CHAPTER 24 Derivatives of Inverse Functions and Logarithms

[Pages:9]CHAPTER 24

Derivatives of Inverse Functions and Logarithms

??

We know that

d dx

x

e

= x. e

But what about derivatives of exponen? tia?l

function wit?h bas?es other? than

e??

In other words, what is

d dx

x

a

?

And what about d

and d

? The main goal of this chapter is

ln(x)

dx

dx loga(x)

to answer these quest?ion?s and thus expand our list of derivative rules.

Let's start with d x . Since

is the inverse of x, we know = ln(a).

a

ln(x)

e

ae

dx

We can thus convert the power x to a power of :

a

e

?

x = ln(a) x = ln(a)x

ae

e.

With

this,

we

can

get

the

derivative

of

x

a

with

the

chain

rule:

hi

h

i

h

i

d

x =d

ln(a)x

=

d

ln(a)x

= ln(a)x

=x

a

e

e

ln(a)x e ln(a) a ln(a).

dx

|d x

{z dx

}

chain rule

So

the

derivative

of

x

a

is

just

x

a

times

the

constant

ln(a).

This

is

a

new

rule.

hi

Rule 16

d

x

a

=

x

ln(a)a

dx

hi

hi

For example,

d

x

10

=

x

ln(10)10

?

2.302

?

x

10

.

Also

d

x

2

=

x?

ln(2)2

dx

h i dx

0.693

?

x

2

.

Notice how special the base e is:

d

x

e

=

x

ln(e)e

=

?x 1e

=

x

e

.

The base = is the only base for which the dedrixvative of x is 1 times x.

ae

??

a

a

Next we will get a rule for d ln(x) . Our strategy will be to use the fact

dx

that ln(x) is the inverse of ex, that is,

if

f (x) = ex, then

? 1

=

f (x) ln(x).

Our

pl?an

is?to

first

develop

a

general

rule

for

?

d

? 1

?

f (x)

dx

and

then

use

it

to

get d ln(x) . (See Chapter 4 If you need to review inverses.)

dx

278

Derivatives of Inverse Functions and Logarithms

h

i

Thus our immediate question is: What is

d

? 1

f (x)

?

To

answer

this,

think

about

the

relation

bedtxween

f

and its

inverse

f

? 1

:

??

1

? =

f f (x) x.

The two sides of this equation are equal functions, so if we di erentiate both sides the derivatives will be equal:

h d

?

?

1

?i

hi

=d

f f (x)

x

dx

dx

The right-hand side of this equation is 1. The left-hand side is the derivative of a composition, so we can apply the chain rule to it:

0? ?

1

?

h d

?

1

i =

f f (x) f (x) 1

dx

wIthnaeanrpteptbloyeificnnagudt!sheWewcehecaadinnonris'utolkleanwtoeewimtwbuyhltaidptilvididedxd?inff?g0 ?1tf(hx?)e1?(axisb)?.obvByeudetdxqi?utf'as?t1ei(oxxna)?cb.tyWlyfew0 ?shfto?ap1t(pxwe)de?:

h

i

d?

1

=

f (x)

0?

1

?

?

1

dx

f f (x) .

This is our latest rule.

Rule

17

(The

inverse

rule)

If

f

is

a

function

having

a

derivative

0

f

and

an

inverse

? 1

,

then

f

h

i

d?

1

=

f (x)

0?

1

?

?.

1

dx

f f (x)

Let'Tsofiinlldustdra? fte?1t(hxi)s?.ruWlee,

suppose f ( know that

x)

0

f

=

3

x

,

= (x)

which

has

an

inverse

f

? 1

(x)

3x2, so our new rule gives

=

p

3

x.

dx

h

i

d?

1

f (x)

dx

=

0?

1

?

1

?=

?

1

?

?=

2

f f (x)

1

3 f (x)

p1 .

32

3x

Granted, this is not all that impressive, since we can use the power rule to get the same answer:

h d?

1

i

hp i

=d 3 =

hi

d

1 3

=

1

?2

3

=

1

=

f (x)

x

x

x

p1 .

2

dx

dx

dx

3

32

3x 3

3x

279

But the inverse rule can be very useful. We'll now use it to find the

derivative

of

ln(x).

Say

= f (x)

x

e

,

so

f

? 1

(x)

=

ln(x).

