SOLUTION

We have

1

(log4 9) · (log9 15) = · log9 16 = log4 16 = 2.

log9 4

6.4 Calculus with Logs and Exponentials to

Arbitrary Bases

6.4.1 DIFFERENTIATION AND INTEGRATION OF

loga x AND ax

We begin by noting these facts:

If a > 0 then

ax

dx

a = a x · ln a; equivalently, a dx = + C.

x

(i)

ln a

dx

1

d

(loga x) =

(ii)

x · ln a

dx

Math Note: As always, we can state these last formulas more generally as

du du

a = au · · ln a

dx dx

and

1 du 1

d

loga u = · · .

u dx ln a

dx

EXAMPLE 6.25

Calculate

d d d d

(log4 (x · cos x)).

(3cos x ),

(5x ), (log8 x),

dx dx dx dx

SOLUTION

We see that

dx

(5 ) = 5x · ln 5.

dx

For the second problem, we apply our general formulation with a = 3, u =

cos x to obtain

d cos x d

) = 3cos x · cos x · ln 3 = 3cos x · (’ sin x) · ln 3.

(3

dx dx

167

CHAPTER 6 Transcendental Functions

Similarly,

1

d

(log8 x) =

x · ln 8

dx

1

d d

(log4 (x · cos x)) = · (x · cos x)

(x · cos x) · ln 4 dx

dx

cos x + (x · (’ sin x))

= .

(x · cos x) · ln 4

EXAMPLE 6.26

Integrate

3cot x · (’ csc2 x) dx.

SOLUTION

For clarity we set •(x) = cot x, • (x) = ’ csc2 x. Then our integral becomes

1

3•(x) · • (x) dx = · 3•(x) · • (x) · ln 3 dx

ln 3

1

= · 3•(x) + C.

ln 3

Resubstituting the expression for •(x) now gives that

1

3cot x · (’ csc2 x) dx = · 3cot x + C.

ln 3

You Try It: Evaluate (log6 (x 3 )/x) dx.

You Try It: Calculate the integral

2

x · 3x dx.

Our new ideas about arbitrary exponents and bases now allow us to formulate a

general result about derivatives of powers:

For any real exponent a we have

da

x = a · x a’1 .

dx

EXAMPLE 6.27 √

’π, x 3, x e.

Calculate the derivative of x

168 CHAPTER 6 Transcendental Functions

SOLUTION

We have

d ’π

= ’π · x ’π ’1 ,

x

dx

d √3 √ √

x = 3 · x 3’1 ,

dx

de

x = e · x e’1 .

dx

2

You Try It: Calculate (d/dx)5sin x’x . Calculate (d/dx)x 4π .

6.4.2 GRAPHING OF LOGARITHMIC AND

EXPONENTIAL FUNCTIONS

If a > 0 and f (x) = loga x, x > 0, then

Y

1

f (x) =

x · ln a

FL

’1

f (x) = 2

x · ln a

AM

f (1) = 0.

Using this information, we can sketch the graph of f (x) = loga x.

If a > 1 then ln a > 0 so that f (x) > 0 and f (x) < 0. The graph of f is

TE

exhibited in Fig. 6.6.

Fig. 6.6

If 0 < a < 1 then ln a = ’ ln(1/a) < 0 so that f (x) < 0 and f (x) > 0. The

graph of f is sketched in Fig. 6.7.

Since g(x) = a x is the inverse function to f (x) = loga x, the graph of g is the

re¬‚ection in the line y = x of the graph of f (Figs 6.6 and 6.7). See Figs 6.8, 6.9.

169

CHAPTER 6 Transcendental Functions

Fig. 6.7

Fig. 6.8

Fig. 6.9

Figure 6.10 shows the graphs of loga x for several different values of a > 1.

Figure 6.11 shows the graphs of a x for several different values of a > 1.

You Try It: Sketch the graph of y = 4x and y = log4 x on the same set of axes.

170 CHAPTER 6 Transcendental Functions

Fig. 6.10

Fig. 6.11

6.4.3 LOGARITHMIC DIFFERENTIATION

We next show how to use the logarithm as an aid to differentiation. The key idea is

that if F is a function taking positive values then we can exploit the formula

F

[ln F ] = (—)

.

F

EXAMPLE 6.28

Calculate the derivative of the function

F (x) = (cos x)(sin x) , 0 < x < π.

171

CHAPTER 6 Transcendental Functions

SOLUTION

We take the natural logarithm of both sides:

ln F (x) = ln (cos x)(sin x) = (sin x) · (ln(cos x)). (†)

Now we calculate the derivative using the formula (—) preceding this example:

The derivative of the left side of (†) is

F (x)

.

