. Which is better?
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 d 3
= (3 − x ) (x − x + 1) + (3 − x )
2 3 2
(x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41

35. Which is better?
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 d 3
= (3 − x ) (x − x + 1) + (3 − x )
2 3 2
(x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
= −5x4 + 12x2 − 2x − 3
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 14 / 41

36. One more
Example
d
Find x sin x.
dx
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 15 / 41

37. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 15 / 41

38. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 15 / 41

39. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 15 / 41

40. Mnemonic
Let u = “hi” and v = “ho”. Then
(uv)′ = vu′ + uv′ = “ho dee hi plus hi dee ho”
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 16 / 41

41. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 17 / 41

42. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product uvw.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 18 / 41

43. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product uvw.
Solution
(uvw)′ .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 18 / 41

44. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product uvw.
Solution
(uvw)′ = ((uv)w)′ .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 18 / 41

45. Iterating the Product Rule
Example
Use the product rule to. find the derivative of a three-fold product uvw.
Apply the product rule
Solution to uv and w
(uvw)′ = ((uv)w)′ .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 18 / 41

46. Iterating the Product Rule
Example
Use the product rule to. find the derivative of a three-fold product uvw.
Apply the product rule
Solution to uv and w
(uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 18 / 41

47. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product uvw.
.
Apply the product rule
Solution to u and v
(uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 18 / 41

48. Iterating the Product Rule
Example
Use the product rule to find the derivative of a three-fold product uvw.
.
Apply the product rule
Solution to u and v
(uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
= (u′ v + uv′ )w + (uv)w′
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 18 / 41

49. 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′
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 18 / 41

50. 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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 18 / 41

51. Outline
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 Negative Integers
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 19 / 41

52. The Quotient Rule
What about the derivative of a quotient?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 20 / 41

53. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 20 / 41

54. 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′
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 20 / 41

55. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 20 / 41

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
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 20 / 41

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′
u′ − Qv′ u′ u v′
=⇒ Q′ = = − ·
v v v v
( u )′ u′ v − uv′
=⇒ Q′ = =
v v2
This is called the Quotient Rule.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 20 / 41

58. The Quotient Rule
We have discovered
Theorem (The Quotient Rule)
u
Let u and v be differentiable at x, and v(x) ̸= 0. Then is differentiable
v
at x, and
( u )′ u′ (x)v(x) − u(x)v′ (x)
(x) =
v v(x)2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 21 / 41

59. Verifying Example
Example
( )
d x2
Verify the quotient rule by computing and comparing it to
dx x
d
(x).
dx
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 22 / 41

60. Verifying Example
Example
( )
d x2
Verify the quotient rule by computing and comparing it to
dx x
d
(x).
dx
Solution
( ) ( )
d x2 x dx x2 − x2 dx (x)
d d
=
dx x x2
x · 2x − x2 · 1
=
x2
x2 d
= 2 =1= (x)
x dx
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 22 / 41

61. Mnemonic
Let u = “hi” and v = “lo”. Then
( u )′ vu′ − uv′
= = “lo dee hi minus hi dee lo over lo lo”
v v2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 23 / 41

62. Examples
Example
d 2x + 5
1.
dx 3x − 2
d sin x
2.
dx x2
d 1
3. 2
dt t + t + 2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 24 / 41

63. Solution to first example
Solution
d 2x + 5
dx 3x − 2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41

64. Solution to first example
Solution
d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d d
=
dx 3x − 2 (3x − 2)2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41

65. Solution to first example
Solution
d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d d
=
dx 3x − 2 (3x − 2)2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41

66. Solution to first example
Solution
d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d d
=
dx 3x − 2 (3x − 2)2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41

67. Solution to first example
Solution
d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d d
=
dx 3x − 2 (3x − 2)2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41

68. Solution to first example
Solution
d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d d
=
dx 3x − 2 (3x − 2)2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41

69. Solution to first example
Solution
d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d d
=
dx 3x − 2 (3x − 2)2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41

70. Solution to first example
Solution
d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d d
=
dx 3x − 2 (3x − 2)2
(3x − 2)(2) − (2x + 5)(3)
=
(3x − 2)2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41

71. Solution to first example
Solution
d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d d
=
dx 3x − 2 (3x − 2)2
(3x − 2)(2) − (2x + 5)(3)
=
(3x − 2)2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41

