This document provides an overview of the key points covered in a calculus lecture on derivatives of logarithmic and exponential functions: 1) It discusses the derivatives of exponential functions with any base, as well as the derivatives of logarithmic functions with any base. 2) It covers using the technique of logarithmic differentiation to find derivatives of functions involving products, quotients, and/or exponentials. 3) The document provides examples of finding derivatives of various logarithmic and exponential functions.

1. Sec on 3.3
Deriva ves of Logarithmic and
Exponen al Func ons
V63.0121.011: Calculus I
Professor Ma hew Leingang
New York University
March 21, 2011
.

2. Announcements
Quiz 3 next week on 2.6,
2.8, 3.1, 3.2

3. Objectives
Know the deriva ves of the
exponen al func ons (with any
base)
Know the deriva ves of the
logarithmic func ons (with any
base)
Use the technique of logarithmic
differen a on to find deriva ves
of func ons involving roducts,
quo ents, and/or exponen als.

4. Outline
Recall Sec on 3.1–3.2
Deriva ve of the natural exponen al func on
Exponen al Growth
Deriva ve of the natural logarithm func on
Deriva ves of other exponen als and logarithms
Other exponen als
Other logarithms
Logarithmic Differen a on
The power rule for irra onal powers

5. Conventions on power expressions
Let a be a posi ve real number.
If n is a posi ve whole number, then an = a · a · · · · · a
n factors
0
a = 1.
1
For any real number r, a−r = .
ar √
For any posi ve whole number n, a1/n = n a.
There is only one con nuous func on which sa sfies all of the
above. We call it the exponen al func on with base a.

6. Properties of exponentials
Theorem
If a > 0 and a ̸= 1, then f(x) = ax is a con nuous func on with
domain (−∞, ∞) and range (0, ∞). In par cular, ax > 0 for all x.
For any real numbers x and y, and posi ve numbers a and b we have
ax+y = ax ay
x−y ax
a = y
a
(a ) = axy
x y
(ab)x = ax bx

7. Properties of exponentials
Theorem
If a > 0 and a ̸= 1, then f(x) = ax is a con nuous func on with
domain (−∞, ∞) and range (0, ∞). In par cular, ax > 0 for all x.
For any real numbers x and y, and posi ve numbers a and b we have
ax+y = ax ay
x−y ax
a = y (nega ve exponents mean reciprocals)
a
(a ) = axy
x y
(ab)x = ax bx

8. Properties of exponentials
Theorem
If a > 0 and a ̸= 1, then f(x) = ax is a con nuous func on with
domain (−∞, ∞) and range (0, ∞). In par cular, ax > 0 for all x.
For any real numbers x and y, and posi ve numbers a and b we have
ax+y = ax ay
x−y ax
a = y (nega ve exponents mean reciprocals)
a
(a ) = axy (frac onal exponents mean roots)
x y
(ab)x = ax bx

9. Graphs of exponential functions
y
y y =y/=3(1/3)x
= (1(2/x)x
2) y = (1/10y x= 10x 3x = 2x
) y= y y = 1.5x
y = 1x
. x

10. The magic number
Defini on
( )n
1
e = lim 1+ = lim+ (1 + h)1/h
n→∞ n h→0

11. Existence of e
See Appendix B
( )n
1
n 1+
We can experimentally n
verify that this number 1 2
exists and is 2 2.25
3 2.37037
e ≈ 2.718281828459045 . . .
10 2.59374
e is irra onal 100 2.70481
1000 2.71692
e is transcendental
106 2.71828

12. Logarithms
Defini on
The base a logarithm loga x is the inverse of the func on ax
y = loga x ⇐⇒ x = ay
The natural logarithm ln x is the inverse of ex . So
y = ln x ⇐⇒ x = ey .