Then

hi

h

i

d

=d

? 1

ln(x)

f (x)

dx

dx

because

ln(x)

=

f

? 1

(x)

=

0?

1

?

?

1

f f (x)

by inverse rule

= 0? 1 ? f ln(x)

because

? 1

=

f (x) ln(x)

=1

ln(x)

e

becase

0= f (x)

x

e

=1

because

ln(x)

e

=

x.

x

hi

Thus d ln(x) = 1 . Figure 24.1 (left) illustrates this remarkable fact.

It

shows

tdhxe

function

x f

(x)

=

ln(x)

along

with

its

derivative

f

0= (x)

1.

Notice

x

how if x is near 0, the tangent to ln(x) at x is very steep, and indeed the

derivative 1 is very large. But as x gets bigger, the tangent to ln(x) gets

x

closer to horizontal (slope 0) while the derivative 1 approaches zero.

x

= y ln(x)

=1 y

x

= || y ln x

=1 y

x

Figure

24.1.

Left:

the

graphs

of

= f (x) ln(x)

(black)

and

0= f (x)

1

(blue)

with

domain

(0, 1).

Right:

the

graphs

of

= || f (x) ln x

(black)

and

0

x

=

f (x)

1

(blue).

x

Notice

however,

that

the

domain

of

ln(x)

is

1 (0, )

but

the

domain

of

1

is

x

?1 (,

0)

[

(0,

1).

So

when

we

say

that

the

derivative

of

ln(x)

is

1,

we

really

x

mean 1 with its domain restricted to (0,1). Figure 24.1 (right) shows a

x

somewhat more complete scenario. It shows the function ln(|x|), which we

will

abbreviate

as

ln |x|.

This

function

has

domain

?1 (,

0)

[

(0,

1),

and

its

derivative is 1 with its usual domain. So our latest rule has two parts.

x

hi

Rule 18

d

=1

ln(x)

and

d

h ?? ??i ln x

=

1.

dx

x

dx

x

280

Derivatives of Inverse Functions and Logarithms

Here it is understood that in the first formula the domain of ln(x) and 1

x

is

(0, 1).

In

the

second

formula

the

domain

of

|| ln x

and

1

is

all

real

numbers

x

except 0. Do not sweat the di erence between the two versions of this rule ?

they say almost the same thing, and the second implies the first. We will

mostly use the first version in parts 3 and 4 of this book, but the second

version becomes particularly useful in Part 5.

At the begin?nin?g of?this ?chapter ?we said? our main goals were to find

formulas for d

dx

x

a

,

d

dx

ln(x)

and d

dx

loga(x) . We've done all but the last

one. For it we will use the change of base formula (Fact 5.1 in Chapter 5,

page 88) which states

= ln(x)

loga(x)

.

ln(a)

Using this, the constant multiple rule and Rule 18, we get

h

i

hi

d

= d ln(x) = 1 ? d

= 1 ?1

loga(x)

ln(x)

.

dx

dx ln(a) ln(a) dx

ln(a) x

With our prior agreement about domains, we get another two-part formula.

h

i

Rule 19

d

=

loga(x)

1

and

h d

?? ??i =

loga x

1.

dx

x ln(a)

dx

x ln(a)

h

i

h

i

h

i

Example 24.1

d

=d

+

d

log3(x) tan(x)

log3(x) tan(x) log3(x) tan(x)

dx

dx

dx

(product rule)

=

1

+

2

tan(x) log3(x) sec (x)

.

x ln(3)

hp

i

Example 24.2 Find d

+ 3+ 5 x ln(x)

.

dx This is the derivative of a function to a power, so we can use the generalized

power rule:

hp

i

h?

?i

d

+ 3+

= d + 3+

1/2

5 x ln(x)

5 x ln(x)

dx

dx

? = 1 + 3+

??

1/2 1

h d

+

3+

i

5 x ln(x)

5 x ln(x)

2

dx

? = 1 + 3+

?? ?

1/2

2+ 1

5 x ln(x) 3x

2

x

2+ 1

= p 3x x

+ 3+

.

2 5 x ln(x)

281

.

Example 24.3

h

? ?i

? ?h i

d ++

3 = ++

2d

7 x ln(x)

0 1 3 ln(x)

ln(x)

dx

dx

??