72. Solution to first example
Solution
d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d d
=
dx 3x − 2 (3x − 2)2
(3x − 2)(2) − (2x + 5)(3)
=
(3x − 2)2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41

73. Solution to first example
Solution
d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d d
=
dx 3x − 2 (3x − 2)2
(3x − 2)(2) − (2x + 5)(3)
=
(3x − 2)2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41

74. Solution to first example
Solution
d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d d
=
dx 3x − 2 (3x − 2)2
(3x − 2)(2) − (2x + 5)(3)
=
(3x − 2)2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41

75. Solution to first example
Solution
d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d d
=
dx 3x − 2 (3x − 2)2
(3x − 2)(2) − (2x + 5)(3)
=
(3x − 2)2
(6x − 4) − (6x + 15)
=
(3x − 2)2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41

76. Solution to first example
Solution
d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
d d
=
dx 3x − 2 (3x − 2)2
(3x − 2)(2) − (2x + 5)(3)
=
(3x − 2)2
(6x − 4) − (6x + 15) 19
= =−
(3x − 2) 2 (3x − 2)2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 25 / 41

77. Examples
Example Answers
d 2x + 5 19
1. 1. −
dx 3x − 2 (3x − 2)2
d sin x
2.
dx x2
d 1
3. 2
dt t + t + 2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 26 / 41

78. Solution to second example
Solution
d sin x
=
dx x2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41

79. Solution to second example
Solution
d sin x x2
=
dx x2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41

80. Solution to second example
Solution
d sin x x2 d sin x
= dx
dx x2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41

81. Solution to second example
Solution
d sin x x2 d sin x − sin x
= dx
dx x2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41

82. Solution to second example
Solution
d sin x x2 d sin x − sin x dx x2
d
= dx
dx x2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41

83. Solution to second example
Solution
d sin x x2 d sin x − sin x dx x2
d
= dx
dx x2 (x2 )2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41

84. Solution to second example
Solution
d sin x x2 d sin x − sin x dx x2
d
= dx
dx x2 (x2 )2
=
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41

85. Solution to second example
Solution
d sin x x2 d sin x − sin x dx x2
d
= dx
dx x2 (x2 )2
x2
=
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41

86. Solution to second example
Solution
d sin x x2 d sin x − sin x dx x2
d
= dx
dx x2 (x2 )2
x2 cos x
=
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41

87. Solution to second example
Solution
d sin x x2 d sin x − sin x dx x2
d
= dx
dx x2 (x2 )2
x2 cos x − 2x
=
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41

88. Solution to second example
Solution
d sin x x2 d sin x − sin x dx x2
d
= dx
dx x2 (x2 )2
x2 cos x − 2x sin x
=
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41

89. Solution to second example
Solution
d sin x x2 d sin x − sin x dx x2
d
= dx
dx x2 (x2 )2
x2 cos x − 2x sin x
=
x4
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41

90. Solution to second example
Solution
d sin x x2 d sin x − sin x dx x2
d
= dx
dx x2 (x2 )2
x2 cos x − 2x sin x
=
x4
x cos x − 2 sin x
=
x3
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 27 / 41

91. Another way to do it
Find the derivative with the product rule instead.
Solution
d sin x d ( )
= sin x · x−2
dx x2 dx
( ) ( )
d −2 d −2
= sin x · x + sin x · x
dx dx
= cos x · x−2 + sin x · (−2x−3 )
= x−3 (x cos x − 2 sin x)
Notice the technique of factoring out the largest negative power,
leaving positive powers.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 28 / 41

92. Examples
Example Answers
d 2x + 5 19
1. 1. −
dx 3x − 2 (3x − 2)2
d sin x x cos x − 2 sin x
2. 2.
dx x2 x3
d 1
3. 2
dt t + t + 2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 29 / 41

93. Solution to third example
Solution
d 1
dt t2 + t + 2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 30 / 41

94. Solution to third example
Solution
d 1 (t2 + t + 2)(0) − (1)(2t + 1)
=
dt t2 + t + 2 (t2 + t + 2)2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 30 / 41

95. Solution to third example
Solution
d 1 (t2 + t + 2)(0) − (1)(2t + 1)
=
dt t2 + t + 2 (t2 + t + 2)2
2t + 1
=− 2
(t + t + 2)2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 30 / 41