13. Facts about Logarithms
Facts
(i) loga (x1 · x2 ) = loga x1 + loga x2
( )
x1
(ii) loga = loga x1 − loga x2
x2
(iii) loga (xr ) = r loga x

14. Graphs of logarithmic functions
y
y =x ex
y =y10y3= 2x
= x
y = log2 x
yy= log3 x
= ln x
(0, 1)
y = log10 x
.
(1, 0) x

15. Change of base formula
Fact
If a > 0 and a ̸= 1, and the same for b, then
logb x
loga x =
logb a

16. Upshot of changing base
The point of the change of base formula
logb x 1
loga x = = · logb x = (constant) · logb x
logb a logb a
is that all the logarithmic func ons are mul ples of each other. So
just pick one and call it your favorite.
Engineers like the common logarithm log = log10
Computer scien sts like the binary logarithm lg = log2
Mathema cians like natural logarithm ln = loge
Naturally, we will follow the mathema cians. Just don’t pronounce
it “lawn.”

17. Outline
Recall Sec on 3.1–3.2
Deriva ve of the natural exponen al func on
Exponen al Growth
Deriva ve of the natural logarithm func on
Deriva ves of other exponen als and logarithms
Other exponen als
Other logarithms
Logarithmic Differen a on
The power rule for irra onal powers

18. Derivatives of Exponentials
Fact
If f(x) = ax , then f′ (x) = f′ (0)ax .

19. Derivatives of Exponentials
Fact
If f(x) = ax , then f′ (x) = f′ (0)ax .
Proof.
Follow your nose:
′ f(x + h) − f(x) ax+h − ax
f (x) = lim = lim
h→0 h h→0 h
a a −a
x h x
a −1
h
= lim = ax · lim = ax · f′ (0).
h→0 h h→0 h

20. The funny limit in the case of e
Ques on
eh − 1
What is lim ?
h→0 h
Solu on

21. The funny limit in the case of e
Ques on
eh − 1
What is lim ?
h→0 h
Solu on ( )n
1
Recall e = lim 1 + = lim (1 + h)1/h . If h is small enough,
n→∞ n h→0
e ≈ (1 + h) . So
1/h
[ ]h
eh − 1 (1 + h)1/h − 1 (1 + h) − 1 h
≈ = = =1
h h h h

22. The funny limit in the case of e
Ques on
eh − 1
What is lim ?
h→0 h
Solu on
So in the limit we get equality:
eh − 1
lim =1
h→0 h

23. Derivative of the natural
exponential function
From
( )
d x ah − 1 eh − 1
a = lim ax and lim =1
dx h→0 h h→0 h
we get:
Theorem
d x
e = ex
dx

24. Exponential Growth
Commonly misused term to say something grows exponen ally
It means the rate of change (deriva ve) is propor onal to the
current value
Examples: Natural popula on growth, compounded interest,
social networks

25. Examples
Example
d
Find e3x .
dx

26. Examples
Example
d
Find e3x .
dx
Solu on
d 3x d
e = e3x (3x) = 3e3x
dx dx

27. Examples
Example
d 2
Find ex .
dx

28. Examples
Example
d 2
Find ex .
dx
Solu on
d x2 2 d 2
e = ex (x2 ) = 2xex
dx dx

29. Examples
Example
d
Find x2 ex .
dx

30. Examples
Example
d
Find x2 ex .
dx
Solu on
d 2 x
x e = 2xex + x2 ex
dx

31. Outline
Recall Sec on 3.1–3.2
Deriva ve of the natural exponen al func on
Exponen al Growth
Deriva ve of the natural logarithm func on
Deriva ves of other exponen als and logarithms
Other exponen als
Other logarithms
Logarithmic Differen a on
The power rule for irra onal powers

32. Derivative ofy the natural logarithm
Let y = ln x. Then x = e so

33. Derivative ofy the natural logarithm
Let y = ln x. Then x = e so
dy
ey =1
dx

34. Derivative ofy the natural logarithm
Let y = ln x. Then x = e so
dy
ey =1
dx
dy 1 1
=⇒ = y=
dx e x

35. Derivative ofy the natural logarithm
Let y = ln x. Then x = e so
dy
ey =1
dx
dy 1 1
=⇒ = y=
dx e x
We have discovered:
Fact
d 1
ln x =
dx x

