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The derivative of an exponential is another exponential, but the derivative of a logarithm is a special "missing" power function.
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. . . . . .
Section 3.3Derivatives of Exponential and
Logarithmic Functions
V63.0121, Calculus I
March 10/11, 2009
Announcements
I Quiz 3 this week: Covers Sections 2.1–2.4I Get half of all unearned ALEKS points by March 22
..Image credit: heipei
. . . . . .
Outline
Derivative of the natural exponential functionExponential Growth
Derivative of the natural logarithm function
Derivatives of other exponentials and logarithmsOther exponentialsOther logarithms
Logarithmic DifferentiationThe power rule for irrational powers
. . . . . .
Derivatives of Exponential Functions
FactIf f(x) = ax, then f′(x) = f′(0)ax.
Proof.Follow your nose:
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
ax+h − ax
h
= limh→0
axah − ax
h= ax · lim
h→0
ah − 1h
= ax · f′(0).
To reiterate: the derivative of an exponential function is a constanttimes that function. Much different from polynomials!
. . . . . .
Derivatives of Exponential Functions
FactIf f(x) = ax, then f′(x) = f′(0)ax.
Proof.Follow your nose:
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
ax+h − ax
h
= limh→0
axah − ax
h= ax · lim
h→0
ah − 1h
= ax · f′(0).
To reiterate: the derivative of an exponential function is a constanttimes that function. Much different from polynomials!
. . . . . .
Derivatives of Exponential Functions
FactIf f(x) = ax, then f′(x) = f′(0)ax.
Proof.Follow your nose:
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
ax+h − ax
h
= limh→0
axah − ax
h= ax · lim
h→0
ah − 1h
= ax · f′(0).
To reiterate: the derivative of an exponential function is a constanttimes that function. Much different from polynomials!
. . . . . .
The funny limit in the case of e
Remember the definition of e:
e = limn→∞
(1 +
1n
)n
= limh→0
(1 + h)1/h
Question
What is limh→0
eh − 1h
?
AnswerIf h is small enough, e ≈ (1 + h)1/h. So
eh − 1h
≈[(1 + h)1/h
]h − 1h
=(1 + h) − 1
h=
hh
= 1
So in the limit we get equality: limh→0
eh − 1h
= 1
. . . . . .
The funny limit in the case of e
Remember the definition of e:
e = limn→∞
(1 +
1n
)n
= limh→0
(1 + h)1/h
Question
What is limh→0
eh − 1h
?
AnswerIf h is small enough, e ≈ (1 + h)1/h. So
eh − 1h
≈[(1 + h)1/h
]h − 1h
=(1 + h) − 1
h=
hh
= 1
So in the limit we get equality: limh→0
eh − 1h
= 1
. . . . . .
The funny limit in the case of e
Remember the definition of e:
e = limn→∞
(1 +
1n
)n
= limh→0
(1 + h)1/h
Question
What is limh→0
eh − 1h
?
AnswerIf h is small enough, e ≈ (1 + h)1/h. So
eh − 1h
≈[(1 + h)1/h
]h − 1h
=(1 + h) − 1
h=
hh
= 1
So in the limit we get equality: limh→0
eh − 1h
= 1
. . . . . .
Derivative of the natural exponential function
From
ddx
ax =
(limh→0
ah − 1h
)ax and lim
h→0
eh − 1h
= 1
we get:
Theorem
ddx
ex = ex
. . . . . .
Exponential Growth
I Commonly misused term to say something grows exponentiallyI It means the rate of change (derivative) is proportional to the
current valueI Examples: Natural population growth, compounded interest,
social networks
. . . . . .
Examples
ExamplesFind these derivatives:
I e3x
I ex2
I x2ex
Solution
I ddx
e3x = 3ex
I ddx
ex2= ex2 d
dx(x2) = 2xex2
I ddx
x2ex = 2xex + x2ex
. . . . . .
Examples
ExamplesFind these derivatives:
I e3x
I ex2
I x2ex
Solution
I ddx
e3x = 3ex
I ddx
ex2= ex2 d
dx(x2) = 2xex2
I ddx
x2ex = 2xex + x2ex
. . . . . .
