We show the the derivative of the exponential function is itself! And the derivative of the natural logarithm function is the reciprocal function. We also show how logarithms can make complicated differentiation problems easier.
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Lesson 16: Derivatives of Logarithmic and Exponential Functions
1. Section 3.3
Derivatives of Exponential and
Logarithmic Functions
V63.0121.034, Calculus I
October 21, 2009
Announcements
. .
Image credit: heipei
. . . . . .
2. Outline
Derivative of the natural exponential function
Exponential Growth
Derivative of the natural logarithm function
Derivatives of other exponentials and logarithms
Other exponentials
Other logarithms
Logarithmic Differentiation
The power rule for irrational powers
. . . . . .
4. Derivatives of Exponential Functions
Fact
If f(x) = ax , then fβ² (x) = fβ² (0)ax .
Proof.
Follow your nose:
f(x + h) β f(x) a x+ h β a x
fβ² (x) = lim = lim
hβ0 h hβ0 h
a x a h β ax a hβ1
= lim = ax Β· lim = ax Β· fβ² (0).
hβ0 h hβ0 h
. . . . . .
5. Derivatives of Exponential Functions
Fact
If f(x) = ax , then fβ² (x) = fβ² (0)ax .
Proof.
Follow your nose:
f(x + h) β f(x) a x+ h β a x
fβ² (x) = lim = lim
hβ0 h hβ0 h
a x a h β ax a hβ1
= lim = ax Β· lim = ax Β· fβ² (0).
hβ0 h hβ0 h
To reiterate: the derivative of an exponential function is a
constant times that function. Much different from polynomials!
. . . . . .
6. The funny limit in the case of e
Remember the deο¬nition of e:
( )
1 n
e = lim 1 + = lim (1 + h)1/h
nββ n hβ0
Question
eh β 1
What is lim ?
hβ0 h
. . . . . .
7. The funny limit in the case of e
Remember the deο¬nition of e:
( )
1 n
e = lim 1 + = lim (1 + h)1/h
nββ n hβ0
Question
eh β 1
What is lim ?
hβ0 h
Answer
If h is small enough, e β (1 + h)1/h . So
[ ]h
eh β 1 (1 + h)1/h β 1 (1 + h) β 1 h
β = = =1
h h h h
. . . . . .
8. The funny limit in the case of e
Remember the deο¬nition of e:
( )
1 n
e = lim 1 + = lim (1 + h)1/h
nββ n hβ0
Question
eh β 1
What is lim ?
hβ0 h
Answer
If h is small enough, e β (1 + h)1/h . So
[ ]h
eh β 1 (1 + h)1/h β 1 (1 + h) β 1 h
β = = =1
h h h h
eh β 1
So in the limit we get equality: lim =1
hβ0 h
. . . . . .
10. Exponential Growth
Commonly misused term to say something grows
exponentially
It means the rate of change (derivative) is proportional to the
current value
Examples: Natural population growth, compounded interest,
social networks
. . . . . .
11. Examples
Examples
Find these derivatives:
e3x
2
ex
x 2 ex
. . . . . .
12. Examples
Examples
Find these derivatives:
e3x
2
ex
x 2 ex
Solution
d 3x
e = 3e3x
dx
. . . . . .
13. Examples
Examples
Find these derivatives:
e3x
2
ex
x 2 ex
Solution
d 3x
e = 3e3x
dx
d x2 2 d 2
e = ex (x2 ) = 2xex
dx dx
. . . . . .
14. Examples
Examples
Find these derivatives:
e3x
2
ex
x 2 ex
Solution
d 3x
e = 3e3x
dx
d x2 2 d 2
e = ex (x2 ) = 2xex
dx dx
d 2 x
x e = 2xex + x2 ex
dx
. . . . . .
15. Outline
Derivative of the natural exponential function
Exponential Growth
Derivative of the natural logarithm function
Derivatives of other exponentials and logarithms
Other exponentials
Other logarithms
Logarithmic Differentiation
The power rule for irrational powers
. . . . . .