36. Derivative ofy the natural logarithm
Let y = ln x. Then x = e so
y
dy
ey =1
dx
dy 1 1
=⇒ = y= ln x
dx e x
We have discovered: . x
Fact
d 1
ln x =
dx x

37. Derivative ofy the natural logarithm
Let y = ln x. Then x = e so
y
dy
ey =1
dx
dy 1 1
=⇒ = y= ln x
1
dx e x
x
We have discovered: . x
Fact
d 1
ln x =
dx x

38. The Tower of Powers
y y′
x3 3x2
x2 2x1 The deriva ve of a power func on is a
power func on of one lower power
x1 1x0
x0 0
? ?
x−1 −1x−2
x−2 −2x−3

39. The Tower of Powers
y y′
x3 3x2
x2 2x1 The deriva ve of a power func on is a
power func on of one lower power
x1 1x0 Each power func on is the deriva ve of
x0 0 another power func on, except x−1
? x−1
x−1 −1x−2
x−2 −2x−3

40. The Tower of Powers
y y′
x3 3x2
x2 2x1 The deriva ve of a power func on is a
power func on of one lower power
x1 1x0 Each power func on is the deriva ve of
x0 0 another power func on, except x−1
ln x x−1 ln x fills in this gap precisely.
x−1 −1x−2
x−2 −2x−3

41. Examples
Examples
Find deriva ves of these func ons:
ln(3x)
x ln x
√
ln x

42. Examples
Example
d
Find ln(3x).
dx

43. Examples
Example
d
Find ln(3x).
dx
Solu on (chain rule way)
d 1 1
ln(3x) = ·3=
dx 3x x

44. Examples
Example
d
Find ln(3x).
dx
Solu on (proper es of logarithms way)
d d 1 1
ln(3x) = (ln(3) + ln(x)) = 0 + =
dx dx x x
The first answer might be surprising un l you see the second solu on.

45. Examples
Example
d
Find x ln x
dx

46. Examples
Example
d
Find x ln x
dx
Solu on
The product rule is in play here:
( ) ( )
d d d 1
x ln x = x ln x + x ln x = 1 · ln x + x · = ln x + 1
dx dx dx x

47. Examples
Example
d √
Find ln x.
dx

48. Examples
Example
d √
Find ln x.
dx
Solu on (chain rule way)
d √ 1 d√ 1 1 1
ln x = √ x=√ √ =
dx x dx x 2 x 2x

49. Examples
Example
d √
Find ln x.
dx
Solu on (proper es of logarithms way)
( )
d √ d 1 1d 1 1
ln x = ln x = ln x = ·
dx dx 2 2 dx 2 x
The first answer might be surprising un l you see the second solu on.

50. Outline
Recall Sec on 3.1–3.2
Deriva ve of the natural exponen al func on
Exponen al Growth
Deriva ve of the natural logarithm func on
Deriva ves of other exponen als and logarithms
Other exponen als
Other logarithms
Logarithmic Differen a on
The power rule for irra onal powers

51. Other logarithms
Example
d x
Use implicit differen a on to find a.
dx

52. Other logarithms
Example
d x
Use implicit differen a on to find a.
dx
Solu on
Let y = ax , so
ln y = ln ax = x ln a

53. Other logarithms
Example
d x
Use implicit differen a on to find a.
dx
Solu on
Let y = ax , so
ln y = ln ax = x ln a
Differen ate implicitly:
1 dy dy
= ln a =⇒ = (ln a)y = (ln a)ax
y dx dx

54. The funny limit in the case of a
x ′ ′
Let y = e . Before we showed y = y (0)y, and now we know
y′ = (ln a)y. So
Corollary
ah − 1
lim = ln a
h→0 h
In par cular
2h − 1 3h − 1
ln 2 = lim ≈ 0.693 ln 3 = lim ≈ 1.10
h→0 h h→0 h

