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. . . . . .
Section3.3DerivativesofExponentialand
LogarithmicFunctions
V63.0121.006/016, CalculusI
March2, 2010
Announcements
I Reviewsessions: tonight, 7:30inCIWW 202and517;tomorrow, 7:00inCIWW 109
I MidtermisMarch4, covering§§1.1–2.5I Recitationthisweekwillcover§§3.1–3.2
. . . . . .
Announcements
I Reviewsessions: tonight, 7:30inCIWW 202and517;tomorrow, 7:00inCIWW 109
I MidtermisMarch4, covering§§1.1–2.5I Recitationthisweekwillcover§§3.1–3.2
. . . . . .
Outline
“Recall”Section3.1–3.2
DerivativeofthenaturalexponentialfunctionExponentialGrowth
Derivativeofthenaturallogarithmfunction
DerivativesofotherexponentialsandlogarithmsOtherexponentialsOtherlogarithms
LogarithmicDifferentiationThepowerruleforirrationalpowers
. . . . . .
Conventionsonexponentialfunctions
Let a beapositiverealnumber.I If n isapositivewholenumber, then an = a · a · · · · · a︸ ︷︷ ︸
n factors
I a0 = 1.
I Foranyrealnumber r, a−r =1ar.
I Foranypositivewholenumber n, a1/n = n√a.
Thereisonlyonecontinuousfunctionwhichsatisfiesalloftheabove. Wecallitthe exponentialfunction withbase a.
. . . . . .
Propertiesofexponentialfunctions
TheoremIf a > 0 and a ̸= 1, then f(x) = ax isacontinuousfunctionwithdomain R andrange (0,∞). Inparticular, ax > 0 forall x. Ifa,b > 0 and x, y ∈ R, then
I ax+y = axay
I ax−y =ax
ayI (ax)y = axy
I (ab)x = axbx
. . . . . .
Graphsofvariousexponentialfunctions
. .x
.y
.y = 1x
.y = 2x.y = 3x.y = 10x .y = 1.5x.y = (1/2)x.y = (1/3)x .y = (1/10)x.y = (2/3)x
. . . . . .
Graphsofvariousexponentialfunctions
. .x
.y
.y = 1x
.y = 2x.y = 3x.y = 10x .y = 1.5x.y = (1/2)x.y = (1/3)x .y = (1/10)x.y = (2/3)x
. . . . . .
Graphsofvariousexponentialfunctions
. .x
.y
.y = 1x
.y = 2x
.y = 3x.y = 10x .y = 1.5x.y = (1/2)x.y = (1/3)x .y = (1/10)x.y = (2/3)x
. . . . . .
Graphsofvariousexponentialfunctions
. .x
.y
.y = 1x
.y = 2x.y = 3x
.y = 10x .y = 1.5x.y = (1/2)x.y = (1/3)x .y = (1/10)x.y = (2/3)x
. . . . . .
Graphsofvariousexponentialfunctions
. .x
.y
.y = 1x
.y = 2x.y = 3x.y = 10x
.y = 1.5x.y = (1/2)x.y = (1/3)x .y = (1/10)x.y = (2/3)x
. . . . . .
Graphsofvariousexponentialfunctions
. .x
.y
.y = 1x
.y = 2x.y = 3x.y = 10x .y = 1.5x
.y = (1/2)x.y = (1/3)x .y = (1/10)x.y = (2/3)x
. . . . . .
Graphsofvariousexponentialfunctions
. .x
.y
.y = 1x
.y = 2x.y = 3x.y = 10x .y = 1.5x.y = (1/2)x
.y = (1/3)x .y = (1/10)x.y = (2/3)x
. . . . . .
Graphsofvariousexponentialfunctions
. .x
.y
.y = 1x
.y = 2x.y = 3x.y = 10x .y = 1.5x.y = (1/2)x.y = (1/3)x
.y = (1/10)x.y = (2/3)x
. . . . . .
Graphsofvariousexponentialfunctions
. .x
.y
.y = 1x
.y = 2x.y = 3x.y = 10x .y = 1.5x.y = (1/2)x.y = (1/3)x .y = (1/10)x
.y = (2/3)x
. . . . . .
Graphsofvariousexponentialfunctions
. .x
.y
.y = 1x
.y = 2x.y = 3x.y = 10x .y = 1.5x.y = (1/2)x.y = (1/3)x .y = (1/10)x.y = (2/3)x
. . . . . .
Themagicnumber
Definition
e = limn→∞
(1+
1n
)n
= limh→0+
(1+ h)1/h
. . . . . .
Existenceof eSeeAppendixB
I Wecanexperimentallyverifythatthisnumberexistsandis
e ≈ 2.718281828459045 . . .
I e isirrationalI e is transcendental
n(1+
1n
)n
1 22 2.25
3 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
. . . . . .
Existenceof eSeeAppendixB
I Wecanexperimentallyverifythatthisnumberexistsandis
e ≈ 2.718281828459045 . . .
I e isirrationalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.37037
10 2.59374100 2.704811000 2.71692106 2.71828
. . . . . .
Existenceof eSeeAppendixB
I Wecanexperimentallyverifythatthisnumberexistsandis
e ≈ 2.718281828459045 . . .
I e isirrationalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374
100 2.704811000 2.71692106 2.71828
. . . . . .
Existenceof eSeeAppendixB
I Wecanexperimentallyverifythatthisnumberexistsandis
e ≈ 2.718281828459045 . . .
I e isirrationalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.70481
1000 2.71692106 2.71828
. . . . . .
Existenceof eSeeAppendixB
I Wecanexperimentallyverifythatthisnumberexistsandis
e ≈ 2.718281828459045 . . .
