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The exponential function is pretty much the only function whose derivative is itself. The derivative of the natural logarithm function is also beautiful as it fills in an important gap. Finally, the technique of logarithmic differentiation allows us to find derivatives without the product rule.
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Section 3.3Derivatives of Exponential and
Logarithmic Functions
V63.0121.021, Calculus I
New York University
October 25, 2010
Announcements
I Midterm is graded. Please see FAQ.
I Quiz 3 next week on 2.6, 2.8, 3.1, 3.2
Announcements
I Midterm is graded. Pleasesee FAQ.
I Quiz 3 next week on 2.6,2.8, 3.1, 3.2
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 2 / 38
Objectives
I Know the derivatives of theexponential functions (withany base)
I Know the derivatives of thelogarithmic functions (withany base)
I Use the technique oflogarithmic differentiation tofind derivatives of functionsinvolving roducts, quotients,and/or exponentials.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 3 / 38
Outline
Recall Section 3.1–3.2
Derivative of the natural exponential functionExponential Growth
Derivative of the natural logarithm function
Derivatives of other exponentials and logarithmsOther exponentialsOther logarithms
Logarithmic DifferentiationThe power rule for irrational powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 4 / 38
Conventions on power expressions
Let a be a positive real number.
I If n is a positive whole number, then an = a · a · · · · · a︸ ︷︷ ︸n factors
I a0 = 1.
I For any real number r , a−r =1
ar.
I For any positive whole number n, a1/n = n√
a.
There is only one continuous function which satisfies all of the above. Wecall it the exponential function with base a.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 5 / 38
Properties of exponential Functions
Theorem
If a > 0 and a 6= 1, then f (x) = ax is a continuous function with domain(−∞,∞) and range (0,∞). In particular, ax > 0 for all x. For any realnumbers x and y, and positive numbers a and b we have
I ax+y = axay
I ax−y =ax
ay
(negative exponents mean reciprocals)
I (ax)y = axy
(fractional exponents mean roots)
I (ab)x = axbx
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 6 / 38
Properties of exponential Functions
Theorem
If a > 0 and a 6= 1, then f (x) = ax is a continuous function with domain(−∞,∞) and range (0,∞). In particular, ax > 0 for all x. For any realnumbers x and y, and positive numbers a and b we have
I ax+y = axay
I ax−y =ax
ay(negative exponents mean reciprocals)
I (ax)y = axy
(fractional exponents mean roots)
I (ab)x = axbx
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 6 / 38
Properties of exponential Functions
Theorem
If a > 0 and a 6= 1, then f (x) = ax is a continuous function with domain(−∞,∞) and range (0,∞). In particular, ax > 0 for all x. For any realnumbers x and y, and positive numbers a and b we have
I ax+y = axay
I ax−y =ax
ay(negative exponents mean reciprocals)
I (ax)y = axy (fractional exponents mean roots)
I (ab)x = axbx
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 6 / 38
Graphs of various exponential functions
x
y
y = 1x
y = 2xy = 3xy = 10x y = 1.5xy = (1/2)xy = (1/3)x y = (1/10)xy = (2/3)x
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 7 / 38
The magic number
Definition
e = limn→∞
(1 +
1
n
)n
= limh→0+
(1 + h)1/h
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 8 / 38
Existence of eSee Appendix B
I We can experimentally verifythat this number exists andis
e ≈ 2.718281828459045 . . .
I e is irrational
I e is transcendental
n
(1 +
1
n
)n
1 2
2 2.25
3 2.37037
10 2.59374
100 2.70481
1000 2.71692
106 2.71828
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 9 / 38
Logarithms
Definition
I The base a logarithm loga x is the inverse of the function ax
y = loga x ⇐⇒ x = ay
I The natural logarithm ln x is the inverse of ex . Soy = ln x ⇐⇒ x = ey .