55. Other logarithms
Example
d
Find log x.
dx a

56. Other logarithms
Example
d
Find log x.
dx a
Solu on
Let y = loga x, so ay = x.

57. Other logarithms
Example
d
Find log x.
dx a
Solu on
Let y = loga x, so ay = x. Now differen ate implicitly:
dy dy 1 1
(ln a)ay = 1 =⇒ = y =
dx dx a ln a x ln a

58. Other logarithms
Example
d
Find log x.
dx a
Solu on
Or we can use the change of base formula:
ln x dy 1 1
y= =⇒ =
ln a dx ln a x

59. More examples
Example
d
Find log2 (x2 + 1)
dx

60. More examples
Example
d
Find log2 (x2 + 1)
dx
Answer
dy 1 1 2x
= (2x) =
dx ln 2 x2 + 1 (ln 2)(x2 + 1)

61. Outline
Recall Sec on 3.1–3.2
Deriva ve of the natural exponen al func on
Exponen al Growth
Deriva ve of the natural logarithm func on
Deriva ves of other exponen als and logarithms
Other exponen als
Other logarithms
Logarithmic Differen a on
The power rule for irra onal powers

62. A nasty derivative
Example
√
(x2 + 1) x + 3
Let y = . Find y′ .
x−1

63. A nasty derivative
Example
√
(x2 + 1) x + 3
Let y = . Find y′ .
x−1
Solu on
We use the quo ent rule, and the product rule in the numerator:
[ √ ] √
(x − 1) 2x x + 3 + (x2 + 1) 1 (x + 3)−1/2 − (x2 + 1) x + 3(1)
y′ = 2
(x − 1)2
√ √
2x x + 3 (x2 + 1) (x2 + 1) x + 3
= + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2

64. Another way
√
(x2 + 1) x + 3
y=
x−1

65. Another way
√
(x2 + 1) x + 3
y=
x−1
1
ln y = ln(x2 + 1) + ln(x + 3) − ln(x − 1)
2

66. Another way
√
(x2 + 1) x + 3
y=
x−1
1
ln y = ln(x2 + 1) + ln(x + 3) − ln(x − 1)
2
1 dy 2x 1 1
= 2 + −
y dx x + 1 2(x + 3) x − 1

67. Another way
√
(x2 + 1) x + 3
y=
x−1
1
ln y = ln(x2 + 1) + ln(x + 3) − ln(x − 1)
2
1 dy 2x 1 1
= 2 + −
y dx x + 1 2(x + 3) x − 1
So
( )
dy 2x 1 1
= + − y
dx x2 + 1 2(x + 3) x − 1

68. Another way
√
(x2 + 1) x + 3
y=
x−1
1
ln y = ln(x2 + 1) + ln(x + 3) − ln(x − 1)
2
1 dy 2x 1 1
= 2 + −
y dx x + 1 2(x + 3) x − 1
So
( ) √
dy 2x 1 1 (x2 + 1) x + 3
= + −
dx x 2+1 2(x + 3) x − 1 x−1

69. Compare and contrast
Using the product, quo ent, and power rules:
√ √
2x x + 3 (x2 + 1) (x2 + 1) x + 3
y′ = + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2
Using logarithmic differen a on:
( ) 2 √
′ 2x 1 1 (x + 1) x + 3
y = + −
x2 + 1 2(x + 3) x − 1 (x − 1)

70. Compare and contrast
Using the product, quo ent, and power rules:
√ √
2x x + 3 (x2 + 1) (x2 + 1) x + 3
y′ = + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2
Using logarithmic differen a on:
( ) 2 √
′ 2x 1 1 (x + 1) x + 3
y = + −
x2 + 1 2(x + 3) x − 1 (x − 1)
Are these the same?

