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Sec on 3.3Deriva ves of Logarithmic and
Exponen al Func ons
V63.0121.011: Calculus IProfessor Ma hew Leingang
New York University
March 21, 2011
Announcements
I Quiz 3 next week on 2.6,2.8, 3.1, 3.2
ObjectivesI Know the deriva ves of theexponen al func ons (with anybase)
I Know the deriva ves of thelogarithmic func ons (with anybase)
I Use the technique of logarithmicdifferen a on to find deriva vesof func ons involving roducts,quo ents, and/or exponen als.
OutlineRecall Sec on 3.1–3.2
Deriva ve of the natural exponen al func onExponen al Growth
Deriva ve of the natural logarithm func on
Deriva ves of other exponen als and logarithmsOther exponen alsOther logarithms
Logarithmic Differen a onThe power rule for irra onal powers
Conventions on power expressionsLet a be a posi ve real number.
I If n is a posi ve whole number, then an = a · a · · · · · a︸ ︷︷ ︸n factors
I a0 = 1.I For any real number r, a−r =
1ar .
I For any posi ve whole number n, a1/n = n√a.
There is only one con nuous func on which sa sfies all of theabove. We call it the exponen al func on with base a.
Properties of exponentialsTheoremIf a > 0 and a ̸= 1, then f(x) = ax is a con nuous func on withdomain (−∞,∞) and range (0,∞). In par cular, ax > 0 for all x.For any real numbers x and y, and posi ve numbers a and b we have
I ax+y = axay
I ax−y =ax
ay
(nega ve exponents mean reciprocals)
I (ax)y = axy
(frac onal exponents mean roots)
I (ab)x = axbx
Properties of exponentialsTheoremIf a > 0 and a ̸= 1, then f(x) = ax is a con nuous func on withdomain (−∞,∞) and range (0,∞). In par cular, ax > 0 for all x.For any real numbers x and y, and posi ve numbers a and b we have
I ax+y = axay
I ax−y =ax
ay (nega ve exponents mean reciprocals)
I (ax)y = axy
(frac onal exponents mean roots)
I (ab)x = axbx
Properties of exponentialsTheoremIf a > 0 and a ̸= 1, then f(x) = ax is a con nuous func on withdomain (−∞,∞) and range (0,∞). In par cular, ax > 0 for all x.For any real numbers x and y, and posi ve numbers a and b we have
I ax+y = axay
I ax−y =ax
ay (nega ve exponents mean reciprocals)
I (ax)y = axy (frac onal exponents mean roots)I (ab)x = axbx
Graphs of exponential functions
.. 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
The magic number
Defini on
e = limn→∞
(1+
1n
)n
= limh→0+
(1+ h)1/h
Existence of eSee Appendix B
I We can experimentallyverify that this numberexists and is
e ≈ 2.718281828459045 . . .
I e is irra onalI e is transcendental
n(1+
1n
)n
1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828
Logarithms
Defini on
I The base a logarithm loga x is the inverse of the func on ax
y = loga x ⇐⇒ x = ay
I The natural logarithm ln x is the inverse of ex. Soy = ln x ⇐⇒ x = ey.
Facts about Logarithms
Facts
(i) loga(x1 · x2) = loga x1 + loga x2
(ii) loga
(x1x2
)= loga x1 − loga x2
(iii) loga(xr) = r loga x
Graphs of logarithmic functions
.. 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
Change of base formula
FactIf a > 0 and a ̸= 1, and the same for b, then
loga x =logb xlogb a
Upshot of changing baseThe point of the change of base formula
loga x =logb xlogb a
=1
logb a· logb x = (constant) · logb x
is that all the logarithmic func ons are mul ples of each other. Sojust pick one and call it your favorite.
I Engineers like the common logarithm log = log10I Computer scien sts like the binary logarithm lg = log2I Mathema cians like natural logarithm ln = loge
Naturally, we will follow the mathema cians. Just don’t pronounceit “lawn.”
