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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.
Citation preview
. . . . . .
Section3.3DerivativesofExponentialand
LogarithmicFunctions
V63.0121.034, CalculusI
October21, 2009
Announcements
I
..Imagecredit: heipei
. . . . . .
Outline
DerivativeofthenaturalexponentialfunctionExponentialGrowth
Derivativeofthenaturallogarithmfunction
DerivativesofotherexponentialsandlogarithmsOtherexponentialsOtherlogarithms
LogarithmicDifferentiationThepowerruleforirrationalpowers
. . . . . .
DerivativesofExponentialFunctions
FactIf f(x) = ax, then f′(x) = f′(0)ax.
Proof.Followyournose:
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).
Toreiterate: thederivativeofanexponentialfunctionisaconstant times thatfunction. Muchdifferentfrompolynomials!
. . . . . .
DerivativesofExponentialFunctions
FactIf f(x) = ax, then f′(x) = f′(0)ax.
Proof.Followyournose:
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).
Toreiterate: thederivativeofanexponentialfunctionisaconstant times thatfunction. Muchdifferentfrompolynomials!
. . . . . .
DerivativesofExponentialFunctions
FactIf f(x) = ax, then f′(x) = f′(0)ax.
Proof.Followyournose:
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).
Toreiterate: thederivativeofanexponentialfunctionisaconstant times thatfunction. Muchdifferentfrompolynomials!
. . . . . .
Thefunnylimitinthecaseof eRememberthedefinitionof e:
e = limn→∞
(1 +
1n
)n
= limh→0
(1 + h)1/h
Question
Whatis limh→0
eh − 1h
?
AnswerIf h issmallenough, e ≈ (1 + h)1/h. So
eh − 1h
≈[(1 + h)1/h
]h − 1h
=(1 + h) − 1
h=
hh
= 1
Sointhelimitwegetequality: limh→0
eh − 1h
= 1
. . . . . .
Thefunnylimitinthecaseof eRememberthedefinitionof e:
e = limn→∞
(1 +
1n
)n
= limh→0
(1 + h)1/h
Question
Whatis limh→0
eh − 1h
?
AnswerIf h issmallenough, e ≈ (1 + h)1/h. So
eh − 1h
≈[(1 + h)1/h
]h − 1h
=(1 + h) − 1
h=
hh
= 1
Sointhelimitwegetequality: limh→0
eh − 1h
= 1
. . . . . .
Thefunnylimitinthecaseof eRememberthedefinitionof e:
e = limn→∞
(1 +
1n
)n
= limh→0
(1 + h)1/h
Question
Whatis limh→0
eh − 1h
?
AnswerIf h issmallenough, e ≈ (1 + h)1/h. So
eh − 1h
≈[(1 + h)1/h
]h − 1h
=(1 + h) − 1
h=
hh
= 1
Sointhelimitwegetequality: limh→0
eh − 1h
= 1
. . . . . .
Derivativeofthenaturalexponentialfunction
From
ddx
ax =
(limh→0
ah − 1h
)ax and lim
h→0
eh − 1h
= 1
weget:
Theorem
ddx
ex = ex
. . . . . .
ExponentialGrowth
I Commonlymisusedtermtosaysomethinggrowsexponentially
I Itmeanstherateofchange(derivative)isproportionaltothecurrentvalue
I Examples: Naturalpopulationgrowth, compoundedinterest,socialnetworks
. . . . . .
Examples
ExamplesFindthesederivatives:
I e3x
I ex2
I x2ex
Solution
I ddx
e3x = 3e3x
I ddx
ex2
= ex2 ddx
(x2) = 2xex2
I ddx
x2ex = 2xex + x2ex
. . . . . .
Examples
ExamplesFindthesederivatives:
I e3x
I ex2
I x2ex
Solution
I ddx
e3x = 3e3x
I ddx
ex2
= ex2 ddx
(x2) = 2xex2
I ddx
x2ex = 2xex + x2ex
. . . . . .
Examples
ExamplesFindthesederivatives:
I e3x
I ex2
I x2ex
Solution
I ddx
e3x = 3e3x
I ddx
ex2
= ex2 ddx
(x2) = 2xex2
I ddx
x2ex = 2xex + x2ex
. . . . . .