71. Compare and contrast
Using the product, quo ent, and power rules:
√ √
2x x + 3 (x2 + 1) (x2 + 1) x + 3
y′ = + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2
Using logarithmic differen a on:
( ) 2 √
′ 2x 1 1 (x + 1) x + 3
y = + −
x2 + 1 2(x + 3) x − 1 (x − 1)
Are these the same?

72. Compare and contrast
Using the product, quo ent, and power rules:
√ √
2x x + 3 (x2 + 1) (x2 + 1) x + 3
y′ = + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2
Using logarithmic differen a on:
( ) 2 √
′ 2x 1 1 (x + 1) x + 3
y = + −
x2 + 1 2(x + 3) x − 1 (x − 1)
Are these the same?

73. Compare and contrast
Using the product, quo ent, and power rules:
√ √
2x x + 3 (x2 + 1) (x2 + 1) x + 3
y′ = + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2
Using logarithmic differen a on:
( ) 2 √
′ 2x 1 1 (x + 1) x + 3
y = + −
x2 + 1 2(x + 3) x − 1 (x − 1)
Are these the same?

74. Compare and contrast
Using the product, quo ent, and power rules:
√ √
2x x + 3 (x2 + 1) (x2 + 1) x + 3
y′ = + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2
Using logarithmic differen a on:
( ) 2 √
′ 2x 1 1 (x + 1) x + 3
y = + −
x2 + 1 2(x + 3) x − 1 (x − 1)
Are these the same?

75. Compare and contrast
Using the product, quo ent, and power rules:
√ √
2x x + 3 (x2 + 1) (x2 + 1) x + 3
y′ = + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2
Using logarithmic differen a on:
( ) 2 √
′ 2x 1 1 (x + 1) x + 3
y = + −
x2 + 1 2(x + 3) x − 1 (x − 1)
Are these the same?

76. Compare and contrast
Using the product, quo ent, and power rules:
√ √
2x x + 3 (x2 + 1) (x2 + 1) x + 3
y′ = + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2
Using logarithmic differen a on:
( ) 2 √
′ 2x 1 1 (x + 1) x + 3
y = + −
x2 + 1 2(x + 3) x − 1 (x − 1)
Are these the same?

77. Compare and contrast
Using the product, quo ent, and power rules:
√ √
2x x + 3 (x2 + 1) (x2 + 1) x + 3
y′ = + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2
Using logarithmic differen a on:
( ) 2 √
′ 2x 1 1 (x + 1) x + 3
y = + −
x2 + 1 2(x + 3) x − 1 (x − 1)
Are these the same?

78. Compare and contrast
Using the product, quo ent, and power rules:
√ √
2x x + 3 (x2 + 1) (x2 + 1) x + 3
y′ = + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2
Using logarithmic differen a on:
( ) 2 √
′ 2x 1 1 (x + 1) x + 3
y = + −
x2 + 1 2(x + 3) x − 1 (x − 1)
Are these the same?

79. Compare and contrast
Using the product, quo ent, and power rules:
√ √
2x x + 3 (x2 + 1) (x2 + 1) x + 3
y′ = + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2
Using logarithmic differen a on:
( ) 2 √
′ 2x 1 1 (x + 1) x + 3
y = + −
x2 + 1 2(x + 3) x − 1 (x − 1)
Are these the same?

80. Compare and contrast
Using the product, quo ent, and power rules:
√ √
2x x + 3 (x2 + 1) (x2 + 1) x + 3
y′ = + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2
Using logarithmic differen a on:
( ) 2 √
′ 2x 1 1 (x + 1) x + 3
y = + −
x2 + 1 2(x + 3) x − 1 (x − 1)
Are these the same?

81. Compare and contrast
Using the product, quo ent, and power rules:
√ √
2x x + 3 (x2 + 1) (x2 + 1) x + 3
y′ = + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2
Using logarithmic differen a on:
( ) 2 √
′ 2x 1 1 (x + 1) x + 3
y = + −
x2 + 1 2(x + 3) x − 1 (x − 1)
Are these the same?