F (x)

Using the product rule, we see that the derivative of the far right side of (†) is

’ sin x

(cos x) · (ln(cos x)) + (sin x) · .

cos x

We conclude that

’ sin x

F (x)

= (cos x) · (ln(cos x)) + (sin x) · .

cos x

F (x)

Thus

sin2 x

F (x) = (cos x) · (ln(cos x)) ’ · F (x)

cos x

sin2 x

= (cos x) · ln(cos x) ’ · (cos x)(sin x)

cos x

You Try It: Differentiate log9 | cos x|.

You Try It: Differentiate 3sin(3x) . Differentiate x sin 3x .

EXAMPLE 6.29

Calculate the derivative of F (x) = x 2 · (sin x) · 5x .

SOLUTION

We have

[ln F (x)] = [ln(x 2 · (sin x) · 5x )]

= [(2 · ln x) + ln(sin x) + (x · ln 5)]

2 cos x

=+ + ln 5.

sin x

x

Using formula (—), we conclude that

2 cos x

F (x)

=+ + ln 5

sin x

F (x) x

172 CHAPTER 6 Transcendental Functions

hence

2 cos x

F (x) = + + ln 5 · [x 2 · (sin x) · 5x ].

sin x

x

You Try It: Calculate (d/dx)[(ln x)ln x ].

6.5 Exponential Growth and Decay

Many processes of nature and many mathematical applications involve logarithmic

and exponential functions. For example, if we examine a population of bacteria,

we notice that the rate at which the population grows is proportional to the number

of bacteria present. To see that this makes good sense, suppose that a bacterium

reproduces itself every 4 hours. If we begin with 5 thousand bacteria then

after 4 hours there are 10 thousand bacteria

after 8 hours there are 20 thousand bacteria

after 12 hours there are 40 thousand bacteria

after 16 hours there are 80 thousand bacteria . . .

etc.

The point is that each new generation of bacteria also reproduces, and the older

generations reproduce as well.Asketch (Fig. 6.12) of the bacteria population against

time shows that the growth is certainly not linear”indeed the shape of the curve

appears to be of exponential form.

Fig. 6.12

173

CHAPTER 6 Transcendental Functions

Notice that when the number of bacteria is large, then different generations of

bacteria will be reproducing at different times. So, averaging out, it makes sense

to hypothesize that the growth of the bacteria population varies continuously as

in Fig. 6.13. Here we are using a standard device of mathematical analysis: even

though the number of bacteria is always an integer, we represent the graph of the

population of bacteria by a smooth curve. This enables us to apply the tools of

calculus to the problem.

Fig. 6.13

6.5.1 A DIFFERENTIAL EQUATION

If B(t) represents the number of bacteria present in a given population at time t,

then the preceding discussion suggests that

dB

= K · B(t),

dt

where K is a constant of proportionality. This equation expresses quantitatively

the assertion that the rate of change of B(t) (that is to say, the quantity dB/dt) is

proportional to B(t). To solve this equation, we rewrite it as

1 dB

· = K.

B(t) dt

We integrate both sides with respect to the variable t:

1 dB

· dt = K dt.

B(t) dt

The left side is

ln |B(t)| + C

and the right side is

Kt + C,

174 CHAPTER 6 Transcendental Functions

where C and C are constants of integration. We thus obtain

ln |B(t)| = Kt + D,

where we have amalgamated the two constants into a single constant D. Exponen-

tiating both sides gives

|B(t)| = eKt+D

or

B(t) = eD · eKt = P · eKt . ()

Notice that we have omitted the absolute value signs since the number of bacteria is

always positive. Also we have renamed the constant eD with the simpler symbol P .

Equation ( ) will be our key to solving exponential growth and decay problems.

We motivated our calculation by discussing bacteria, but in fact the calculation

applies to any function which grows at a rate proportional to the size of the function.

Next we turn to some examples.

6.5.2 BACTERIAL GROWTH

EXAMPLE 6.30

A population of bacteria tends to double every four hours. If there are 5000

bacteria at 9:00 a.m., then how many will there be at noon?

SOLUTION

To answer this question, let B(t) be the number of bacteria at time t. For con-

venience, let t = 0 correspond to 9:00 a.m. and suppose that time is measured

in hours. Thus noon corresponds to t = 3.

Equation ( ) guarantees that

B(t) = P · eKt

for some undetermined constants P and K. We also know that

5000 = B(0) = P · eK·0 = P .

We see that P = 5000 and B(t) = 5000 · eKt . We still need to solve for K.

Since the population tends to double in four hours, there will be 10,000

bacteria at time t = 4; hence

10000 = B(4) = 5000 · eK·4 .