Facts
(i) loga(x1 · x2) = loga x1 + loga x2
(ii) loga
(x1x2
)= loga x1 − loga x2
(iii) loga(x r ) = r loga x
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 10 / 38
Graphs of logarithmic functions
x
yy = 2x
y = log2 x
(0, 1)
(1, 0)
y = 3x
y = log3 x
y = 10x
y = log10 x
y = ex
y = ln x
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 11 / 38
Change of base formula for logarithms
Fact
If a > 0 and a 6= 1, and the same for b, then
loga x =logb x
logb a
Proof.
I If y = loga x , then x = ay
I So logb x = logb(ay ) = y logb a
I Therefore
y = loga x =logb x
logb a
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 12 / 38
Upshot of changing base
The point of the change of base formula
loga x =logb x
logb a=
1
logb a· logb x = (constant) · logb x
is that all the logarithmic functions are multiples of each other. So justpick one and call it your favorite.
I Engineers like the common logarithm log = log10I Computer scientists like the binary logarithm lg = log2I Mathematicians like natural logarithm ln = loge
Naturally, we will follow the mathematicians. Just don’t pronounce it“lawn.”
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 13 / 38
Outline
Recall Section 3.1–3.2
Derivative of the natural exponential functionExponential Growth
Derivative of the natural logarithm function
Derivatives of other exponentials and logarithmsOther exponentialsOther logarithms
Logarithmic DifferentiationThe power rule for irrational powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 14 / 38
Derivatives of Exponential Functions
Fact
If f (x) = ax , then f ′(x) = f ′(0)ax .
Proof.
Follow your nose:
f ′(x) = limh→0
f (x + h)− f (x)
h= lim
h→0
ax+h − ax
h
= limh→0
axah − ax
h= ax · lim
h→0
ah − 1
h= ax · f ′(0).
To reiterate: the derivative of an exponential function is a constant timesthat function. Much different from polynomials!
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 15 / 38
Derivatives of Exponential Functions
Fact
If f (x) = ax , then f ′(x) = f ′(0)ax .
Proof.
Follow your nose:
f ′(x) = limh→0
f (x + h)− f (x)
h= lim
h→0
ax+h − ax
h
= limh→0
axah − ax
h= ax · lim
h→0
ah − 1
h= ax · f ′(0).
To reiterate: the derivative of an exponential function is a constant timesthat function. Much different from polynomials!
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 15 / 38
Derivatives of Exponential Functions
Fact
If f (x) = ax , then f ′(x) = f ′(0)ax .
Proof.
Follow your nose:
f ′(x) = limh→0
f (x + h)− f (x)
h= lim
h→0
ax+h − ax
h
= limh→0
axah − ax
h= ax · lim
h→0
ah − 1
h= ax · f ′(0).
To reiterate: the derivative of an exponential function is a constant timesthat function. Much different from polynomials!
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 15 / 38
The funny limit in the case of e
Remember the definition of e:
e = limn→∞
(1 +
1
n
)n
= limh→0
(1 + h)1/h
Question
What is limh→0
eh − 1
h?
Answer
If h is small enough, e ≈ (1 + h)1/h. So
eh − 1
h≈[(1 + h)1/h
]h − 1
h=
(1 + h)− 1
h=
h
h= 1
So in the limit we get equality: limh→0
eh − 1
h= 1
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 16 / 38
The funny limit in the case of e
Remember the definition of e:
e = limn→∞
(1 +
1
n
)n
= limh→0
(1 + h)1/h
Question
What is limh→0
eh − 1
h?
Answer
If h is small enough, e ≈ (1 + h)1/h. So
eh − 1
h≈[(1 + h)1/h
]h − 1
h=
(1 + h)− 1
h=
h
h= 1
So in the limit we get equality: limh→0
eh − 1
h= 1
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 16 / 38
The funny limit in the case of e
Remember the definition of e:
e = limn→∞
(1 +
1
n
)n
= limh→0
(1 + h)1/h
Question
What is limh→0
eh − 1
h?