82. Compare and contrast
Using the product, quo ent, and power rules:
√ √
2x x + 3 (x2 + 1) (x2 + 1) x + 3
y′ = + √ −
(x − 1) 2 x + 3(x − 1) (x − 1)2
Using logarithmic differen a on:
( ) 2 √
′ 2x 1 1 (x + 1) x + 3
y = + −
x2 + 1 2(x + 3) x − 1 (x − 1)
Are these the same? Yes.

83. Derivatives of powers
Ques on y
Let y = xx . Which of these is true?
(A) Since y is a power func on,
y′ = x · xx−1 = xx .
(B) Since y is an exponen al 1
func on, y′ = (ln x) · xx
.
(C) Neither x
1

84. Derivatives of powers
Ques on y
Let y = xx . Which of these is true?
(A) Since y is a power func on,
y′ = x · xx−1 = xx .
(B) Since y is an exponen al 1
func on, y′ = (ln x) · xx
.
(C) Neither x
1

85. Why not?
Answer y
(A) y′ ̸= xx because xx > 0 for all
x > 0, and this func on
decreases at some places
1
.
x
1

86. Why not?
Answer y
(A) y′ ̸= xx because xx > 0 for all
x > 0, and this func on
decreases at some places
(B) y′ ̸= (ln x)xx because (ln x)xx = 0
when x = 1, and this func on 1
does not have a horizontal .
tangent at x = 1. x
1

87. It’s neither!
Solu on
If y = xx , then

88. It’s neither!
Solu on
If y = xx , then
ln y = x ln x

89. It’s neither!
Solu on
If y = xx , then
ln y = x ln x
1 dy 1
= x · + ln x = 1 + ln x
y dx x

90. It’s neither!
Solu on
If y = xx , then
ln y = x ln x
1 dy 1
= x · + ln x = 1 + ln x
y dx x
dy
= (1 + ln x)xx = xx + (ln x)xx
dx

91. Or both?
Solu on
d x y
x = xx + (ln x)xx = (1 + ln x)xx
dx
1
.
x
1

92. Or both?
Solu on
d x y
x = xx + (ln x)xx = (1 + ln x)xx
dx
Remarks
Each of these terms is one of the 1
wrong answers! .
x
1

93. Or both?
Solu on
d x y
x = xx + (ln x)xx = (1 + ln x)xx
dx
Remarks
Each of these terms is one of the 1
wrong answers! .
x
1

94. Or both?
Solu on
d x y
x = xx + (ln x)xx = (1 + ln x)xx
dx
Remarks
Each of these terms is one of the 1
wrong answers! .
x
1

95. Or both?
Solu on
d x y
x = xx + (ln x)xx = (1 + ln x)xx
dx
Remarks
Each of these terms is one of the 1
wrong answers! .
x
1

96. Or both?
Solu on
d x y
x = xx + (ln x)xx = (1 + ln x)xx
dx
Remarks
Each of these terms is one of the 1
wrong answers! .
y′ < 0 on the interval (0, e−1 ) x
1

97. Or both?
Solu on
d x y
x = xx + (ln x)xx = (1 + ln x)xx
dx
Remarks
Each of these terms is one of the 1
wrong answers! .
y′ < 0 on the interval (0, e−1 ) x
1
y′ = 0 when x = e−1

98. Derivatives of power functions
with any exponent
Fact (The power rule)
Let y = xr . Then y′ = rxr−1 .

99. Derivatives of power functions
with any exponent
Fact (The power rule)
Let y = xr . Then y′ = rxr−1 .
Proof.
y = xr =⇒ ln y = r ln x
Now differen ate:
1 dy r dy y
= =⇒ = r = rxr−1
y dx x dx x

100. Summary
Deriva ves of y y′
Logarithmic and
Exponen al Func ons ex ex
Logarithmic
Differen a on can allow ax (ln a) · ax
us to avoid the product 1
and quo ent rules. ln x
x
We are finally done with 1 1
loga x ·
the Power Rule! ln a x