Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 021 slides)

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.


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


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.


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



Theorem


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



Theorem


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


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


The magic number

Definition

e = limn→∞

(1 +

1

n

)n

= limh→0+

(1 + h)1/h


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


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


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


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


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.”


Outline







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!



Fact


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).




Fact


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).



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




e = limn→∞

(1 +

1

n

)n

= limh→0

(1 + h)1/h

Question

What is limh→0

eh − 1

h?

Answer


eh − 1

h≈[(1 + h)1/h

]h − 1

h=

(1 + h)− 1

h=

h

h= 1


eh − 1

h= 1




e = limn→∞

(1 +

1

n

)n

= limh→0

(1 + h)1/h

Question

What is limh→0

eh − 1

h?

Answer


eh − 1

h≈[(1 + h)1/h

]h − 1

h=

(1 + h)− 1

h=

h

h= 1


eh − 1

h= 1


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


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


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


Examples

Examples


I e3x

I ex2

I x2ex

Solution

Id

dxe3x = 3e3x

Id

dxex

2= ex

2 d

dx(x2) = 2xex

2

Id



Examples

Examples


I e3x

I ex2

I x2ex

Solution

Id

dxe3x = 3e3x

Id

dxex

2= ex

2 d

dx(x2) = 2xex

2

Id



Examples

Examples


I e3x

I ex2

I x2ex

Solution

Id

dxe3x = 3e3x

Id

dxex

2= ex

2 d

dx(x2) = 2xex

2

Id



Outline








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




eydy

dx= 1

=⇒ dy

dx=

1

ey=

1

x

We have discovered:

Fact

d

dxln x =

1

x

x

y

ln x1

x




eydy

dx= 1

=⇒ dy

dx=

1

ey=

1

x

We have discovered:

Fact

d

dxln x =

1

x

x

y

ln x1

x




eydy

dx= 1

=⇒ dy

dx=

1

ey=

1

x

We have discovered:

Fact

d

dxln x =

1

x

x

y

ln x1

x




eydy

dx= 1

=⇒ dy

dx=

1

ey=

1

x

We have discovered:

Fact

d

dxln x =

1

x

x

y

ln x

1

x




eydy

dx= 1

=⇒ dy

dx=

1

ey=

1

x

We have discovered:

Fact

d

dxln x =

1

x

x

y

ln x1

x


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.


The Tower of Powers

y y ′

x3 3x2

x2 2x1

x1 1x0

x0 0

? x−1

x−1 −1x−2

x−2 −2x−3





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





Examples

Examples


I ln(3x)

I x ln x

I ln√

x


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.


Examples

Example

Findd

dxln(3x).


d

dxln(3x) =

1

3x· 3 =

1

x


d

dxln(3x) =

d

dx(ln(3) + ln(x)) = 0 +

1

x=

1

x



Examples

Example

Findd

dxln(3x).


d

dxln(3x) =

1

3x· 3 =

1

x


d

dxln(3x) =

d

dx(ln(3) + ln(x)) = 0 +

1

x=

1

x



Examples

Example

Findd

dxln(3x).


d

dxln(3x) =

1

3x· 3 =

1

x


d

dxln(3x) =

d

dx(ln(3) + ln(x)) = 0 +

1

x=

1

x



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


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


Examples

Example

Findd

dxln√

x .


d

dxln√

x =1√x

d

dx

√x =

1√x

1

2√

x=

1

2x


d

dxln√

x =d

dx

(1

2ln x

)=

1

2

d

dxln x =

1

2· 1

x



Examples

Example

Findd

dxln√

x .


d

dxln√

x =1√x

d

dx

√x =

1√x

1

2√

x=

1

2x


d

dxln√

x =d

dx

(1

2ln x

)=

1

2

d

dxln x =

1

2· 1

x



Examples

Example

Findd

dxln√

x .


d

dxln√

x =1√x

d

dx

√x =

1√x

1

2√

x=

1

2x


d

dxln√

x =d

dx

(1

2ln x

)=

1

2

d

dxln x =

1

2· 1

x



Examples

Example

Findd

dxln√

x .


d

dxln√

x =1√x

d

dx

√x =

1√x

1

2√

x=

1

2x


d

dxln√

x =d

dx

(1

2ln x

)=

1

2

d

dxln x =

1

2· 1

x



Outline







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


Other logarithms

Example


dxax .

Solution



1

y

dy

dx= ln a =⇒ dy



ln a = limh→0

ah − 1

h


Other logarithms

Example


dxax .

Solution



1

y

dy

dx= ln a =⇒ dy



ln a = limh→0

ah − 1

h


Other logarithms

Example


dxax .

Solution



1

y

dy

dx= ln a =⇒ dy



ln a = limh→0

ah − 1

h


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.


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



ln a

Sody

dx=

1

ln a

1

x.


Other logarithms

Example

Findd

dxloga x .

Solution


(ln a)aydy

dx= 1 =⇒ dy

dx=

1

ay ln a=

1

x ln a



ln a

Sody

dx=

1

ln a

1

x.


Other logarithms

Example

Findd

dxloga x .

Solution


(ln a)aydy

dx= 1 =⇒ dy

dx=

1

ay ln a=

1

x ln a



ln a

Sody

dx=

1

ln a

1

x.