22. The Tower of Powers
y yβ²
The derivative of a
x3 3x2 power function is a
power function of one
x2 2x1
lower power
x1 1x0
x0 0
? ?
xβ1 β1xβ2
xβ2 β2xβ3
. . . . . .
23. The Tower of Powers
y yβ²
The derivative of a
x3 3x2 power function is a
power function of one
x2 2x1
lower power
x1 1x0 Each power function is
x 0
0 the derivative of another
power function, except
? x β1 xβ1
xβ1 β1xβ2
xβ2 β2xβ3
. . . . . .
24. The Tower of Powers
y yβ²
The derivative of a
x3 3x2 power function is a
power function of one
x2 2x1
lower power
x1 1x0 Each power function is
x 0
0 the derivative of another
power function, except
ln x x β1 xβ1
xβ1 β1xβ2 ln x ο¬lls in this gap
precisely.
xβ2 β2xβ3
. . . . . .
25. Outline
Derivative of the natural exponential function
Exponential Growth
Derivative of the natural logarithm function
Derivatives of other exponentials and logarithms
Other exponentials
Other logarithms
Logarithmic Differentiation
The power rule for irrational powers
. . . . . .
26. Other logarithms
Example
d x
Use implicit differentiation to ο¬nd a.
dx
. . . . . .
27. Other logarithms
Example
d x
Use implicit differentiation to ο¬nd a.
dx
Solution
Let y = ax , so
ln y = ln ax = x ln a
. . . . . .
28. Other logarithms
Example
d x
Use implicit differentiation to ο¬nd a.
dx
Solution
Let y = ax , so
ln y = ln ax = x ln a
Differentiate implicitly:
1 dy dy
= ln a =β = (ln a)y = (ln a)ax
y dx dx
. . . . . .
29. Other logarithms
Example
d x
Use implicit differentiation to ο¬nd a.
dx
Solution
Let y = ax , so
ln y = ln ax = x ln a
Differentiate implicitly:
1 dy dy
= ln a =β = (ln a)y = (ln a)ax
y dx dx
Before we showed yβ² = yβ² (0)y, so now we know that
2h β 1 3h β 1
ln 2 = lim β 0.693 ln 3 = lim β 1.10
hβ0 h hβ0 h
. . . . . .
31. Other logarithms
Example
d
Find loga x.
dx
Solution
Let y = loga x, so ay = x.
. . . . . .
32. Other logarithms
Example
d
Find loga x.
dx
Solution
Let y = loga x, so ay = x. Now differentiate implicitly:
dy dy 1 1
(ln a)ay = 1 =β = y =
dx dx a ln a x ln a
. . . . . .
33. Other logarithms
Example
d
Find loga x.
dx
Solution
Let y = loga x, so ay = x. Now differentiate implicitly:
dy dy 1 1
(ln a)ay = 1 =β = y =
dx dx a ln a x ln a
Another way to see this is to take the natural logarithm:
ln x
ay = x =β y ln a = ln x =β y =
ln a
dy 1 1
So = .
dx ln a x
. . . . . .
35. More examples
Example
d
Find log2 (x2 + 1)
dx
Answer
dy 1 1 2x
= 2+1
(2x) =
dx ln 2 x (ln 2)(x2 + 1)
. . . . . .
36. Outline
Derivative of the natural exponential function
Exponential Growth
Derivative of the natural logarithm function
Derivatives of other exponentials and logarithms
Other exponentials
Other logarithms
Logarithmic Differentiation
The power rule for irrational powers
. . . . . .
37. A nasty derivative
Example β
(x2 + 1) x + 3
Let y = . Find yβ² .
xβ1
. . . . . .