Examples
ExamplesFindthesederivatives:
I e3x
I ex2
I x2ex
Solution
I ddx
e3x = 3e3x
I ddx
ex2
= ex2 ddx
(x2) = 2xex2
I ddx
x2ex = 2xex + x2ex
. . . . . .
Outline
DerivativeofthenaturalexponentialfunctionExponentialGrowth
Derivativeofthenaturallogarithmfunction
DerivativesofotherexponentialsandlogarithmsOtherexponentialsOtherlogarithms
LogarithmicDifferentiationThepowerruleforirrationalpowers
. . . . . .
Derivativeofthenaturallogarithmfunction
Let y = ln x. Thenx = ey so
eydydx
= 1
=⇒ dydx
=1ey
=1x
So:
Fact
ddx
ln x =1x
. .x
.y
.ln x
.1x
. . . . . .
Derivativeofthenaturallogarithmfunction
Let y = ln x. Thenx = ey so
eydydx
= 1
=⇒ dydx
=1ey
=1x
So:
Fact
ddx
ln x =1x
. .x
.y
.ln x
.1x
. . . . . .
Derivativeofthenaturallogarithmfunction
Let y = ln x. Thenx = ey so
eydydx
= 1
=⇒ dydx
=1ey
=1x
So:
Fact
ddx
ln x =1x
. .x
.y
.ln x
.1x
. . . . . .
Derivativeofthenaturallogarithmfunction
Let y = ln x. Thenx = ey so
eydydx
= 1
=⇒ dydx
=1ey
=1x
So:
Fact
ddx
ln x =1x
. .x
.y
.ln x
.1x
. . . . . .
Derivativeofthenaturallogarithmfunction
Let y = ln x. Thenx = ey so
eydydx
= 1
=⇒ dydx
=1ey
=1x
So:
Fact
ddx
ln x =1x
. .x
.y
.ln x
.1x
. . . . . .
Derivativeofthenaturallogarithmfunction
Let y = ln x. Thenx = ey so
eydydx
= 1
=⇒ dydx
=1ey
=1x
So:
Fact
ddx
ln x =1x
. .x
.y
.ln x
.1x
. . . . . .
TheTowerofPowers
y y′
x3 3x2
x2 2x1
x1 1x0
x0 0
? ?
x−1 −1x−2
x−2 −2x−3
I Thederivativeofapowerfunctionisapowerfunctionofonelowerpower
I Eachpowerfunctionisthederivativeofanotherpowerfunction, exceptx−1
I ln x fillsinthisgapprecisely.
. . . . . .
TheTowerofPowers
y y′
x3 3x2
x2 2x1
x1 1x0
x0 0
? x−1
x−1 −1x−2
x−2 −2x−3
I Thederivativeofapowerfunctionisapowerfunctionofonelowerpower
I Eachpowerfunctionisthederivativeofanotherpowerfunction, exceptx−1
I ln x fillsinthisgapprecisely.
. . . . . .
TheTowerofPowers
y y′
x3 3x2
x2 2x1
x1 1x0
x0 0
ln x x−1
x−1 −1x−2
x−2 −2x−3
I Thederivativeofapowerfunctionisapowerfunctionofonelowerpower
I Eachpowerfunctionisthederivativeofanotherpowerfunction, exceptx−1
I ln x fillsinthisgapprecisely.
. . . . . .
Outline
DerivativeofthenaturalexponentialfunctionExponentialGrowth
Derivativeofthenaturallogarithmfunction
DerivativesofotherexponentialsandlogarithmsOtherexponentialsOtherlogarithms
LogarithmicDifferentiationThepowerruleforirrationalpowers
. . . . . .
Otherlogarithms
Example
Useimplicitdifferentiationtofindddx
ax.
SolutionLet y = ax, so
ln y = ln ax = x ln a
Differentiateimplicitly:
1ydydx
= ln a =⇒ dydx
= (ln a)y = (ln a)ax
Beforeweshowed y′ = y′(0)y, sonowweknowthat
ln 2 = limh→0
2h − 1h
≈ 0.693 ln 3 = limh→0
3h − 1h
≈ 1.10
. . . . . .