I e isirrationalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692
106 2.71828
. . . . . .
Existenceof eSeeAppendixB
I Wecanexperimentallyverifythatthisnumberexistsandis
e ≈ 2.718281828459045 . . .
I e isirrationalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
. . . . . .
Existenceof eSeeAppendixB
I Wecanexperimentallyverifythatthisnumberexistsandis
e ≈ 2.718281828459045 . . .
I e isirrationalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
. . . . . .
Existenceof eSeeAppendixB
I Wecanexperimentallyverifythatthisnumberexistsandis
e ≈ 2.718281828459045 . . .
I e isirrational
I e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
. . . . . .
Existenceof eSeeAppendixB
I Wecanexperimentallyverifythatthisnumberexistsandis
e ≈ 2.718281828459045 . . .
I e isirrationalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
. . . . . .
Logarithms
Definition
I Thebase a logarithm loga x istheinverseofthefunction ax
y = loga x ⇐⇒ x = ay
I Thenaturallogarithm ln x istheinverseof ex. Soy = ln x ⇐⇒ x = ey.
Facts
(i) loga(x · x′) = loga x+ loga x
′
(ii) loga( xx′
)= loga x− loga x
′
(iii) loga(xr) = r loga x
. . . . . .
Logarithms
Definition
I Thebase a logarithm loga x istheinverseofthefunction ax
y = loga x ⇐⇒ x = ay
I Thenaturallogarithm ln x istheinverseof ex. Soy = ln x ⇐⇒ x = ey.
Facts
(i) loga(x · x′) = loga x+ loga x
′
(ii) loga( xx′
)= loga x− loga x
′
(iii) loga(xr) = r loga x
. . . . . .
Logarithms
Definition
I Thebase a logarithm loga x istheinverseofthefunction ax
y = loga x ⇐⇒ x = ay
I Thenaturallogarithm ln x istheinverseof ex. Soy = ln x ⇐⇒ x = ey.
Facts
(i) loga(x · x′) = loga x+ loga x
′
(ii) loga( xx′
)= loga x− loga x
′
(iii) loga(xr) = r loga x
. . . . . .
Logarithms
Definition
I Thebase a logarithm loga x istheinverseofthefunction ax
y = loga x ⇐⇒ x = ay
I Thenaturallogarithm ln x istheinverseof ex. Soy = ln x ⇐⇒ x = ey.
Facts
(i) loga(x · x′) = loga x+ loga x
′
(ii) loga( xx′
)= loga x− loga x
′
(iii) loga(xr) = r loga x
. . . . . .
Logarithmsconvertproductstosums
I Suppose y = loga x and y′ = loga x′
I Then x = ay and x′ = ay′
I So xx′ = ayay′= ay+y′
I Therefore
loga(xx′) = y+ y′ = loga x+ loga x
′
. . . . . .
Graphsoflogarithmicfunctions
. .x
.y.y = 2x
.y = log2 x
. .(0, 1)
..(1, 0)
.y = 3x
.y = log3 x
.y = 10x
.y = log10 x
.y = ex
.y = ln x
. . . . . .
Graphsoflogarithmicfunctions
. .x
.y.y = 2x
.y = log2 x
. .(0, 1)
..(1, 0)
.y = 3x
.y = log3 x
.y = 10x
.y = log10 x
.y = ex
.y = ln x
. . . . . .
Graphsoflogarithmicfunctions
. .x
.y.y = 2x
.y = log2 x
. .(0, 1)
..(1, 0)
.y = 3x
.y = log3 x
.y = 10x
.y = log10 x
.y = ex
.y = ln x
. . . . . .
Graphsoflogarithmicfunctions
. .x
.y.y = 2x
.y = log2 x
. .(0, 1)
..(1, 0)
.y = 3x
.y = log3 x
.y = 10x
.y = log10 x
.y = ex
.y = ln x
. . . . . .
Changeofbaseformulaforexponentials
FactIf a > 0 and a ̸= 1, then
loga x =ln xln a
Proof.
I If y = loga x, then x = ay
I So ln x = ln(ay) = y ln aI Therefore
y = loga x =ln xln a
. . . . . .
Changeofbaseformulaforexponentials
FactIf a > 0 and a ̸= 1, then
loga x =ln xln a
Proof.
I If y = loga x, then x = ay
I So ln x = ln(ay) = y ln aI Therefore
y = loga x =ln xln a
. . . . . .
Outline
“Recall”Section3.1–3.2
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
Iddx
e3x = 3e3x
Iddx
ex2= ex
2 ddx
(x2) = 2xex2
Iddx
x2ex = 2xex + x2ex
. . . . . .
Examples
ExamplesFindthesederivatives:
I e3x
I ex2
I x2ex
Solution
Iddx
e3x = 3e3x
Iddx
ex2= ex
2 ddx
(x2) = 2xex2
Iddx
x2ex = 2xex + x2ex
. . . . . .
Examples
ExamplesFindthesederivatives:
I e3x
I ex2
I x2ex
Solution
Iddx
e3x = 3e3x
Iddx
ex2= ex
2 ddx
(x2) = 2xex2
Iddx
x2ex = 2xex + x2ex
. . . . . .
Examples
ExamplesFindthesederivatives:
I e3x
I ex2
I x2ex
Solution
Iddx
e3x = 3e3x
Iddx
ex2= ex
2 ddx
(x2) = 2xex2
Iddx
x2ex = 2xex + x2ex
. . . . . .
Outline
“Recall”Section3.1–3.2
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
“Recall”Section3.1–3.2
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
“Recall”Section3.1–3.2
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′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 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′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 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′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 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′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 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