38. A nasty derivative
Example β
(x2 + 1) x + 3
Let y = . Find yβ² .
xβ1
Solution
We use the quotient rule, and the product rule in the numerator:
[ β ] β
β² (x β 1) 2x x + 3 + (x2 + 1) 1 (x + 3)β1/2 β (x2 + 1) x + 3(1)
2
y =
(x β 1)2
β β
2x x + 3 (x2 + 1) (x 2 + 1 ) x + 3
= + β β
(x β 1 ) 2 x + 3(x β 1) (x β 1)2
. . . . . .
39. Another way
β
(x 2 + 1 ) x + 3
y=
xβ1
1
ln y = ln(x2 + 1) + ln(x + 3) β ln(x β 1)
2
1 dy 2x 1 1
= 2 + β
y dx x + 1 2(x + 3) x β 1
So
( )
dy 2x 1 1
= + β y
dx x2 + 1 2(x + 3) x β 1
( ) β
2x 1 1 (x2 + 1) x + 3
= + β
x2 + 1 2(x + 3) x β 1 xβ1
. . . . . .
40. Compare and contrast
Using the product, quotient, and power rules:
β β
β² 2x x + 3 (x2 + 1) (x2 + 1) x + 3
y = + β β
(x β 1) 2 x + 3(x β 1) (x β 1)2
Using logarithmic differentiation:
( ) β
β² 2x 1 1 (x2 + 1) x + 3
y = + β
x2 + 1 2(x + 3) x β 1 xβ1
. . . . . .
41. Compare and contrast
Using the product, quotient, and power rules:
β β
β² 2x x + 3 (x2 + 1) (x2 + 1) x + 3
y = + β β
(x β 1) 2 x + 3(x β 1) (x β 1)2
Using logarithmic differentiation:
( ) β
β² 2x 1 1 (x2 + 1) x + 3
y = + β
x2 + 1 2(x + 3) x β 1 xβ1
Are these the same?
. . . . . .
42. Compare and contrast
Using the product, quotient, and power rules:
β β
β² 2x x + 3 (x2 + 1) (x2 + 1) x + 3
y = + β β
(x β 1) 2 x + 3(x β 1) (x β 1)2
Using logarithmic differentiation:
( ) β
β² 2x 1 1 (x2 + 1) x + 3
y = + β
x2 + 1 2(x + 3) x β 1 xβ1
Are these the same?
Which do you like better?
. . . . . .
43. Compare and contrast
Using the product, quotient, and power rules:
β β
β² 2x x + 3 (x2 + 1) (x2 + 1) x + 3
y = + β β
(x β 1) 2 x + 3(x β 1) (x β 1)2
Using logarithmic differentiation:
( ) β
β² 2x 1 1 (x2 + 1) x + 3
y = + β
x2 + 1 2(x + 3) x β 1 xβ1
Are these the same?
Which do you like better?
What kinds of expressions are well-suited for logarithmic
differentiation?
. . . . . .
44. Derivatives of powers
Let y = xx . Which of these is true?
(A) Since y is a power function, yβ² = x Β· xxβ1 = xx .
(B) Since y is an exponential function, yβ² = (ln x) Β· xx
(C) Neither
. . . . . .
45. Derivatives of powers
Let y = xx . Which of these is true?
(A) Since y is a power function, yβ² = x Β· xxβ1 = xx .
(B) Since y is an exponential function, yβ² = (ln x) Β· xx
(C) Neither
. . . . . .
46. Itβs neither! Or both?
If y = xx , then
ln y = x ln x
1 dy 1
= x Β· + ln x = 1 + ln x
y dx x
dy
= xx + (ln x)xx
dx
Each of these terms is one of the wrong answers!
. . . . . .
48. Derivative of arbitrary powers
Fact (The power rule)
Let y = xr . Then yβ² = rxrβ1 .
Proof.
y = xr =β ln y = r ln x
Now differentiate:
1 dy r
=
y dx x
dy y
=β = r = rxrβ1
dx x
. . . . . .