Answer
If h is small enough, e ≈ (1 + h)1/h. So
eh − 1
h≈[(1 + h)1/h
]h − 1
h=
(1 + h)− 1
h=
h
h= 1
So in the limit we get equality: limh→0
eh − 1
h= 1
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 16 / 38
Derivative of the natural exponential function
Fromd
dxax =
(limh→0
ah − 1
h
)ax and lim
h→0
eh − 1
h= 1
we get:
Theorem
d
dxex = ex
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 17 / 38
Exponential Growth
I Commonly misused term to say something grows exponentially
I It means the rate of change (derivative) is proportional to the currentvalue
I Examples: Natural population growth, compounded interest, socialnetworks
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 18 / 38
Examples
Examples
Find derivatives of these functions:
I e3x
I ex2
I x2ex
Solution
Id
dxe3x = 3e3x
Id
dxex
2= ex
2 d
dx(x2) = 2xex
2
Id
dxx2ex = 2xex + x2ex
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 19 / 38
Examples
Examples
Find derivatives of these functions:
I e3x
I ex2
I x2ex
Solution
Id
dxe3x = 3e3x
Id
dxex
2= ex
2 d
dx(x2) = 2xex
2
Id
dxx2ex = 2xex + x2ex
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 19 / 38
Examples
Examples
Find derivatives of these functions:
I e3x
I ex2
I x2ex
Solution
Id
dxe3x = 3e3x
Id
dxex
2= ex
2 d
dx(x2) = 2xex
2
Id
dxx2ex = 2xex + x2ex
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 19 / 38
Examples
Examples
Find derivatives of these functions:
I e3x
I ex2
I x2ex
Solution
Id
dxe3x = 3e3x
Id
dxex
2= ex
2 d
dx(x2) = 2xex
2
Id
dxx2ex = 2xex + x2ex
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 19 / 38
Outline
Recall Section 3.1–3.2
Derivative of the natural exponential functionExponential Growth
Derivative of the natural logarithm function
Derivatives of other exponentials and logarithmsOther exponentialsOther logarithms
Logarithmic DifferentiationThe power rule for irrational powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 20 / 38
Derivative of the natural logarithm function
Let y = ln x . Thenx = ey so
eydy
dx= 1
=⇒ dy
dx=
1
ey=
1
x
We have discovered:
Fact
d
dxln x =
1
x
x
y
ln x1
x
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 21 / 38
Derivative of the natural logarithm function
Let y = ln x . Thenx = ey so
eydy
dx= 1
=⇒ dy
dx=
1
ey=
1
x
We have discovered:
Fact
d
dxln x =
1
x
x
y
ln x1
x
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 21 / 38
Derivative of the natural logarithm function
Let y = ln x . Thenx = ey so
eydy
dx= 1
=⇒ dy
dx=
1
ey=
1
x
We have discovered:
Fact
d
dxln x =
1
x
x
y
ln x1
x
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 21 / 38
Derivative of the natural logarithm function
Let y = ln x . Thenx = ey so
eydy
dx= 1
=⇒ dy
dx=
1
ey=
1
x
We have discovered:
Fact
d
dxln x =
1
x
x
y
ln x1
x
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 21 / 38
Derivative of the natural logarithm function
Let y = ln x . Thenx = ey so
eydy
dx= 1
=⇒ dy
dx=
1
ey=
1
x
We have discovered:
Fact
d
dxln x =
1
x
x
y
ln x
1
x
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 21 / 38
Derivative of the natural logarithm function
Let y = ln x . Thenx = ey so
eydy
dx= 1
=⇒ dy
dx=
1
ey=
1
x
We have discovered:
Fact
d
dxln x =
1
x
x
y
ln x1
x
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 21 / 38
The Tower of Powers
y y ′
x3 3x2
x2 2x1
x1 1x0
x0 0
? ?
