lesson 14: derivatives of exponential and logarithmic functions (section 041 slides)

Section 3.3Derivatives of Exponential and

Logarithmic Functions

V63.0121.041, 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

V63.0121.041, Calculus I (NYU) Section 3.3 Derivs of Exp and Log October 25, 2010 2 / 34

. . . . . .

Objectives

I Know the derivatives of theexponential functions (withany base)

I Know the derivatives of thelogarithmic functions (withany base)

I Use the technique oflogarithmic differentiationto find derivatives offunctions involving roducts,quotients, and/orexponentials.


. . . . . .

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 =

1ar.

I For any positive whole number n, a1/n = n√a.

There is only one continuous function which satisfies all of the above.We call it the exponential function with base a.


. . . . . .

Properties of exponential Functions

TheoremIf a > 0 and a ̸= 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


. . . . . .



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


. . . . . .



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 = 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

Definition

e = limn→∞

(1+

1n

)n= lim

h→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 irrationalI e is transcendental

n(1+

1n

)n

1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 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(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 for logarithms

FactIf a > 0 and a ̸= 1, and the same for b, then

loga x =logb xlogb a

Proof.

I If y = loga x, then x = ay

I So logb x = logb(ay) = y logb a

I Thereforey = loga x =

logb xlogb a


. . . . . .

Upshot of changing base

The 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 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

FactIf 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).

To reiterate: the derivative of an exponential function is a constanttimes that function. Much different from polynomials!


. . . . . .

The funny limit in the case of e

Remember the definition of e:

e = limn→∞

(1+

1n

)n= lim

h→0(1+ h)1/h

Question

What is limh→0

eh − 1h

?

AnswerIf h is small enough, e ≈ (1+ h)1/h. So

eh − 1h

≈[(1+ h)1/h

]h − 1h

=(1+ h)− 1

h=

hh= 1

So in the limit we get equality: limh→0

eh − 1h

= 1


. . . . . .

Derivative of the natural exponential function

Fromddx

ax =(limh→0

ah − 1h

)ax and lim

h→0

eh − 1h

= 1

we get:

Theorem

ddx

ex = ex


. . . . . .

Exponential Growth

I Commonly misused term to say something grows exponentiallyI It means the rate of change (derivative) is proportional to the

current valueI Examples: Natural population growth, compounded interest,

social networks


. . . . . .

Examples

Examples

Find derivatives of these functions:I e3x

I ex2

I x2ex

Solution

Iddx

e3x = 3e3x

Iddx

ex2= ex

2 ddx

(x2) = 2xex2

Iddx

x2ex = 2xex + x2ex


. . . . . .

Outline







. . . . . .


Let 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 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 gapprecisely.


. . . . . .

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





. . . . . .

Outline







. . . . . .

Other logarithms

Example

Use implicit differentiation to findddx

ax.

SolutionLet y = ax, so

ln y = lnax = x ln a

Differentiate implicitly:

1ydydx

= ln a =⇒ dydx

= (ln a)y = (ln a)ax

Before we showed y′ = y′(0)y, so now we know that

ln a = limh→0

ah − 1h


. . . . . .

Other logarithms

Example

Findddx

loga x.

SolutionLet y = loga x, so ay = x. Now differentiate implicitly:

(ln a)aydydx

= 1 =⇒ dydx

=1

ay ln a=

1x ln a

Another way to see this is to take the natural logarithm:

ay = x =⇒ y ln a = ln x =⇒ y =ln xln a

Sodydx

=1ln a

1x.


. . . . . .

Other logarithms

Example

Findddx

loga x.

SolutionLet y = loga x, so ay = x.

Now differentiate implicitly:

(ln a)aydydx

= 1 =⇒ dydx

=1

ay ln a=

1x ln a



Sodydx

=1ln a

1x.


. . . . . .

Other logarithms

Example

Findddx

loga x.

SolutionLet y = loga x, so ay = x. Now differentiate implicitly:

(ln a)aydydx

= 1 =⇒ dydx

=1

ay ln a=

1x ln a



Sodydx

=1ln a

1x.


. . . . . .

More examples

Example

Findddx

log2(x2 + 1)

Answer

dydx

=1ln 2

1x2 + 1

(2x) =2x

(ln 2)(x2 + 1)


. . . . . .

Outline







. . . . . .

A nasty derivative

Example

Let y =(x2 + 1)

√x+ 3

x− 1. Find y′.

SolutionWe 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) +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


. . . . . .

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′ =(

2xx2 + 1

+1

2(x+ 3)− 1

x− 1

)(x2 + 1)

√x+ 3

x− 1

I Are these the same?I Which do you like better?I What kinds of expressions are well-suited for logarithmic

differentiation?


. . . . . .



y′ =2x

√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2


y′ =(

2xx2 + 1

+1

2(x+ 3)− 1

x− 1

)(x2 + 1)

√x+ 3

x− 1

I Are these the same?

I Which do you like better?I What kinds of expressions are well-suited for logarithmic

differentiation?


. . . . . .



y′ =2x

√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2


y′ =(

2xx2 + 1

+1

2(x+ 3)− 1

x− 1

)(x2 + 1)

√x+ 3

x− 1

I Are these the same?I Which do you like better?

I What kinds of expressions are well-suited for logarithmicdifferentiation?


. . . . . .



y′ =2x

√x+ 3

(x− 1)+

(x2 + 1)2√x+ 3(x− 1)

− (x2 + 1)√x+ 3

(x− 1)2


y′ =(

2xx2 + 1

+1

2(x+ 3)− 1

x− 1

)(x2 + 1)

√x+ 3

x− 1

I Are these the same?I Which do you like better?I What kinds of expressions are well-suited for logarithmic

differentiation?


. . . . . .

Derivatives of powers.

.

QuestionLet 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) Neither ..x

.y

..1

..1

Answer

(A) This can’t be y′ because xx > 0 for all x > 0, and this function decreasesat some places

(B) This can’t be y′ because (ln x)xx = 0 when x = 1, and this function doesnot have a horizontal tangent at x = 1.


. . . . . .

It's neither! Or both?

SolutionIf y = xx, then

ln y = x ln x1ydydx

= x · 1x+ ln x = 1+ ln x

dydx

= (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


. . . . . .



ln y = x ln x1ydydx

= x · 1x+ ln x = 1+ ln x

dydx


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


. . . . . .



ln y = x ln x1ydydx

= x · 1x+ ln x = 1+ ln x

dydx


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


. . . . . .

Derivatives of power functions with any exponent

Fact (The power rule)

Let y = xr. Then y′ = rxr−1.

Proof.

y = xr =⇒ ln y = r ln x

Now differentiate:

1ydydx

=rx

=⇒ dydx

= ryx= rxr−1


. . . . . .

Summary

I Derivatives of logarithmic and exponential functions:

y y′

ex ex

ax (ln a) · ax

ln x1x

loga x1ln a

· 1x

I Logarithmic Differentiation can allow us to avoid the product andquotient rules.


lesson 14: derivatives of exponential and logarithmic functions (section 041 slides)

Technology

loga x x

ln x x

thenlogb x loga x

logarithm loga x

nyu section

realnumbers x

derivs of exp

ayso logb x