Examples
ExamplesFind these derivatives:
I e3x
I ex2
I x2ex
Solution
I ddx
e3x = 3ex
I ddx
ex2= ex2 d
dx(x2) = 2xex2
I ddx
x2ex = 2xex + x2ex
. . . . . .
Examples
ExamplesFind these derivatives:
I e3x
I ex2
I x2ex
Solution
I ddx
e3x = 3ex
I ddx
ex2= ex2 d
dx(x2) = 2xex2
I ddx
x2ex = 2xex + x2ex
. . . . . .
Outline
Derivative of the natural exponential functionExponential Growth
Derivative of the natural logarithm function
Derivatives of other exponentials and logarithmsOther exponentialsOther logarithms
Logarithmic DifferentiationThe power rule for irrational powers
. . . . . .
Derivative of the natural logarithm function
Let y = ln x. Thenx = ey so
ey dydx
= 1
=⇒ dydx
=1ey =
1x
So:
Fact
ddx
ln x =1x
. .x
.y
.ln x
.1x
. . . . . .
Derivative of the natural logarithm function
Let y = ln x. Thenx = ey so
ey dydx
= 1
=⇒ dydx
=1ey =
1x
So:
Fact
ddx
ln x =1x
. .x
.y
.ln x
.1x
. . . . . .
Derivative of the natural logarithm function
Let y = ln x. Thenx = ey so
ey dydx
= 1
=⇒ dydx
=1ey =
1x
So:
Fact
ddx
ln x =1x
. .x
.y
.ln x
.1x
. . . . . .
Derivative of the natural logarithm function
Let y = ln x. Thenx = ey so
ey dydx
= 1
=⇒ dydx
=1ey =
1x
So:
Fact
ddx
ln x =1x
. .x
.y
.ln x
.1x
. . . . . .
Derivative of the natural logarithm function
Let y = ln x. Thenx = ey so
ey dydx
= 1
=⇒ dydx
=1ey =
1x
So:
Fact
ddx
ln x =1x
. .x
.y
.ln x
.1x
. . . . . .
Derivative of the natural logarithm function
Let y = ln x. Thenx = ey so
ey dydx
= 1
=⇒ dydx
=1ey =
1x
So:
Fact
ddx
ln x =1x
. .x
.y
.ln x
.1x
. . . . . .
The Tower of Powers
y y′
x3 3x2
x2 2x1
x1 1x0
x0 0
? ?
x−1 −1x−2
x−2 −2x−3
I The derivative of a powerfunction is a powerfunction of one lowerpower
I Each power function isthe derivative of anotherpower function, exceptx−1
I ln x fills in this gapprecisely.
. . . . . .
The Tower of Powers
y y′
x3 3x2
x2 2x1
x1 1x0
x0 0
? x−1
x−1 −1x−2
x−2 −2x−3
I The derivative of a powerfunction is a powerfunction of one lowerpower
I Each power function isthe derivative of anotherpower function, exceptx−1
I ln x fills in this gapprecisely.
. . . . . .
The Tower of Powers
y y′
x3 3x2
x2 2x1
x1 1x0
x0 0
ln x x−1
x−1 −1x−2
x−2 −2x−3
I The derivative of a powerfunction is a powerfunction of one lowerpower
I Each power function isthe derivative of anotherpower function, exceptx−1
I ln x fills in this gapprecisely.
. . . . . .
Outline
Derivative of the natural exponential functionExponential Growth
Derivative of the natural logarithm function
Derivatives of other exponentials and logarithmsOther exponentialsOther logarithms
Logarithmic DifferentiationThe power rule for irrational powers
. . . . . .
Other logarithms
Example
Use implicit differentiation to findddx
ax.
SolutionLet y = ax, so
ln y = ln ax = x ln a
Differentiate implicitly:
1y
dydx
= ln a =⇒ dydx
= (ln a)y = (ln a)ax
Before we showed y′ = y′(0)y, so now we know that
ln 2 = limh→0
2h − 1h
≈ 0.693 ln 3 = limh→0
3h − 1h
≈ 1.10
. . . . . .
Other logarithms
Example
Use implicit differentiation to findddx
ax.