x−1 −1x−2
x−2 −2x−3
I The derivative of a powerfunction is a power functionof one lower power
I Each power function is thederivative of another powerfunction, except x−1
I ln x fills in this gap precisely.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 22 / 38
The Tower of Powers
y y ′
x3 3x2
x2 2x1
x1 1x0
x0 0
? x−1
x−1 −1x−2
x−2 −2x−3
I The derivative of a powerfunction is a power functionof one lower power
I Each power function is thederivative of another powerfunction, except x−1
I ln x fills in this gap precisely.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 22 / 38
The Tower of Powers
y y ′
x3 3x2
x2 2x1
x1 1x0
x0 0
ln x x−1
x−1 −1x−2
x−2 −2x−3
I The derivative of a powerfunction is a power functionof one lower power
I Each power function is thederivative of another powerfunction, except x−1
I ln x fills in this gap precisely.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 22 / 38
Examples
Examples
Find derivatives of these functions:
I ln(3x)
I x ln x
I ln√
x
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 23 / 38
Examples
Example
Findd
dxln(3x).
Solution (chain rule way)
d
dxln(3x) =
1
3x· 3 =
1
x
Solution (properties of logarithms way)
d
dxln(3x) =
d
dx(ln(3) + ln(x)) = 0 +
1
x=
1
x
The first answer might be surprising until you see the second solution.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 24 / 38
Examples
Example
Findd
dxln(3x).
Solution (chain rule way)
d
dxln(3x) =
1
3x· 3 =
1
x
Solution (properties of logarithms way)
d
dxln(3x) =
d
dx(ln(3) + ln(x)) = 0 +
1
x=
1
x
The first answer might be surprising until you see the second solution.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 24 / 38
Examples
Example
Findd
dxln(3x).
Solution (chain rule way)
d
dxln(3x) =
1
3x· 3 =
1
x
Solution (properties of logarithms way)
d
dxln(3x) =
d
dx(ln(3) + ln(x)) = 0 +
1
x=
1
x
The first answer might be surprising until you see the second solution.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 24 / 38
Examples
Example
Findd
dxln(3x).
Solution (chain rule way)
d
dxln(3x) =
1
3x· 3 =
1
x
Solution (properties of logarithms way)
d
dxln(3x) =
d
dx(ln(3) + ln(x)) = 0 +
1
x=
1
x
The first answer might be surprising until you see the second solution.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 24 / 38
Examples
Example
Findd
dxx ln x
Solution
The product rule is in play here:
d
dxx ln x =
(d
dxx
)ln x + x
(d
dxln x
)= 1 · ln x + x · 1
x= ln x + 1
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 25 / 38
Examples
Example
Findd
dxx ln x
Solution
The product rule is in play here:
d
dxx ln x =
(d
dxx
)ln x + x
(d
dxln x
)= 1 · ln x + x · 1
x= ln x + 1
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 25 / 38
Examples
Example
Findd
dxln√
x .
Solution (chain rule way)
d
dxln√
x =1√x
d
dx
√x =
1√x
1
2√
x=
1
2x
Solution (properties of logarithms way)
d
dxln√
x =d
dx
(1
2ln x
)=
1
2
d
dxln x =
1
2· 1
x
The first answer might be surprising until you see the second solution.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 26 / 38
Examples
Example
Findd
dxln√
x .
Solution (chain rule way)
d
dxln√
x =1√x
d
dx
√x =
1√x
1
2√
x=
1
2x
Solution (properties of logarithms way)
d
dxln√
x =d
dx
(1
2ln x
)=
1
2
d
dxln x =
1
2· 1
x
The first answer might be surprising until you see the second solution.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 26 / 38
Examples
Example
Findd
dxln√
x .
Solution (chain rule way)
d
dxln√
x =1√x
d
dx
√x =
1√x
1
2√
x=
1
2x
Solution (properties of logarithms way)
d
dxln√
x =d
dx
(1
2ln x
)=
1
2
d
dxln x =
1
2· 1
x
The first answer might be surprising until you see the second solution.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 26 / 38
Examples
Example
Findd
dxln√
x .