Otherlogarithms
Example
Useimplicitdifferentiationtofindddx
ax.
SolutionLet y = ax, so
ln y = ln ax = x ln a
Differentiateimplicitly:
1ydydx
= ln a =⇒ dydx
= (ln a)y = (ln a)ax
Beforeweshowed y′ = y′(0)y, sonowweknowthat
ln 2 = limh→0
2h − 1h
≈ 0.693 ln 3 = limh→0
3h − 1h
≈ 1.10
. . . . . .
Otherlogarithms
Example
Useimplicitdifferentiationtofindddx
ax.
SolutionLet y = ax, so
ln y = ln ax = x ln a
Differentiateimplicitly:
1ydydx
= ln a =⇒ dydx
= (ln a)y = (ln a)ax
Beforeweshowed y′ = y′(0)y, sonowweknowthat
ln 2 = limh→0
2h − 1h
≈ 0.693 ln 3 = limh→0
3h − 1h
≈ 1.10
. . . . . .
Otherlogarithms
Example
Useimplicitdifferentiationtofindddx
ax.
SolutionLet y = ax, so
ln y = ln ax = x ln a
Differentiateimplicitly:
1ydydx
= ln a =⇒ dydx
= (ln a)y = (ln a)ax
Beforeweshowed y′ = y′(0)y, sonowweknowthat
ln 2 = limh→0
2h − 1h
≈ 0.693 ln 3 = limh→0
3h − 1h
≈ 1.10
. . . . . .
Otherlogarithms
Example
Findddx
loga x.
SolutionLet y = loga x, so ay = x. Nowdifferentiateimplicitly:
(ln a)aydydx
= 1 =⇒ dydx
=1
ay ln a=
1x ln a
Anotherwaytoseethisistotakethenaturallogarithm:
ay = x =⇒ y ln a = ln x =⇒ y =ln xln a
Sodydx
=1ln a
1x.
. . . . . .
Otherlogarithms
Example
Findddx
loga x.
SolutionLet y = loga x, so ay = x.
Nowdifferentiateimplicitly:
(ln a)aydydx
= 1 =⇒ dydx
=1
ay ln a=
1x ln a
Anotherwaytoseethisistotakethenaturallogarithm:
ay = x =⇒ y ln a = ln x =⇒ y =ln xln a
Sodydx
=1ln a
1x.
. . . . . .
Otherlogarithms
Example
Findddx
loga x.
SolutionLet y = loga x, so ay = x. Nowdifferentiateimplicitly:
(ln a)aydydx
= 1 =⇒ dydx
=1
ay ln a=
1x ln a
Anotherwaytoseethisistotakethenaturallogarithm:
ay = x =⇒ y ln a = ln x =⇒ y =ln xln a
Sodydx
=1ln a
1x.
. . . . . .
Otherlogarithms
Example
Findddx
loga x.
SolutionLet y = loga x, so ay = x. Nowdifferentiateimplicitly:
(ln a)aydydx
= 1 =⇒ dydx
=1
ay ln a=
1x ln a
Anotherwaytoseethisistotakethenaturallogarithm:
ay = x =⇒ y ln a = ln x =⇒ y =ln xln a
Sodydx
=1ln a
1x.
. . . . . .
Moreexamples
Example
Findddx
log2(x2 + 1)
Answer
dydx
=1ln 2
1x2 + 1
(2x) =2x
(ln 2)(x2 + 1)
. . . . . .
Moreexamples
Example
Findddx
log2(x2 + 1)
Answer
dydx
=1ln 2
1x2 + 1
(2x) =2x
(ln 2)(x2 + 1)
. . . . . .
Outline
DerivativeofthenaturalexponentialfunctionExponentialGrowth
Derivativeofthenaturallogarithmfunction
DerivativesofotherexponentialsandlogarithmsOtherexponentialsOtherlogarithms
LogarithmicDifferentiationThepowerruleforirrationalpowers
. . . . . .
A nastyderivative
Example
Let y =(x2 + 1)
√x + 3
x− 1. Find y′.