We divide by 5000 to obtain

2 = eK·4 .

175

CHAPTER 6 Transcendental Functions

Taking the natural logarithm of both sides yields

ln 2 = ln(eK·4 ) = 4K.

We conclude that K = [ln 2]/4. As a result,

B(t) = 5000 · e([ln 2]/4)t .

We simplify this equation by noting that

e([ln 2]/4)t = (eln 2 )t/4 = 2t/4 .

In conclusion,

B(t) = 5000 · 2t/4 .

The number of bacteria at noon (time t = 3) is then given by

B(3) = 5000 · 23/4 ≈ 8409.

It is important to realize that population growth problems cannot be described

using just arithmetic. Exponential growth is nonlinear, and advanced analytical

ideas (such as calculus) must be used to understand it.

EXAMPLE 6.31

Suppose that a certain petri dish contains 6000 bacteria at 9:00 p.m. and

10000 bacteria at 11:00 p.m. How many of the bacteria were there at

7:00 p.m.?

SOLUTION

We know that

B(t) = P · eKt .

The algebra is always simpler if we take one of the times in the initial data to

correspond to t = 0. So let us say that 9:00 p.m. is t = 0. Then 11:00 p.m. is

t = 2 and 7:00 p.m. is t = ’2. The initial data then tell us that

6000 = P · eK·0 (—)

10000 = P · eK·2 . (——)

From equation (—) we may immediately conclude that P = 6000. Substituting

this into (——) gives

10000 = 6000 · (eK )2 .

We conclude that

√

5

=√ .

eK

3

176 CHAPTER 6 Transcendental Functions

As a result,

√ t

5

B(t) = 6000 · √ .

3

At time t = ’2 (7:00 p.m.) the number of bacteria was therefore

√ ’2

5 3

B(’2) = 6000 · √ = · 6000 = 3600.

5

3

You Try It: A petri dish has 5000 bacteria at 1:00 p.m. on a certain day and 8000

bacteria at 5:00 p.m. that same day. How many bacteria were there at noon?

6.5.3 RADIOACTIVE DECAY

Another natural phenomenon which ¬ts into our theoretical framework is radio-

active decay. Radioactive material, such as C14 (radioactive carbon), has a half life.

Saying that the half life of a material is h years means that if A grams of material is

present at time t then A/2 grams will be present at time t + h. In other words, half

of the material decays every h years. But this is another way of saying that the rate

at which the radioactive material vanishes is proportional to the amount present.

So equation ( ) will apply to problems about radioactive decay.

EXAMPLE 6.32

Five grams of a certain radioactive isotope decay to three grams in 100

years. After how many more years will there be just one gram?

SOLUTION

First note that the answer is not “we lose two grams every hundred years

so ¦.” The rate of decay depends on the amount of material present. That is

the key.

Instead, we let R(t) denote the amount of radioactive material at time t.

Equation ( ) guarantees that R has the form

R(t) = P · eKt .

Letting t = 0 denote the time at which there are 5 grams of isotope, and

measuring time in years, we have

R(0) = 5 R(100) = 3.

and

From the ¬rst piece of information we learn that

5 = P · eK·0 = P .

Hence P = 5 and

R(t) = 5 · eKt = 5 · (eK )t .

177

CHAPTER 6 Transcendental Functions

The second piece of information yields

3 = R(100) = 5 · (eK )100 .

We conclude that

3

(eK )100 =

5

or

1/100

3

=

K

e .

5

Thus the formula for the amount of isotope present at time t is

t/100

3

R(t) = 5 · .

5

Thus we have complete information about the function R, and we can answer

the original question.

There will be 1 gram of material present when

t/100

3

1 = R(t) = 5 ·

5

or

t/100

1 3

= .

5 5

We solve for t by taking the natural logarithm of both sides:

t/100

3 t

ln(1/5) = ln = · ln(3/5).

5 100

We conclude that there is 1 gram of radioactive material remaining when

ln(1/5)

t = 100 · ≈ 315.066.

ln(3/5)

So at time t = 315.066, or after 215.066 more years, there will be 1 gram of

the isotope remaining.

You Try It: Our analysis of exponential growth and decay is derived from the

hypothesis that the rate of growth is proportional to the amount of matter present.

Suppose instead that we are studying a system in which the rate of decay is propor-

tional to the square of the amount of matter. Let M(t) denote the amount of matter

at time t. Then our physical law is expressed as

dM

= C · M 2.

dt

178 CHAPTER 6 Transcendental Functions

Here C is a (negative) constant of proportionality. We apply the method of

“separation of variables” described earlier in the section. Thus

dM/dt

=C

M2

so that

dM/dt

dt = C dt.