Solution (chain rule way)
d
dxln√
x =1√x
d
dx
√x =
1√x
1
2√
x=
1
2x
Solution (properties of logarithms way)
d
dxln√
x =d
dx
(1
2ln x
)=
1
2
d
dxln x =
1
2· 1
x
The first answer might be surprising until you see the second solution.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 26 / 38
Outline
Recall Section 3.1–3.2
Derivative of the natural exponential functionExponential Growth
Derivative of the natural logarithm function
Derivatives of other exponentials and logarithmsOther exponentialsOther logarithms
Logarithmic DifferentiationThe power rule for irrational powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 27 / 38
Other logarithms
Example
Use implicit differentiation to findd
dxax .
Solution
Let y = ax , soln y = ln ax = x ln a
Differentiate implicitly:
1
y
dy
dx= ln a =⇒ dy
dx= (ln a)y = (ln a)ax
Before we showed y ′ = y ′(0)y , so now we know that
ln a = limh→0
ah − 1
h
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 28 / 38
Other logarithms
Example
Use implicit differentiation to findd
dxax .
Solution
Let y = ax , soln y = ln ax = x ln a
Differentiate implicitly:
1
y
dy
dx= ln a =⇒ dy
dx= (ln a)y = (ln a)ax
Before we showed y ′ = y ′(0)y , so now we know that
ln a = limh→0
ah − 1
h
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 28 / 38
Other logarithms
Example
Use implicit differentiation to findd
dxax .
Solution
Let y = ax , soln y = ln ax = x ln a
Differentiate implicitly:
1
y
dy
dx= ln a =⇒ dy
dx= (ln a)y = (ln a)ax
Before we showed y ′ = y ′(0)y , so now we know that
ln a = limh→0
ah − 1
h
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 28 / 38
Other logarithms
Example
Use implicit differentiation to findd
dxax .
Solution
Let y = ax , soln y = ln ax = x ln a
Differentiate implicitly:
1
y
dy
dx= ln a =⇒ dy
dx= (ln a)y = (ln a)ax
Before we showed y ′ = y ′(0)y , so now we know that
ln a = limh→0
ah − 1
h
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 28 / 38
Other logarithms
Example
Findd
dxloga x .
Solution
Let y = loga x, so ay = x. Now differentiate implicitly:
(ln a)aydy
dx= 1 =⇒ dy
dx=
1
ay ln a=
1
x ln a
Another way to see this is to take the natural logarithm:
ay = x =⇒ y ln a = ln x =⇒ y =ln x
ln a
Sody
dx=
1
ln a
1
x.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 29 / 38
Other logarithms
Example
Findd
dxloga x .
Solution
Let y = loga x, so ay = x.
Now differentiate implicitly:
(ln a)aydy
dx= 1 =⇒ dy
dx=
1
ay ln a=
1
x ln a
Another way to see this is to take the natural logarithm:
ay = x =⇒ y ln a = ln x =⇒ y =ln x
ln a
Sody
dx=
1
ln a
1
x.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 29 / 38
Other logarithms
Example
Findd
dxloga x .
Solution
Let y = loga x, so ay = x. Now differentiate implicitly:
(ln a)aydy
dx= 1 =⇒ dy
dx=
1
ay ln a=
1
x ln a
Another way to see this is to take the natural logarithm:
ay = x =⇒ y ln a = ln x =⇒ y =ln x
ln a
Sody
dx=
1
ln a
1
x.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 29 / 38
Other logarithms
Example
Findd
dxloga x .
Solution
Let y = loga x, so ay = x. Now differentiate implicitly:
(ln a)aydy
dx= 1 =⇒ dy
dx=
1
ay ln a=
1
x ln a
Another way to see this is to take the natural logarithm:
ay = x =⇒ y ln a = ln x =⇒ y =ln x
ln a
Sody
dx=
1
ln a
1
x.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 29 / 38
More examples
Example
Findd
dxlog2(x2 + 1)
Answer
dy
dx=
1
ln 2
1
x2 + 1(2x) =
2x
(ln 2)(x2 + 1)
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 30 / 38
More examples
Example
Findd
dxlog2(x2 + 1)
Answer
dy
dx=
1
ln 2
1
x2 + 1(2x) =
2x
(ln 2)(x2 + 1)
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 30 / 38
Outline
Recall Section 3.1–3.2
Derivative of the natural exponential functionExponential Growth
Derivative of the natural logarithm function
Derivatives of other exponentials and logarithmsOther exponentialsOther logarithms
Logarithmic DifferentiationThe power rule for irrational powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 31 / 38
A nasty derivative
Example
Let y =(x2 + 1)
√x + 3
x − 1. Find y ′.
