lesson 14: derivatives of logarithmic and exponential functions

Sections 3.1–3.3Derivatives of Exponential and

Logarithmic Functions

V63.0121.002.2010Su, Calculus I

New York University

June 1, 2010

Announcements

I Today: Homework 2 dueI Tomorrow: Section 3.4, reviewI Thursday: Midterm in class

. . . . . .

. . . . . .

Announcements

I Today: Homework 2 dueI Tomorrow: Section 3.4,

reviewI Thursday: Midterm in class

V63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 2 / 54

. . . . . .

Objectives for Sections 3.1 and 3.2

I Know the definition of anexponential function

I Know the properties ofexponential functions

I Understand and apply thelaws of logarithms,including the change ofbase formula.


. . . . . .

Objectives for Section 3.3

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

Definition of exponential functionsProperties of exponential FunctionsThe number e and the natural exponential function

Compound InterestThe number eA limit

Logarithmic FunctionsDerivatives of Exponential Functions

Exponential GrowthDerivative of the natural logarithm functionDerivatives of other exponentials and logarithms

Other exponentialsOther logarithms

Logarithmic DifferentiationThe power rule for irrational powers


. . . . . .

Derivation of exponential functions

DefinitionIf a is a real number and n is a positive whole number, then

an = a · a · · · · · a︸︷︷︸n factors

Examples

I 23 = 2 · 2 · 2 = 8I 34 = 3 · 3 · 3 · 3 = 81I (−1)5 = (−1)(−1)(−1)(−1)(−1) = −1


. . . . . .

FactIf a is a real number, then

I ax+y = axay

I ax−y =ax

ayI (ax)y = axy

I (ab)x = axbx

whenever all exponents are positive whole numbers.

Proof.Check for yourself:

ax+y = a · a · · · · · a︸︷︷︸x+ y factors

= a · a · · · · · a︸︷︷︸x factors

·a · a · · · · · a︸︷︷︸y factors

= axay


. . . . . .

Let's be conventional

I The desire that these properties remain true gives us conventionsfor ax when x is not a positive whole number.

I For example:an = an+0 !

= ana0

DefinitionIf a ̸= 0, we define a0 = 1.

I Notice 00 remains undefined (as a limit form, it’s indeterminate).


. . . . . .

Conventions for negative exponents

If n ≥ 0, we wantan · a−n !

= an+(−n) = a0 = 1

Definition

If n is a positive integer, we define a−n =1an

.

Fact

I The convention that a−n =1an

“works” for negative n as well.

I If m and n are any integers, then am−n =am

an.


. . . . . .

Conventions for fractional exponents

If q is a positive integer, we want

(a1/q)q != a1 = a

DefinitionIf q is a positive integer, we define a1/q = q

√a. We must have a ≥ 0 if q

is even.

Notice that q√ap =( q√a)p. So we can unambiguously say

ap/q = (ap)1/q = (a1/q)p


. . . . . .

Conventions for irrational powers

I So ax is well-defined if x is rational.I What about irrational powers?

DefinitionLet a > 0. Then

ax = limr→x

r rational

ar

In other words, to approximate ax for irrational x, take r close to x butrational and compute ar.


. . . . . .

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


. . . . . .


. .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


. . . . . .


. .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


. . . . . .


. .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


. . . . . .


. .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


. . . . . .


. .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


. . . . . .


. .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


. . . . . .


. .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


. . . . . .


. .x

.y

.y = 1x



. . . . . .

Outline








. . . . . .

Properties of exponential Functions.

.

TheoremIf a > 0 and a ̸= 1, then f(x) = ax is a continuous function with domain R andrange (0,∞). In particular, ax > 0 for all x. If a,b > 0 and x, y ∈ R, then

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

Proof.

I This is true for positive integer exponents by natural definitionI Our conventional definitions make these true for rational exponentsI Our limit definition make these for irrational exponents, too


. . . . . .


.


I ax+y = axay

I ax−y =ax

aynegative exponents mean reciprocals.

I (ax)y = axy

fractional exponents mean roots

I (ab)x = axbx

Proof.



. . . . . .


.


I ax+y = axay

I ax−y =ax

aynegative exponents mean reciprocals.

I (ax)y = axy fractional exponents mean rootsI (ab)x = axbx

Proof.



. . . . . .

Simplifying exponential expressions

Example

Simplify: 82/3

Solution

I 82/3 =3√82 =

3√64 = 4

I Or,(

3√8)2

= 22 = 4.

Example

Simplify:√8

21/2

Answer2


. . . . . .

Limits of exponential functions

Fact (Limits of exponentialfunctions)

I If a > 1, then limx→∞

ax = ∞and lim

x→−∞ax = 0

I If 0 < a < 1, thenlimx→∞

ax = 0 andlim

x→−∞ax = ∞ . .x

.y

.y = 1x



. . . . . .

Outline








. . . . . .

Compounded Interest

QuestionSuppose you save $100 at 10% annual interest, with interestcompounded once a year. How much do you have

I After one year?I After two years?I after t years?

Answer

I $100+ 10% = $110I $110+ 10% = $110+ $11 = $121I $100(1.1)t.


. . . . . .

Compounded Interest



Answer

I $100+ 10% = $110

I $110+ 10% = $110+ $11 = $121I $100(1.1)t.


. . . . . .

Compounded Interest



Answer

I $100+ 10% = $110I $110+ 10% = $110+ $11 = $121

I $100(1.1)t.


. . . . . .

Compounded Interest



Answer

I $100+ 10% = $110I $110+ 10% = $110+ $11 = $121I $100(1.1)t.


. . . . . .

Compounded Interest: quarterly

QuestionSuppose you save $100 at 10% annual interest, with interestcompounded four times a year. How much do you have


Answer

I $100(1.025)4 = $110.38, not $100(1.1)4!I $100(1.025)8 = $121.84I $100(1.025)4t.