SolutionLet y = ax, so
ln y = ln ax = x ln a
Differentiate implicitly:
1y
dydx
= ln a =⇒ dydx
= (ln a)y = (ln a)ax
Before we showed y′ = y′(0)y, so now we know that
ln 2 = limh→0
2h − 1h
≈ 0.693 ln 3 = limh→0
3h − 1h
≈ 1.10
. . . . . .
Other logarithms
Example
Use implicit differentiation to findddx
ax.
SolutionLet y = ax, so
ln y = ln ax = x ln a
Differentiate implicitly:
1y
dydx
= ln a =⇒ dydx
= (ln a)y = (ln a)ax
Before we showed y′ = y′(0)y, so now we know that
ln 2 = limh→0
2h − 1h
≈ 0.693 ln 3 = limh→0
3h − 1h
≈ 1.10
. . . . . .
Other logarithms
Example
Use implicit differentiation to findddx
ax.
SolutionLet y = ax, so
ln y = ln ax = x ln a
Differentiate implicitly:
1y
dydx
= ln a =⇒ dydx
= (ln a)y = (ln a)ax
Before we showed y′ = y′(0)y, so now we know that
ln 2 = limh→0
2h − 1h
≈ 0.693 ln 3 = limh→0
3h − 1h
≈ 1.10
. . . . . .
Other logarithms
Example
Findddx
loga x.
SolutionLet y = loga x, so ay = x. Now differentiate implicitly:
(ln a)ay dydx
= 1 =⇒ dydx
=1
ay ln a=
1x ln a
Another way to see this is to take the natural logarithm:
ay = x =⇒ y ln a = ln x =⇒ y =ln xln a
Sodydx
=1
ln a1x
.
. . . . . .
Other logarithms
Example
Findddx
loga x.
SolutionLet y = loga x, so ay = x.
Now differentiate implicitly:
(ln a)ay dydx
= 1 =⇒ dydx
=1
ay ln a=
1x ln a
Another way to see this is to take the natural logarithm:
ay = x =⇒ y ln a = ln x =⇒ y =ln xln a
Sodydx
=1
ln a1x
.
. . . . . .
Other logarithms
Example
Findddx
loga x.
SolutionLet y = loga x, so ay = x. Now differentiate implicitly:
(ln a)ay dydx
= 1 =⇒ dydx
=1
ay ln a=
1x ln a
Another way to see this is to take the natural logarithm:
ay = x =⇒ y ln a = ln x =⇒ y =ln xln a
Sodydx
=1
ln a1x
.
. . . . . .
Other logarithms
Example
Findddx
loga x.
SolutionLet y = loga x, so ay = x. Now differentiate implicitly:
(ln a)ay dydx
= 1 =⇒ dydx
=1
ay ln a=
1x ln a
Another way to see this is to take the natural logarithm:
ay = x =⇒ y ln a = ln x =⇒ y =ln xln a
Sodydx
=1
ln a1x
.
. . . . . .
More examples
Example
Findddx
log2(x2 + 1)
Answer
dydx
=1
ln 21
x2 + 1(2x) =
2x(ln 2)(x2 + 1)
. . . . . .
More examples
Example
Findddx
log2(x2 + 1)
Answer
dydx
=1
ln 21
x2 + 1(2x) =
2x(ln 2)(x2 + 1)
. . . . . .
Outline
Derivative of the natural exponential functionExponential Growth
Derivative of the natural logarithm function
Derivatives of other exponentials and logarithmsOther exponentialsOther logarithms
Logarithmic DifferentiationThe power rule for irrational powers
. . . . . .
A nasty derivative
Example
Let y =(x2 + 1)
√x + 3
x − 1. Find y′.
SolutionWe use the quotient rule, and the product rule in the numerator:
y′ =(x − 1)
[2x√
x + 3 + (x2 + 1) 12(x + 3)−1/2
]− (x2 + 1)
√x + 3(1)
(x − 1)2
=2x√
x + 3(x − 1)
+(x2 + 1)
2√
x + 3(x − 1)− (x2 + 1)
√x + 3
(x − 1)2
. . . . . .
A nasty derivative
Example
Let y =(x2 + 1)
√x + 3
x − 1. Find y′.