Solution
We use the quotient 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
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 32 / 38
A nasty derivative
Example
Let y =(x2 + 1)
√x + 3
x − 1. Find y ′.
Solution
We use the quotient 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
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 32 / 38
Another way
y =(x2 + 1)
√x + 3
x − 1
ln y = ln(x2 + 1) +1
2ln(x + 3)− ln(x − 1)
1
y
dy
dx=
2x
x2 + 1+
1
2(x + 3)− 1
x − 1
So
dy
dx=
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)y
=
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)(x2 + 1)
√x + 3
x − 1
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 33 / 38
Compare and contrast
I Using the product, quotient, 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 differentiation:
y ′ =
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)(x2 + 1)
√x + 3
(x − 1)
I Are these the same?
Yes.
I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation?
Products, quotients, and powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38
Compare and contrast
I Using the product, quotient, 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 differentiation:
y ′ =
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)(x2 + 1)
√x + 3
(x − 1)
I Are these the same?
Yes.
I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation?
Products, quotients, and powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38
Compare and contrast
I Using the product, quotient, 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 differentiation:
y ′ =
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)(x2 + 1)
√x + 3
(x − 1)
I Are these the same?
Yes.
I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation?
Products, quotients, and powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38
Compare and contrast
I Using the product, quotient, 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 differentiation:
y ′ =
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)(x2 + 1)
√x + 3
(x − 1)
I Are these the same?
Yes.
I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation?
Products, quotients, and powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38
Compare and contrast
I Using the product, quotient, 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 differentiation:
y ′ =
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)(x2 + 1)
√x + 3
(x − 1)
I Are these the same?
Yes.
I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation?
Products, quotients, and powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38
Compare and contrast
I Using the product, quotient, 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 differentiation:
y ′ =
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)(x2 + 1)
√x + 3
(x − 1)
I Are these the same?
Yes.
I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation?
Products, quotients, and powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38
Compare and contrast
I Using the product, quotient, 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 differentiation:
y ′ =
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)(x2 + 1)
√x + 3
(x − 1)
I Are these the same?
Yes.
I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation?
Products, quotients, and powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38
Compare and contrast
I Using the product, quotient, 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 differentiation:
y ′ =
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)(x2 + 1)
√x + 3
(x − 1)
I Are these the same?
Yes.
I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation?
Products, quotients, and powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38
Compare and contrast
I Using the product, quotient, 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 differentiation:
y ′ =
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)(x2 + 1)
√x + 3
(x − 1)
I Are these the same?
Yes.
I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation?
Products, quotients, and powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38
Compare and contrast
I Using the product, quotient, 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 differentiation:
y ′ =
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)(x2 + 1)
√x + 3
(x − 1)
I Are these the same?
Yes.
I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation?
Products, quotients, and powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38
Compare and contrast
I Using the product, quotient, 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 differentiation:
y ′ =
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)(x2 + 1)
√x + 3
(x − 1)
I Are these the same?
Yes.
I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation?
Products, quotients, and powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38
Compare and contrast
I Using the product, quotient, 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 differentiation:
y ′ =
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)(x2 + 1)
√x + 3
(x − 1)
I Are these the same?
Yes.
I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation?
Products, quotients, and powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38
Compare and contrast
I Using the product, quotient, 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 differentiation:
y ′ =
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)(x2 + 1)
√x + 3
(x − 1)
I Are these the same?
Yes.
I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation?
Products, quotients, and powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38
Compare and contrast
I Using the product, quotient, 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 differentiation:
y ′ =
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)(x2 + 1)
√x + 3
(x − 1)
I Are these the same? Yes.
