. . . . . .

Section3.3DerivativesofExponentialand

LogarithmicFunctions

V63.0121.034, CalculusI

October21, 2009

Announcements

I

..Imagecredit: heipei

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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