OutlineRecall Sec on 3.1–3.2
Deriva ve of the natural exponen al func onExponen al Growth
Deriva ve of the natural logarithm func on
Deriva ves of other exponen als and logarithmsOther exponen alsOther logarithms
Logarithmic Differen a onThe power rule for irra onal powers
Derivatives of ExponentialsFactIf 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).
Derivatives of ExponentialsFactIf 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).
The funny limit in the case of eQues on
What is limh→0
eh − 1h
?
Solu on
The funny limit in the case of eQues on
What is limh→0
eh − 1h
?
Solu on
Recall e = limn→∞
(1+
1n
)n
= limh→0
(1+ h)1/h. If h is small enough,
e ≈ (1+ h)1/h. So
eh − 1h
≈[(1+ h)1/h
]h − 1h
=(1+ h)− 1
h=
hh= 1
The funny limit in the case of eQues on
What is limh→0
eh − 1h
?
Solu onSo in the limit we get equality:
limh→0
eh − 1h
= 1
Derivative of the naturalexponential function
From
ddx
ax =
(limh→0
ah − 1h
)ax and lim
h→0
eh − 1h
= 1
we get:
Theorem
ddx
ex = ex
Exponential GrowthI Commonly misused term to say something grows exponen allyI It means the rate of change (deriva ve) is propor onal to thecurrent value
I Examples: Natural popula on growth, compounded interest,social networks
Examples
Example
Findddx
e3x.
Solu on
ddx
e3x = e3xddx
(3x) = 3e3x
Examples
Example
Findddx
e3x.
Solu on
ddx
e3x = e3xddx
(3x) = 3e3x
Examples
Example
Findddx
ex2.
Solu on
ddx
ex2= ex
2 ddx
(x2) = 2xex2
Examples
Example
Findddx
ex2.
Solu on
ddx
ex2= ex
2 ddx
(x2) = 2xex2
Examples
Example
Findddx
x2ex.
Solu on
ddx
x2ex = 2xex + x2ex
Examples
Example
Findddx
x2ex.
Solu on
ddx
x2ex = 2xex + x2ex
OutlineRecall Sec on 3.1–3.2
Deriva ve of the natural exponen al func onExponen al Growth
Deriva ve of the natural logarithm func on
Deriva ves of other exponen als and logarithmsOther exponen alsOther logarithms
Logarithmic Differen a onThe power rule for irra onal powers
Derivative of the natural logarithmLet y = ln x. Then x = ey so
eydydx
= 1
=⇒ dydx
=1ey
=1x
We have discovered:Fact
ddx
ln x =1x
.. x.
y
.
ln x
.
1x
Derivative of the natural logarithmLet y = ln x. Then x = ey so
eydydx
= 1
=⇒ dydx
=1ey
=1x
We have discovered:Fact
ddx
ln x =1x
.. x.
y
.
ln x
.
1x
Derivative of the natural logarithmLet y = ln x. Then x = ey so
eydydx
= 1
=⇒ dydx
=1ey
=1x
We have discovered:Fact
ddx
ln x =1x
.. x.
y
.
ln x
.
1x
Derivative of the natural logarithmLet y = ln x. Then x = ey so
eydydx
= 1
=⇒ dydx
=1ey
=1x
We have discovered:Fact
ddx
ln x =1x
.. x.
y
.
ln x
.
1x
Derivative of the natural logarithmLet y = ln x. Then x = ey so
eydydx
= 1
=⇒ dydx
=1ey
=1x
We have discovered:Fact
ddx
ln x =1x
.. x.
y
.
ln x
.
1x
Derivative of the natural logarithmLet y = ln x. Then x = ey so
eydydx
= 1
=⇒ dydx
=1ey
=1x
We have discovered:Fact
ddx
ln x =1x
.. x.
y
.
ln x
.
1x
The Tower of Powersy y′
x3 3x2
x2 2x1
x1 1x0
x0 0
? ?
x−1 −1x−2
x−2 −2x−3
I The deriva ve of a power func on is apower func on of one lower power
I Each power func on is the deriva ve ofanother power func on, except x−1
I ln x fills in this gap precisely.