??

2

=+

21

1 3 ln(x)

=

+ 3 ln(x) 1

.

x

x

Example 24.4 (quotient rule)

hi

hi

h

i

d

? + ?d

ln(x) x ln(x) x

1? + ? x ln(x) 1

d ln(x) = dx

dx = x

2

2

dx x

x

x

=

+ 1 ln(x)

.

2

x

Example 24.5

Find the derivative of

2+ x3

+ x2

.

10

(

The composition =

2+ + x 3x 2

can

be

broken

up

as

y 10

=u y 10

= 2+

+

u x 3x 2.

The chain rule then gives d y = d y ? du

dx du dx

=

u? + +

ln(10) 10 (2x 3 0)

=

2+ + x 3x 2

+

ln(10) 10

(2x 3) .

In Example 24.5 we di

erentiated

a

function

of

form

a

g(x)

?

.?

Let's

repeat

our steps to get a chain rule generalization for the rule

d

x

a

=

ln(a)

x

a

.

dx

Example 24.6 Find the derivative of ag(x). (

The

composition

y

=

g(x)

a

can

be

broken

up

as

=u ya

=

u g(x).

The chain rule then gives d y = d y ? du

dx du dx

=

u? 0

ln(a) a g (x)

=

0 g(x)

ln(a) a g (x) .

d

? This?

g(x)

e

examples shws

d

??

g(x)

a

=

ln(a)ag(x) g0(x),

a

companion

to

the

rule

=

e

g(x)

0

g

(

x).

We

dx

will

summarize

these

and

chain

rule

generalizations

dx

of the other rules from this chapter on the bottom of the next page.

282

Derivatives of Inverse Functions and Logarithms

Example 24.7 Find the derivative of y = ln ?? sin(x)??.

(

This is a composition, and the function can be broken up as

= || y ln u

=

u sin(x)

The chain rule gives

dy

=

dy

du

=

1 cos(x)

=

1

= cos(x)

cos(x) .

dx du dx u

sin(x)

sin(x)

funEctxioanmopflefr2o4m.7lnil??lgu(sxt)r??aotreslna?

com?mon pattern, which is to di erentiate a g(x) . Let's redo the example in this setting.

Example 24.8 Find the derivative of y = ln??g(x)??.

(

This is a composition, and the function can be broken up as

= || y ln u

=

u g(x)

0

The chain rule gives

dy

=

dy

du

=

10 g (x)

=

1

0= g (x)

g (x) .

dx du dx u

g(x)

g(x)

Example 24.8 has shown that

h d

??

??i =

h 1 ?d

i

0

= g (x)

ln g(x)

g(x)

.

dx

g(x) dx

g(x)

??

This is the chain rule generalization of the rule d

? ? ?? 0

dx

|| ln x

= 1 , and it is worth

x

remember?ing.? It implies

d dx

ln

g(x)

= g (x) , and we often us it this way.

g(x)

(fRoreca??ll

ln

??

g(x) is not defined when g(x) is more all-encompassing.)

is

negative,

so

the

rule

as

stated

ln g(x)

Here is a summary of this chapter's main rules, along side their chain

rule generalizations. Remember them and internalize them.

Di erentiation rules for exponential and log functions

Rule

hi

d x=x ee

dxh i

d x=

x

a ln(a) a

dx

h d

?? ?? i = 1

ln x

dx

x

h d

?? ?? i = 1

loga x

dx

x ln(a)

Chain rule generalization

hi

d

g(x)

=

0 g(x)

e

e g (x)

dx

hi

d g(x) =

0 g(x)

a

ln(a) a g (x)

dx

h d

??

??i

=

0

g (x)

ln g(x)

dx

g(x)

h d

??

??i =

0

g (x)

loga g(x)

dx

g(x) ln(a)

We

prefer

the

base

e,

so

you

should

expect

to

that

the

formulas

for

x

a

and

loga to play less of a role. (Though in computer science, log2 is significant!)

283

Example 24.9

Find

the

derivative

of

y

=

?? 5 ln 4x

+ 3+ 6x

x + 3??.

We will do this in two di erent then we will use the formula d

w? ay??s. Fi??r?s=t ln g(x)

dx

we will use the chain rule, and

0

g (x) g(x)

from (the

previous

page.

