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. . . . . .
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