96. A nice little takeaway
Fact
1
Let v be differentiable at x, and v(x) ̸= 0. Then is differentiable at 0,
v
and ( )′
1 v′
=− 2
v v
Proof.
( )
d 1 v· d
dx (1) −1· d
dx v v · 0 − 1 · v′ v′
= = =− 2
dx v v2 v2 v
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 31 / 41

97. Examples
Example Answers
d 2x + 5 19
1. 1. −
dx 3x − 2 (3x − 2)2
d sin x x cos x − 2 sin x
2. 2.
dx x2 x3
d 1 2t + 1
3. 2
3. − 2
dt t + t + 2 (t + t + 2)2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 32 / 41

98. Outline
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 Negative Integers
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 33 / 41

99. Derivative of Tangent
Example
d
Find tan x
dx
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 34 / 41

100. Derivative of Tangent
Example
d
Find tan x
dx
Solution
( )
d d sin x
tan x =
dx dx cos x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 34 / 41

101. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 34 / 41

102. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 34 / 41

103. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 34 / 41

104. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 34 / 41

105. Derivative of Cotangent
Example
d
Find cot x
dx
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 35 / 41

106. Derivative of Cotangent
Example
d
Find cot x
dx
Answer
d 1
cot x = − 2 = − csc2 x
dx sin x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 35 / 41

107. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 35 / 41

108. 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
2
− sin x − cos2 x
=
sin2 x . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 35 / 41

109. 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
2
− sin x − cos2 x 1
= 2
=− 2
sin x sin x . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 35 / 41

110. 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
2
− sin x − cos2 x 1
= 2
= − 2 = − csc2 x
sin x sin x . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 35 / 41

111. Derivative of Secant
Example
d
Find sec x
dx
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 36 / 41

112. Derivative of Secant
Example
d
Find sec x
dx
Solution
( )
d d 1
sec x =
dx dx cos x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 36 / 41

113. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 36 / 41

114. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 36 / 41

115. 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
= = ·
cos2 x cos x cos x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 36 / 41

116. 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
= = · = sec x tan x
cos2 x cos x cos x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 36 / 41

117. Derivative of Cosecant
Example
d
Find csc x
dx
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 37 / 41

118. Derivative of Cosecant
Example
d
Find csc x
dx
Answer
d
csc x = − csc x cot x
dx
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 37 / 41

119. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 37 / 41

120. 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 . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 37 / 41

121. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 37 / 41

122. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 37 / 41

123. Recap: Derivatives of trigonometric functions
y y′
sin x cos x
Functions come in pairs
cos x − sin x (sin/cos, tan/cot, sec/csc)
tan x sec2 x Derivatives of pairs follow
similar patterns, with
cot x − csc2 x functions and co-functions
sec x sec x tan x switched and an extra sign.
csc x − csc x cot x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 38 / 41

124. Outline
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 Negative Integers
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 39 / 41

125. Power Rule for Negative Integers
We will use the quotient rule to prove
Theorem
d −n
x = (−n)x−n−1
dx
for positive integers n.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 40 / 41

126. Power Rule for Negative Integers
We will use the quotient rule to prove
Theorem
d −n
x = (−n)x−n−1
dx
for positive integers n.
Proof.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 40 / 41

127. Power Rule for Negative Integers
We will 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 40 / 41

128. Power Rule for Negative Integers
We will use the quotient rule to prove
Theorem
d −n
x = (−n)x−n−1
dx
for positive integers n.
Proof.
d n
d −n d 1 x
x = n
= − dxn 2
dx dx x (x )
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 40 / 41

129. Power Rule for Negative Integers
We will use the quotient rule to prove
Theorem
d −n
x = (−n)x−n−1
dx
for positive integers n.
Proof.
d n
d −n d 1 x
x = n
= − dxn 2
dx dx x (x )
nxn−1
=−
x2n
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 40 / 41

130. Power Rule for Negative Integers
We will use the quotient rule to prove
Theorem
d −n
x = (−n)x−n−1
dx
for positive integers n.
Proof.
d n
d −n d 1 x
x = n
= − dxn 2
dx dx x (x )
nxn−1
=− = −nxn−1−2n
x2n
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 40 / 41

131. Power Rule for Negative Integers
We will use the quotient rule to prove
Theorem
d −n
x = (−n)x−n−1
dx
for positive integers n.
Proof.
d n
d −n d 1 x
x = n
= − dxn 2
dx dx x (x )
nxn−1
=− = −nxn−1−2n = −nx−n−1
x2n
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 40 / 41

132. Summary
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 41 / 41