More examples

Example

Findd

dxlog2(x2 + 1)

Answer

dy

dx=

1

ln 2

1

x2 + 1(2x) =

2x

(ln 2)(x2 + 1)


More examples

Example

Findd

dxlog2(x2 + 1)

Answer

dy

dx=

1

ln 2

1

x2 + 1(2x) =

2x

(ln 2)(x2 + 1)


Outline







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


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


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


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




y ′ =2x√

x + 3

(x − 1)+

(x2 + 1)

2√

x + 3(x − 1)− (x2 + 1)

√x + 3

(x − 1)2


y ′ =

(2x

x2 + 1+

1

2(x + 3)− 1

x − 1

)(x2 + 1)

√x + 3

(x − 1)


Yes.







y ′ =2x√

x + 3

(x − 1)+

(x2 + 1)

2√

x + 3(x − 1)− (x2 + 1)

√x + 3

(x − 1)2


y ′ =

(2x

x2 + 1+

1

2(x + 3)− 1

x − 1

)(x2 + 1)

√x + 3

(x − 1)


Yes.







y ′ =2x√

x + 3

(x − 1)+

(x2 + 1)

2√

x + 3(x − 1)− (x2 + 1)

√x + 3

(x − 1)2


y ′ =

(2x

x2 + 1+

1

2(x + 3)− 1

x − 1

)(x2 + 1)

√x + 3

(x − 1)


Yes.







y ′ =2x√

x + 3

(x − 1)+

(x2 + 1)

2√

x + 3(x − 1)− (x2 + 1)

√x + 3

(x − 1)2


y ′ =

(2x

x2 + 1+

1

2(x + 3)− 1

x − 1

)(x2 + 1)

√x + 3

(x − 1)


Yes.







y ′ =2x√

x + 3

(x − 1)+

(x2 + 1)

2√

x + 3(x − 1)− (x2 + 1)

√x + 3

(x − 1)2


y ′ =

(2x

x2 + 1+

1

2(x + 3)− 1

x − 1

)(x2 + 1)

√x + 3

(x − 1)


Yes.







y ′ =2x√

x + 3

(x − 1)+

(x2 + 1)

2√

x + 3(x − 1)− (x2 + 1)

√x + 3

(x − 1)2


y ′ =

(2x

x2 + 1+

1

2(x + 3)− 1

x − 1

)(x2 + 1)

√x + 3

(x − 1)


Yes.







y ′ =2x√

x + 3

(x − 1)+

(x2 + 1)

2√

x + 3(x − 1)− (x2 + 1)

√x + 3

(x − 1)2


y ′ =

(2x

x2 + 1+

1

2(x + 3)− 1

x − 1

)(x2 + 1)

√x + 3

(x − 1)


Yes.







y ′ =2x√

x + 3

(x − 1)+

(x2 + 1)

2√

x + 3(x − 1)− (x2 + 1)

√x + 3

(x − 1)2


y ′ =

(2x

x2 + 1+

1

2(x + 3)− 1

x − 1

)(x2 + 1)

√x + 3

(x − 1)


Yes.







y ′ =2x√

x + 3

(x − 1)+

(x2 + 1)

2√

x + 3(x − 1)− (x2 + 1)

√x + 3

(x − 1)2


y ′ =

(2x

x2 + 1+

1

2(x + 3)− 1

x − 1

)(x2 + 1)

√x + 3

(x − 1)


Yes.







y ′ =2x√

x + 3

(x − 1)+

(x2 + 1)

2√

x + 3(x − 1)− (x2 + 1)

√x + 3

(x − 1)2


y ′ =

(2x

x2 + 1+

1

2(x + 3)− 1

x − 1

)(x2 + 1)

√x + 3

(x − 1)


Yes.







y ′ =2x√

x + 3

(x − 1)+

(x2 + 1)

2√

x + 3(x − 1)− (x2 + 1)

√x + 3

(x − 1)2


y ′ =

(2x

x2 + 1+

1

2(x + 3)− 1

x − 1

)(x2 + 1)

√x + 3

(x − 1)


Yes.







y ′ =2x√

x + 3

(x − 1)+

(x2 + 1)

2√

x + 3(x − 1)− (x2 + 1)

√x + 3

(x − 1)2


y ′ =

(2x

x2 + 1+

1

2(x + 3)− 1

x − 1

)(x2 + 1)

√x + 3

(x − 1)


Yes.







y ′ =2x√

x + 3

(x − 1)+

(x2 + 1)

2√

x + 3(x − 1)− (x2 + 1)

√x + 3

(x − 1)2


y ′ =

(2x

x2 + 1+

1

2(x + 3)− 1

x − 1

)(x2 + 1)

√x + 3

(x − 1)

I Are these the same? Yes.







y ′ =2x√

x + 3

(x − 1)+

(x2 + 1)

2√

x + 3(x − 1)− (x2 + 1)

√x + 3

(x − 1)2


y ′ =

(2x

x2 + 1+

1

2(x + 3)− 1

x − 1

)(x2 + 1)

√x + 3

(x − 1)








y ′ =2x√

x + 3

(x − 1)+

(x2 + 1)

2√

x + 3(x − 1)− (x2 + 1)

√x + 3

(x − 1)2


y ′ =

(2x

x2 + 1+

1

2(x + 3)− 1

x − 1

)(x2 + 1)

√x + 3

(x − 1)








y ′ =2x√

x + 3

(x − 1)+

(x2 + 1)

2√

x + 3(x − 1)− (x2 + 1)

√x + 3

(x − 1)2


y ′ =

(2x

x2 + 1+

1

2(x + 3)− 1

x − 1

)(x2 + 1)

√x + 3

(x − 1)



I What kinds of expressions are well-suited for logarithmicdifferentiation? Products, quotients, and powers


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.


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.


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



Solution

If y = xx , then

ln y = x ln x

1

y

dy

dx= x · 1

x+ ln x = 1 + ln x

dy


Remarks



I y ′ = 0 when x = e−1



Solution

If y = xx , then

ln y = x ln x

1

y

dy

dx= x · 1

x+ ln x = 1 + ln x

dy


Remarks



I y ′ = 0 when x = e−1


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


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


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!


Technology

Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 021 slides)