The Tower of Powersy y′
x3 3x2
x2 2x1
x1 1x0
x0 0
? x−1
x−1 −1x−2
x−2 −2x−3
I The deriva ve of a power func on is apower func on of one lower power
I Each power func on is the deriva ve ofanother power func on, except x−1
I ln x fills in this gap precisely.
The Tower of Powersy y′
x3 3x2
x2 2x1
x1 1x0
x0 0
ln x x−1
x−1 −1x−2
x−2 −2x−3
I The deriva ve of a power func on is apower func on of one lower power
I Each power func on is the deriva ve ofanother power func on, except x−1
I ln x fills in this gap precisely.
Examples
Examples
Find deriva ves of these func ons:I ln(3x)I x ln xI ln
√x
ExamplesExample
Findddx
ln(3x).
ExamplesExample
Findddx
ln(3x).
Solu on (chain rule way)
ddx
ln(3x) =13x
· 3 =1x
ExamplesExample
Findddx
ln(3x).
Solu on (proper es of logarithms way)
ddx
ln(3x) =ddx
(ln(3) + ln(x)) = 0+1x=
1x
The first answermight be surprising un l you see the second solu on.
ExamplesExample
Findddx
x ln x
Solu onThe product rule is in play here:
ddx
x ln x =(
ddx
x)ln x+ x
(ddx
ln x)
= 1 · ln x+ x · 1x= ln x+ 1
ExamplesExample
Findddx
x ln x
Solu onThe product rule is in play here:
ddx
x ln x =(
ddx
x)ln x+ x
(ddx
ln x)
= 1 · ln x+ x · 1x= ln x+ 1
ExamplesExample
Findddx
ln√x.
ExamplesExample
Findddx
ln√x.
Solu on (chain rule way)
ddx
ln√x =
1√xddx
√x =
1√x
12√x=
12x
ExamplesExample
Findddx
ln√x.
Solu on (proper es of logarithms way)
ddx
ln√x =
ddx
(12ln x
)=
12ddx
ln x =12· 1x
The first answermight be surprising un l you see the second solu on.
OutlineRecall Sec on 3.1–3.2
Deriva ve of the natural exponen al func onExponen al Growth
Deriva ve of the natural logarithm func on
Deriva ves of other exponen als and logarithmsOther exponen alsOther logarithms
Logarithmic Differen a onThe power rule for irra onal powers
Other logarithmsExample
Use implicit differen a on to findddx
ax.
Solu onLet y = ax, so
ln y = ln ax = x ln a
Differen ate implicitly:
1ydydx
= ln a =⇒ dydx
= (ln a)y = (ln a)ax
Other logarithmsExample
Use implicit differen a on to findddx
ax.
Solu onLet y = ax, so
ln y = ln ax = x ln a
Differen ate implicitly:
1ydydx
= ln a =⇒ dydx
= (ln a)y = (ln a)ax
Other logarithmsExample
Use implicit differen a on to findddx
ax.
Solu onLet y = ax, so
ln y = ln ax = x ln a
Differen ate implicitly:
1ydydx
= ln a =⇒ dydx
= (ln a)y = (ln a)ax
The funny limit in the case of aLet y = ex. Before we showed y′ = y′(0)y, and now we knowy′ = (ln a)y. So
Corollary
limh→0
ah − 1h
= ln a
In par cular
ln 2 = limh→0
2h − 1h
≈ 0.693 ln 3 = limh→0
3h − 1h
≈ 1.10
Other logarithmsExample
Findddx
loga x.
Solu on
Other logarithmsExample
Findddx
loga x.
Solu onLet y = loga x, so ay = x.
Now differen ate implicitly:
(ln a)aydydx
= 1 =⇒ dydx
=1
ay ln a=
1x ln a
Other logarithmsExample
Findddx
loga x.
Solu onLet y = loga x, so ay = x. Now differen ate implicitly:
(ln a)aydydx
= 1 =⇒ dydx
=1
ay ln a=
1x ln a
Other logarithmsExample
Findddx
loga x.