Using the chain rule, we first break the function up as

= || y ln u

= 5+ 3+ +

u 4x 6x x 3

The chain rule gives d y = d y du

dx du dx

?

?

=1

4+ 2+

20x 18x 1

u

?

?

=

1

4+ 2+ =

5+ 3+ + 20x 18x 1

4x 6x x 3

4+ 2+ 20x 18x 1

5+ 3+ + . 4x 6x x 3

? Next, using the formula d

??

??

=

0

g (x) ,

from

the

previous

page,

the

ln g(x)

dx

g(x)

h answer comes in one step: d

??

5+

3+

+ ?? i =

4+ 2+ 20x 18x 1

.

ln 4x 6x x 3

5+ 3+ +

dx

4x 6x x 3

So using the formula is quicker. But you should soon reach the point

where the above to approaches are equally automatic. Doing the chain rule

in your head is in essence using the formula.

Example 24.10

Find

the

derivative

of

y

=

ln

???tan

p

2

x

+

3x

?

???.

This has the form chain rule or the

offoarmcoumlaposdit?ilonn??

ln ??g??(x=)??, g(x)

so

1

we can use either the straight

0

g (x)

from

the

previous

page.

dx

g(x)

Let's try the formula.

h d

???

p 2+

? ???i =

ln tan x 3x

p1

h p

?i

?d

2+

tan x 3x

dx

2+ dx

tan x 3x

p

? hp

i

= p1

? 2 2+ d

2+

sec x 3x

x 3x

2+

dx

tan x 3x

= p1

?

p

2

2+

?? 1 2+

?1?

1

2

d

h

2+

i

sec x 3x x 3x

x 3x

2+

2

dx

tan x 3x

= p1

?

p

2

2+

?? 1 2+

?? 1

2

+

sec x 3x x 3x (2x 3)

2+

2

tan x 3x

p

?

+

2 2+

(2x 3) sec x 3x

=

p

?p

2+

2+

2 tan x 3x x 3x

284

Derivatives of Inverse Functions and Logarithms

Exercises for Chapter 24

In exercises 1?20 di erentiate the given function.

p

?

1.

+1+ ln(x)

+ x3

2.

ln

2+ 1 x

x

x

3. ln(w)

w?

?

5.

3

ln sin (x)

p

7.

+ ?? + 5 ln( )

?3

??

9. cos ln(x) p

11.

?2 + ? + ln(5 p)

?9

13.

2+ ln(x 1)

+ 3x 1

??

15.

ln

x

xe

??

17.

tan?

ln(x)

+ x

19.

ln

+1 1

x

21.

Find

+? ln(3 h) ln(3) li!m

h0

h

4.

1 x2?+ ln(x)?

6. ln tan(x) ? ? ??

8.

ln

sec

3

x

? ? ??

10.

3

sec ln x

?

?

12.

ln

3

sec(x )

?

?

14.

sec

3

ln(x )

16.

x

ln (x) e

? ? ??

18.

ln sin

3

x

3

20.

x ln(x) 3+

22.

x Find

1 z?

28 li!m ?

z 3z 3

Exercise Solutions for Chapter 24

In exercises 1?20 di erentiate the given function.

p

1.

d

+1+ ln(x)

+ x3

=1? 1 +

p1

2

dx

x

x x 2x

3.

d

ln(w) =

1 w

??

?

w ln(w) 1

=

? 1 ln(w)

2

2

dw w

w

w

h?

?i

h

i

h

i

5.

d

3

ln sin (x)

=

1

d

3 =1

d

2

= cos(x)

sin (x)

3 sin (x) sin(x) 3

3

3

dx

sin (x) dx

sin (x)

dx

sin(x)

7.

h

pi

h

i

?

p

d ?

5 + ln(??) +

?3

=

d ?

5 + ln(??) + ?3/2

=

+ 0

??

+

3 ?1/2

=

1 ?

+

3

?

d

d

2

2

h ? ?i

??

??

9. d cos ln(x) = ? sin ln(x) 1 = ? sin ln(x)

dx

x

x

h

pi

h

i

p

11.

d

?2

+

? ln(5 )

+

?9

=d

?2

+

? ln(5 )

+

?9/2

=+ 0

5 + 9 ?7/2 = 1 + 9

?

?

?7

dx

dx

52

2

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