SolutionWeusethequotientrule, andtheproductruleinthenumerator:
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 nastyderivative
Example
Let y =(x2 + 1)
√x + 3
x− 1. Find y′.
SolutionWeusethequotientrule, andtheproductruleinthenumerator:
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
. . . . . .
Anotherway
y =(x2 + 1)
√x + 3
x− 1
ln y = ln(x2 + 1) +12ln(x + 3) − ln(x− 1)
1ydydx
=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
. . . . . .
Compareandcontrast
I Usingtheproduct, quotient, andpowerrules:
y′ =2x
√x + 3
(x− 1)+
(x2 + 1)
2√x + 3(x− 1)
− (x2 + 1)√x + 3
(x− 1)2
I Usinglogarithmicdifferentiation:
y′ =
(2x
x2 + 1+
12(x + 3)
− 1x− 1
)(x2 + 1)
√x + 3
x− 1
I Arethesethesame?I Whichdoyoulikebetter?I Whatkindsofexpressionsarewell-suitedforlogarithmicdifferentiation?
. . . . . .
Compareandcontrast
I Usingtheproduct, quotient, andpowerrules:
y′ =2x
√x + 3
(x− 1)+
(x2 + 1)
2√x + 3(x− 1)
− (x2 + 1)√x + 3
(x− 1)2
I Usinglogarithmicdifferentiation:
y′ =
(2x
x2 + 1+
12(x + 3)
− 1x− 1
)(x2 + 1)
√x + 3
x− 1
I Arethesethesame?
I Whichdoyoulikebetter?I Whatkindsofexpressionsarewell-suitedforlogarithmicdifferentiation?
. . . . . .
Compareandcontrast
I Usingtheproduct, quotient, andpowerrules:
y′ =2x
√x + 3
(x− 1)+
(x2 + 1)
2√x + 3(x− 1)
− (x2 + 1)√x + 3
(x− 1)2
I Usinglogarithmicdifferentiation:
y′ =
(2x
x2 + 1+
12(x + 3)
− 1x− 1
)(x2 + 1)
√x + 3
x− 1
I Arethesethesame?I Whichdoyoulikebetter?
I Whatkindsofexpressionsarewell-suitedforlogarithmicdifferentiation?
. . . . . .
Compareandcontrast
I Usingtheproduct, quotient, andpowerrules:
y′ =2x
√x + 3
(x− 1)+
(x2 + 1)
2√x + 3(x− 1)
− (x2 + 1)√x + 3
(x− 1)2
I Usinglogarithmicdifferentiation:
y′ =
(2x
x2 + 1+
12(x + 3)
− 1x− 1
)(x2 + 1)
√x + 3
x− 1
I Arethesethesame?I Whichdoyoulikebetter?I Whatkindsofexpressionsarewell-suitedforlogarithmicdifferentiation?
. . . . . .
Derivativesofpowers
Let y = xx. Whichoftheseistrue?
(A) Since y isapowerfunction, y′ = x · xx−1 = xx.
(B) Since y isanexponentialfunction, y′ = (ln x) · xx
(C) Neither
. . . . . .
Derivativesofpowers
Let y = xx. Whichoftheseistrue?
(A) Since y isapowerfunction, y′ = x · xx−1 = xx.
(B) Since y isanexponentialfunction, y′ = (ln x) · xx
(C) Neither
. . . . . .
It’sneither! Orboth?
If y = xx, then
ln y = x ln x
1ydydx
= x · 1x
+ ln x = 1 + ln x
dydx
= xx + (ln x)xx
Eachofthesetermsisoneofthewronganswers!
. . . . . .
Derivativeofarbitrarypowers
Fact(Thepowerrule)Let y = xr. Then y′ = rxr−1.
Proof.
y = xr =⇒ ln y = r ln x
Nowdifferentiate:
1ydydx
=rx
=⇒ dydx
= ryx
= rxr−1
. . . . . .
Derivativeofarbitrarypowers
Fact(Thepowerrule)Let y = xr. Then y′ = rxr−1.
Proof.
y = xr =⇒ ln y = r ln x
Nowdifferentiate:
1ydydx
=rx
=⇒ dydx
= ryx
= rxr−1