I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation?
Products, quotients, and powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38
Compare and contrast
I Using the product, quotient, 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 differentiation:
y ′ =
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)(x2 + 1)
√x + 3
(x − 1)
I Are these the same? Yes.
I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation?
Products, quotients, and powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38
Compare and contrast
I Using the product, quotient, 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 differentiation:
y ′ =
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)(x2 + 1)
√x + 3
(x − 1)
I Are these the same? Yes.
I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation?
Products, quotients, and powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38
Compare and contrast
I Using the product, quotient, 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 differentiation:
y ′ =
(2x
x2 + 1+
1
2(x + 3)− 1
x − 1
)(x2 + 1)
√x + 3
(x − 1)
I Are these the same? Yes.
I Which do you like better?
I What kinds of expressions are well-suited for logarithmicdifferentiation? Products, quotients, and powers
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 34 / 38
Derivatives of powers
Question
Let y = xx . Which of these is true?
(A) Since y is a power function,y ′ = x · xx−1 = xx .
(B) Since y is an exponentialfunction, y ′ = (ln x) · xx
(C) Neitherx
y
1
1
Answer
(A) This can’t be y ′ because xx > 0 for all x > 0, and this function decreases atsome places
(B) This can’t be y ′ because (ln x)xx = 0 when x = 1, and this function does nothave a horizontal tangent at x = 1.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 35 / 38
Derivatives of powers
Question
Let y = xx . Which of these is true?
(A) Since y is a power function,y ′ = x · xx−1 = xx .
(B) Since y is an exponentialfunction, y ′ = (ln x) · xx
(C) Neitherx
y
1
1
Answer
(A) This can’t be y ′ because xx > 0 for all x > 0, and this function decreases atsome places
(B) This can’t be y ′ because (ln x)xx = 0 when x = 1, and this function does nothave a horizontal tangent at x = 1.
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 35 / 38
It’s neither! Or both?
Solution
If y = xx , then
ln y = x ln x
1
y
dy
dx= x · 1
x+ ln x = 1 + ln x
dy
dx= (1 + ln x)xx = xx + (ln x)xx
Remarks
I Each of these terms is one of the wrong answers!
I y ′ < 0 on the interval (0, e−1)
I y ′ = 0 when x = e−1
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 36 / 38
It’s neither! Or both?
Solution
If y = xx , then
ln y = x ln x
1
y
dy
dx= x · 1
x+ ln x = 1 + ln x
dy
dx= (1 + ln x)xx = xx + (ln x)xx
Remarks
I Each of these terms is one of the wrong answers!
I y ′ < 0 on the interval (0, e−1)
I y ′ = 0 when x = e−1
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 36 / 38
It’s neither! Or both?
Solution
If y = xx , then
ln y = x ln x
1
y
dy
dx= x · 1
x+ ln x = 1 + ln x
dy
dx= (1 + ln x)xx = xx + (ln x)xx
Remarks
I Each of these terms is one of the wrong answers!
I y ′ < 0 on the interval (0, e−1)
I y ′ = 0 when x = e−1
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 36 / 38
Derivatives of power functions with any exponent
Fact (The power rule)
Let y = x r . Then y ′ = rx r−1.
Proof.
y = x r =⇒ ln y = r ln x
Now differentiate:
1
y
dy
dx=
r
x
=⇒ dy
dx= r
y
x= rx r−1
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 37 / 38
Derivatives of power functions with any exponent
Fact (The power rule)
Let y = x r . Then y ′ = rx r−1.
Proof.
y = x r =⇒ ln y = r ln x
Now differentiate:
1
y
dy
dx=
r
x
=⇒ dy
dx= r
y
x= rx r−1
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 37 / 38
Summary
I Derivatives of logarithmic and exponential functions:
y y ′
ex ex
ax (ln a) · ax
ln x1
x
loga x1
ln a· 1
x
I Logarithmic Differentiation can allow us to avoid the product andquotient rules.
I We are finally done with the Power Rule!
V63.0121.021, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 38 / 38