Solu onOr we can use the change of base formula:
y =ln xln a
=⇒ dydx
=1ln a
1x
More examplesExample
Findddx
log2(x2 + 1)
Answer
dydx
=1ln 2
1x2 + 1
(2x) =2x
(ln 2)(x2 + 1)
More examplesExample
Findddx
log2(x2 + 1)
Answer
dydx
=1ln 2
1x2 + 1
(2x) =2x
(ln 2)(x2 + 1)
OutlineRecall Sec on 3.1–3.2
Deriva ve of the natural exponen al func onExponen al Growth
Deriva ve of the natural logarithm func on
Deriva ves of other exponen als and logarithmsOther exponen alsOther logarithms
Logarithmic Differen a onThe power rule for irra onal powers
A nasty derivativeExample
Let y =(x2 + 1)
√x+ 3
x− 1. Find y′.
Solu onWe use the quo ent 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 derivativeExample
Let y =(x2 + 1)
√x+ 3
x− 1. Find y′.
Solu onWe use the quo ent 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) +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
Another way
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
Another way
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
Another way
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
Another way
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
)(x2 + 1)
√x+ 3
x− 1
Compare and contrastI Using the product, quo ent, 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 differen a on:
y′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 1
)(x2 + 1)
√x+ 3
(x− 1)
I Are these the same?
Yes.
Compare and contrastI Using the product, quo ent, 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 differen a on:
y′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 1
)(x2 + 1)
√x+ 3
(x− 1)
I Are these the same?
Yes.
Compare and contrastI Using the product, quo ent, 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 differen a on:
y′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 1
)(x2 + 1)
√x+ 3
(x− 1)
I Are these the same?
Yes.
Compare and contrastI Using the product, quo ent, 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 differen a on:
y′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 1
)(x2 + 1)
√x+ 3
(x− 1)
I Are these the same?
Yes.
Compare and contrastI Using the product, quo ent, 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 differen a on:
y′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 1
)(x2 + 1)
√x+ 3
(x− 1)
I Are these the same?
Yes.
Compare and contrastI Using the product, quo ent, 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 differen a on:
y′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 1
)(x2 + 1)
√x+ 3
(x− 1)
I Are these the same?
Yes.
Compare and contrastI Using the product, quo ent, 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 differen a on:
y′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 1
)(x2 + 1)
√x+ 3
(x− 1)
I Are these the same?
Yes.
Compare and contrastI Using the product, quo ent, 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 differen a on:
y′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 1
)(x2 + 1)
√x+ 3
(x− 1)
I Are these the same?
Yes.
Compare and contrastI Using the product, quo ent, 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 differen a on:
y′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 1
)(x2 + 1)
√x+ 3
(x− 1)
I Are these the same?
Yes.
Compare and contrastI Using the product, quo ent, 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 differen a on:
y′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 1
)(x2 + 1)
√x+ 3
(x− 1)
I Are these the same?
Yes.
Compare and contrastI Using the product, quo ent, 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 differen a on:
y′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 1
)(x2 + 1)
√x+ 3
(x− 1)
I Are these the same?
Yes.
Compare and contrastI Using the product, quo ent, 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 differen a on:
y′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 1
)(x2 + 1)
√x+ 3
(x− 1)
I Are these the same?
Yes.
Compare and contrastI Using the product, quo ent, 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 differen a on:
y′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 1
)(x2 + 1)
√x+ 3
(x− 1)
I Are these the same?
Yes.
Compare and contrastI Using the product, quo ent, 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 differen a on:
y′ =(
2xx2 + 1
+1
2(x+ 3)− 1
x− 1
)(x2 + 1)
√x+ 3
(x− 1)
I Are these the same? Yes.
Derivatives of powers
Ques on
Let y = xx. Which of these is true?(A) Since y is a power func on,
y′ = x · xx−1 = xx.(B) Since y is an exponen al
func on, y′ = (ln x) · xx
(C) Neither ..x
.
y
..1
..
1
Derivatives of powers
Ques on
Let y = xx. Which of these is true?(A) Since y is a power func on,
y′ = x · xx−1 = xx.(B) Since y is an exponen al
func on, y′ = (ln x) · xx
(C) Neither ..x
.
y
..1
..