M2

Evaluating the integrals, we ¬nd that

1

’ = Ct + D.

M

We have combined the constants from the two integrations. In summary,

1

M(t) = ’ .

Ct + D

Y

For the problem to be realistic, we will require that C < 0 (so that M > 0 for

FL

large values of t) and we see that the population decays like the reciprocal of a

linear function when t becomes large.

AM

Re-calculate Example 6.32 using this new law of exponential decay.

6.5.4 COMPOUND INTEREST

TE

Yet a third illustration of exponential growth is in the compounding of interest. If

principal P is put in the bank at p percent simple interest per year then after one

year the account has

p

P · 1+

100

dollars. [Here we assume, of course, that all interest is reinvested in the account.]

But if the interest is compounded n times during the year then the year is divided

into n equal pieces and at each time interval of length 1/n an interest payment of

percent p/n is added to the account. Each time this fraction of the interest is added

to the account, the money in the account is multiplied by

p/n

1+ .

100

Since this is done n times during the year, the result at the end of the year is that

the account holds

pn

P · 1+ (—)

100n

179

CHAPTER 6 Transcendental Functions

dollars at the end of the year. Similarly, at the end of t years, the money accumulated

will be

p nt

P · 1+ .

100n

Let us set

n · 100

k=

p

and rewrite (—) as

k p/100

kp/100

1 1

P · 1+ =P · 1+ .

k k

It is useful to know the behavior of the account if the number of times the inter-

est is compounded per year becomes arbitrarily large (this is called continuous

compounding of interest). Continuous compounding corresponds to calculating the

limit of the last formula as k (or, equivalently, n), tends to in¬nity.

We know from the discussion in Subsection 6.2.3 that the expression (1 + 1/k)k

tends to e. Therefore the size of the account after one year of continuous

compounding of interest is

P · ep/100 .

After t years of continuous compounding of interest the total money is

P · ept/100 . ()

EXAMPLE 6.33

If $6000 is placed in a savings account with 5% annual interest compounded

continuously, then how large is the account after four and one half years?

SOLUTION

If M(t) is the amount of money in the account at time t, then the preceding

discussion guarantees that

M(t) = 6000 · e5t/100 .

After four and one half years the size of the account is therefore

M(9/2) = 6000 · e5·(9/2)/100 ≈ $7513.94.

EXAMPLE 6.34

A wealthy woman wishes to set up an endowment for her nephew. She

wants the endowment to pay the young man $100,000 in cash on the day of

his twenty-¬rst birthday.The endowment is set up on the day of the nephew™s

180 CHAPTER 6 Transcendental Functions

birth and is locked in at 11% interest compounded continuously. How much

principal should be put into the account to yield the desired payoff?

SOLUTION

Let P be the initial principal deposited in the account on the day of the

nephew™s birth. Using our compound interest equation ( ), we have

100000 = P · e11·21/100 ,

expressing the fact that after 21 years at 11% interest compounded continuously

we want the value of the account to be $100,000.

Solving for P gives

P = 100000 · e’0.11·21 = 100000 · e’2.31 = 9926.13.

The aunt needs to endow the fund with an initial $9926.13.

You Try It: Suppose that we want a certain endowment to pay $50,000 in cash

ten years from now. The endowment will be set up today with $5,000 principal and

locked in at a ¬xed interest rate. What interest rate (compounded continuously) is

needed to guarantee the desired payoff?

6.6 Inverse Trigonometric Functions

6.6.1 INTRODUCTORY REMARKS

Figure 6.14 shows the graphs of each of the six trigonometric functions. Notice that

each graph has the property that some horizontal line intersects the graph at least

twice. Therefore none of these functions is invertible. Another way of seeing this

point is that each of the trigonometric functions is 2π -periodic (that is, the function

repeats itself every 2π units: f (x + 2π ) = f (x)), hence each of these functions is

not one-to-one.

If we want to discuss inverses for the trigonometric functions, then we must

restrict their domains (this concept was introduced in Subsection 1.8.5). In this

section we learn the standard methods for performing this restriction operation

with the trigonometric functions.