. . . . . .




Answer

I $100(1.025)4 = $110.38,

not $100(1.1)4!I $100(1.025)8 = $121.84I $100(1.025)4t.


. . . . . .




Answer

I $100(1.025)4 = $110.38, not $100(1.1)4!

I $100(1.025)8 = $121.84I $100(1.025)4t.


. . . . . .




Answer

I $100(1.025)4 = $110.38, not $100(1.1)4!I $100(1.025)8 = $121.84

I $100(1.025)4t.


. . . . . .




Answer

I $100(1.025)4 = $110.38, not $100(1.1)4!I $100(1.025)8 = $121.84I $100(1.025)4t.


. . . . . .

Compounded Interest: monthly

QuestionSuppose you save $100 at 10% annual interest, with interestcompounded twelve times a year. How much do you have after tyears?

Answer$100(1+ 10%/12)12t


. . . . . .

Compounded Interest: general

QuestionSuppose you save P at interest rate r, with interest compounded ntimes a year. How much do you have after t years?

Answer

B(t) = P(1+

rn

)nt


. . . . . .

Compounded Interest: continuous

QuestionSuppose you save P at interest rate r, with interest compounded everyinstant. How much do you have after t years?

Answer

B(t) = limn→∞

P(1+

rn

)nt= lim

n→∞P(1+

1n

)rnt

= P[

limn→∞

(1+

1n

)n

︸︷︷︸independent of P, r, or t

]rt


. . . . . .

The magic number

Definition

e = limn→∞

(1+

1n

)n

So now continuously-compounded interest can be expressed as

B(t) = Pert.


. . . . . .

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 2.25

3 2.3703710 2.59374100 2.704811000 2.71692106 2.71828


. . . . . .



e ≈ 2.718281828459045 . . .


n(1+

1n

)n

1 22 2.253 2.37037

10 2.59374100 2.704811000 2.71692106 2.71828


. . . . . .



e ≈ 2.718281828459045 . . .


n(1+

1n

)n

1 22 2.253 2.3703710 2.59374

100 2.704811000 2.71692106 2.71828


. . . . . .



e ≈ 2.718281828459045 . . .


n(1+

1n

)n

1 22 2.253 2.3703710 2.59374100 2.70481

1000 2.71692106 2.71828


. . . . . .



e ≈ 2.718281828459045 . . .


n(1+

1n

)n

1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692

106 2.71828


. . . . . .



e ≈ 2.718281828459045 . . .


n(1+

1n

)n

1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828


. . . . . .



e ≈ 2.718281828459045 . . .

I e is irrational

I e is transcendental

n(1+

1n

)n

1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828


. . . . . .



e ≈ 2.718281828459045 . . .


n(1+

1n

)n

1 22 2.253 2.3703710 2.59374100 2.704811000 2.71692106 2.71828


. . . . . .

Meet the Mathematician: Leonhard Euler

I Born in Switzerland, livedin Prussia (Germany) andRussia

I Eyesight trouble all his life,blind from 1766 onward

I Hundreds of contributionsto calculus, number theory,graph theory, fluidmechanics, optics, andastronomy

Leonhard Paul EulerSwiss, 1707–1783


. . . . . .

A limit.

.

Question

What is limh→0

eh − 1h

?

Answer

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

eh − 1h

≈ 1

I In fact, limh→0

eh − 1h

= 1.

I This can be used to characterize e: limh→0

2h − 1h

= 0.693 · · · < 1 and

limh→0

3h − 1h

= 1.099 · · · > 1


. . . . . .

Outline








. . . . . .

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(x · x′) = loga x+ loga x′

(ii) loga( xx′)= loga x− loga x

′

(iii) loga(xr) = r loga x


. . . . . .

Logarithms convert products to sums

I Suppose y = loga x and y′ = loga x′

I Then x = ay and x′ = ay′

I So xx′ = ayay′= ay+y′

I Thereforeloga(xx

′) = y+ y′ = loga x+ loga x′


. . . . . .

Example

Write as a single logarithm: 2 ln 4− ln 3.

Solution

I 2 ln 4− ln 3 = ln 42 − ln 3 = ln42

3

I notln 42

ln 3!

Example

Write as a single logarithm: ln34+ 4 ln 2

Answerln 12


. . . . . .

..“lawn”

.

.Image credit: SelvaV63.0121.002.2010Su, Calculus I (NYU) Exponential and Logarithmic June 1, 2010 31 / 54

http://www.flickr.com/photos/selva/27991

. . . . . .

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 exponentials

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








. . . . . .

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 these derivatives:I e3x

I ex2

I x2ex

Solution

Iddx

e3x = 3e3x

Iddx

ex2= ex

2 ddx

(x2) = 2xex2

Iddx

x2ex = 2xex + x2ex


. . . . . .

Outline








. . . . . .

Derivative of the natural logarithm function

Let y = ln x. Then x = ey

so

eydydx

= 1

=⇒ dydx

=1ey

=1x

So:

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 2 = limh→0

2h − 1h

≈ 0.693 ln 3 = limh→0

3h − 1h

≈ 1.10


. . . . . .

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

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 exponential function, y′ = (ln x) · xx

(C) Neither


. . . . . .

It's neither! Or both?

If y = xx, then

ln y = x ln x1ydydx

= x · 1x+ ln x = 1+ ln x

dydx

= xx + (ln x)xx

Each of these terms is one of the wrong answers!


. . . . . .

Derivative of arbitrary powers

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


lesson 14: derivatives of logarithmic and exponential functions

Technology

nyu exponential

positive integer

logarithmic functionsv63

a0 definition

negative n

real number

derivatives of functions

ya ax y