1
Why not?Answer
(A) y′ ̸= xx because xx > 0 for allx > 0, and this func ondecreases at some places
(B) y′ ̸= (ln x)xx because (ln x)xx = 0when x = 1, and this func ondoes not have a horizontaltangent at x = 1.
..x
.
y
..1
..
1
Why not?Answer
(A) y′ ̸= xx because xx > 0 for allx > 0, and this func ondecreases at some places
(B) y′ ̸= (ln x)xx because (ln x)xx = 0when x = 1, and this func ondoes not have a horizontaltangent at x = 1.
..x
.
y
..1
..
1
It’s neither!Solu onIf y = xx, then
ln y = x ln x1ydydx
= x · 1x+ ln x = 1+ ln x
dydx
= (1+ ln x)xx = xx + (ln x)xx
It’s neither!Solu onIf y = xx, then
ln y = x ln x
1ydydx
= x · 1x+ ln x = 1+ ln x
dydx
= (1+ ln x)xx = xx + (ln x)xx
It’s neither!Solu onIf y = xx, then
ln y = x ln x1ydydx
= x · 1x+ ln x = 1+ ln x
dydx
= (1+ ln x)xx = xx + (ln x)xx
It’s neither!Solu onIf y = xx, then
ln y = x ln x1ydydx
= x · 1x+ ln x = 1+ ln x
dydx
= (1+ ln x)xx = xx + (ln x)xx
Or both?Solu on
ddx
xx = xx + (ln x)xx = (1+ ln x)xx
Remarks
I Each of these terms is one of thewrong answers!
I y′ < 0 on the interval (0, e−1)
I y′ = 0 when x = e−1
..x
.
y
..1
..
1
Or both?Solu on
ddx
xx = xx + (ln x)xx = (1+ ln x)xx
Remarks
I Each of these terms is one of thewrong answers!
I y′ < 0 on the interval (0, e−1)
I y′ = 0 when x = e−1
..x
.
y
..1
..
1
Or both?Solu on
ddx
xx = xx + (ln x)xx = (1+ ln x)xx
Remarks
I Each of these terms is one of thewrong answers!
I y′ < 0 on the interval (0, e−1)
I y′ = 0 when x = e−1
..x
.
y
..1
..
1
Or both?Solu on
ddx
xx = xx + (ln x)xx = (1+ ln x)xx
Remarks
I Each of these terms is one of thewrong answers!
I y′ < 0 on the interval (0, e−1)
I y′ = 0 when x = e−1
..x
.
y
..1
..
1
Or both?Solu on
ddx
xx = xx + (ln x)xx = (1+ ln x)xx
Remarks
I Each of these terms is one of thewrong answers!
I y′ < 0 on the interval (0, e−1)
I y′ = 0 when x = e−1
..x
.
y
..1
..
1
Or both?Solu on
ddx
xx = xx + (ln x)xx = (1+ ln x)xx
Remarks
I Each of these terms is one of thewrong answers!
I y′ < 0 on the interval (0, e−1)
I y′ = 0 when x = e−1
..x
.
y
..1
..
1
Or both?Solu on
ddx
xx = xx + (ln x)xx = (1+ ln x)xx
Remarks
I Each of these terms is one of thewrong answers!
I y′ < 0 on the interval (0, e−1)
I y′ = 0 when x = e−1
..x
.
y
..1
..
1
Derivatives of power functionswith any exponent
Fact (The power rule)
Let y = xr. Then y′ = rxr−1.
Proof.
y = xr =⇒ ln y = r ln x
Now differen ate:
1ydydx
=rx
=⇒ dydx
= ryx= rxr−1
Derivatives of power functionswith any exponent
Fact (The power rule)
Let y = xr. Then y′ = rxr−1.
Proof.
y = xr =⇒ ln y = r ln x
Now differen ate:
1ydydx
=rx
=⇒ dydx
= ryx= rxr−1
SummaryI Deriva ves ofLogarithmic andExponen al Func ons
I LogarithmicDifferen a on can allowus to avoid the productand quo ent rules.
I We are finally done withthe Power Rule!
y y′
ex ex
ax (ln a) · ax
ln x1x
loga x1ln a
· 1x