SolutionWe use the quotient rule, and the product rule in the numerator:
y′ =(x − 1)
[2x√
x + 3 + (x2 + 1) 12(x + 3)−1/2
]− (x2 + 1)
√x + 3(1)
(x − 1)2
=2x√
x + 3(x − 1)
+(x2 + 1)
2√
x + 3(x − 1)− (x2 + 1)
√x + 3
(x − 1)2
. . . . . .
Another way
y =(x2 + 1)
√x + 3
x − 1
ln y = ln(x2 + 1) +12
ln(x + 3) − ln(x − 1)
1y
dydx
=2x
x2 + 1+
12(x + 3)
− 1x − 1
So
dydx
=
(2x
x2 + 1+
12(x + 3)
− 1x − 1
)y
=
(2x
x2 + 1+
12(x + 3)
− 1x − 1
)(x2 + 1)
√x + 3
x − 1
. . . . . .
Compare and contrast
I Using the product, quotient, and power rules:
y′ =2x√
x + 3(x − 1)
+(x2 + 1)
2√
x + 3(x − 1)− (x2 + 1)
√x + 3
(x − 1)2
I Using logarithmic differentiation:
y′ =
(2x
x2 + 1+
12(x + 3)
− 1x − 1
)(x2 + 1)
√x + 3
x − 1
I Are these the same?I Which do you like better?I What kinds of expressions are well-suited for logarithmic
differentiation?
. . . . . .
Compare and contrast
I Using the product, quotient, and power rules:
y′ =2x√
x + 3(x − 1)
+(x2 + 1)
2√
x + 3(x − 1)− (x2 + 1)
√x + 3
(x − 1)2
I Using logarithmic differentiation:
y′ =
(2x
x2 + 1+
12(x + 3)
− 1x − 1
)(x2 + 1)
√x + 3
x − 1
I Are these the same?
I Which do you like better?I What kinds of expressions are well-suited for logarithmic
differentiation?
. . . . . .
Compare and contrast
I Using the product, quotient, and power rules:
y′ =2x√
x + 3(x − 1)
+(x2 + 1)
2√
x + 3(x − 1)− (x2 + 1)
√x + 3
(x − 1)2
I Using logarithmic differentiation:
y′ =
(2x
x2 + 1+
12(x + 3)
− 1x − 1
)(x2 + 1)
√x + 3
x − 1
I Are these the same?I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation?
. . . . . .
Compare and contrast
I Using the product, quotient, and power rules:
y′ =2x√
x + 3(x − 1)
+(x2 + 1)
2√
x + 3(x − 1)− (x2 + 1)
√x + 3
(x − 1)2
I Using logarithmic differentiation:
y′ =
(2x
x2 + 1+
12(x + 3)
− 1x − 1
)(x2 + 1)
√x + 3
x − 1
I Are these the same?I Which do you like better?I What kinds of expressions are well-suited for logarithmic
differentiation?
. . . . . .
Derivatives of powers
Let y = xx. Which of these is true?
(A) Since y is a power function, y′ = x · xx−1 = xx.
(B) Since y is an exponential function, y′ = (ln x) · xx
(C) Neither
. . . . . .
Derivatives of powers
Let y = xx. Which of these is true?
(A) Since y is a power function, y′ = x · xx−1 = xx.
(B) Since y is an exponential function, y′ = (ln x) · xx
(C) Neither
. . . . . .
It’s neither! Or both?
If y = xx, then
ln y = x ln x
1y
dydx
= x · 1x
+ ln x = 1 + ln x
dydx
= xx + (ln x)xx
Each of these terms is one of the wrong answers!
. . . . . .
Derivative of arbitrary powers
Fact (The power rule)Let y = xr. Then y′ = rxr−1.
Proof.
y = xr =⇒ ln y = r ln x
Now differentiate:
1y
dydx
=rx
=⇒ dydx
= ryx
= rxr−1
. . . . . .
Derivative of arbitrary powers
Fact (The power rule)Let y = xr. Then y′ = rxr−1.
Proof.
y = xr =⇒ ln y = r ln x
Now differentiate:
1y
dydx
=rx
=⇒ dydx
= ryx
= rxr−1