6.6.2 INVERSE SINE AND COSINE

Consider the sine function with domain restricted to the interval [’π/2, π/2]

(Fig. 6.15). We use the notation Sin x to denote this restricted function. Observe

that

d

Sin x = cos x > 0

dx

181

CHAPTER 6 Transcendental Functions

3

2

sec x

1

cos x sin x

_3 _2 _1 1 2 3

_1

csc x

_2

tan x cot x

_3

Fig. 6.14

y

y = Sin x

1

F/2 x

Fig. 6.15

182 CHAPTER 6 Transcendental Functions

on the interval (’π/2, π/2). At the endpoints of the interval, and only there, the

function Sin x takes the values ’1 and +1. Therefore Sin x is increasing on its

entire domain. So it is one-to-one. Furthermore the Sine function assumes every

value in the interval [’1, 1]. Thus Sin : [’π/2, π/2] ’ [’1, 1] is one-to-one and

onto; therefore f (x) = Sin x is an invertible function.

We can obtain the graph of Sin’1 x by the principle of re¬‚ection in the line

y = x (Fig. 6.16). The function Sin’1 : [’1, 1] ’ [’π/2, π/2] is increasing,

one-to-one, and onto.

y

F/2

_1

y = Sin x

x

1

_ F/2

Fig. 6.16

The study of the inverse of cosine involves similar considerations, but we must

select a different domain for our function. We de¬ne Cos x to be the cosine function

restricted to the interval [0, π]. Then, as Fig. 6.17 shows, g(x) = Cos x is a one-to-

one function. It takes on all the values in the interval [’1, 1]. Thus Cos : [0, π] ’

[’1, 1] is one-to-one and onto; therefore it possesses an inverse.

We re¬‚ect the graph of Cos x in the line y = x to obtain the graph of the function

Cos’1 . The result is shown in Fig. 6.18.

EXAMPLE 6.35

Calculate

√ √

3 2

Sin’1 Sin’1 0, Sin’1 ’

, ,

2 2

√ √

3 2

Cos’1 Cos’1 0, Cos’1

’ , .

2 2

183

CHAPTER 6 Transcendental Functions

y

y = Cos x

1

F/2 F x

Fig. 6.17

y

F

F/2 _1

y = Cos x

x

1

Fig. 6.18

SOLUTION

We have

√

3 π

Sin’1 = ,

2 3

Sin’1 0 = 0,

184 CHAPTER 6 Transcendental Functions

√

2 π

Sin’1 ’ =’ .

2 4

√

Notice that even though the sine function takes the value 3/2 at many different

values of the variable x, the function Sine takes this value only at x = π/3.

Similar comments apply to the other two examples.

We also have

√

3 5π

Cos’1 ’ = ,

2 6

π

Cos’1 0 = ,

2

√

2 π

Cos’1 =.

2 4

We calculate the derivative of f (t) = Sin’1 t by using the usual trick for inverse

functions. The result is

1 1

d

(Sin’1 (x)) = =√ .

1 ’ sin2 (Sin’1 x) 1 ’ x2

dx

The derivative of the function Cos’1 t is calculated much like that of Sin’1 t. We

¬nd that

1

d

(Cos’1 (x)) = ’ √ .

1’x

dx 2

EXAMPLE 6.36

Calculate the following derivatives:

1

d d d

Sin’1 x Sin’1 (x 2 + x) Sin’1

, , √.

√

dx dx dx x

x= 2/2 x=’ 3

x=1/3

SOLUTION

We have

√

1

d

Sin’1 x √ =√ = 2,

√

x= 2/2

1 ’ x2

dx x= 2/2

1 15

d

Sin’1 x 2 + x = · (2x + 1) =√ ,

1 ’ (x 2 + x)2

dx 65

x=1/3 x=1/3

1 1 1

d

Sin’1 (1/x) = ·’ = ’√ .

√ x2 √

1 ’ (1/x)2

dx 6

x=’ 3 x=’ 3

185

CHAPTER 6 Transcendental Functions

You Try It: Calculate (d/dx)Cos’1 [x 2 + x]. Also calculate (d/dx)Sin’1 —

[ln x ’ x 3 ].

EXAMPLE 6.37

Calculate each of the following derivatives:

√

d d d

Cos’1 x Cos’1 (ln x) Cos’1 ( x)

, , .

√

dx dx dx

x=1/2 x= e x=1/2

SOLUTION

We have

1 2

d

Cos’1 x =’ √ = ’√ ,

1 ’ x2

dx 3

x=1/2 x=1/2

’2

1 1

d

Cos’1 (ln x) =’ · = ’√ ,

√

1 ’ (ln x)2 √

dx x 3e

x= e x= e

√ 1 1 ’1/2

d

Cos’1 ( x) =’ · = ’1.

√ x

1 ’ ( x)2 2

dx x=1/2 x=1/2

You Try It: Calculate (d/dx) ln[Cos’1 x] and (d/dx) exp[Sin’1 x].

6.6.3 THE INVERSE TANGENT FUNCTION

De¬ne the function Tan x to be the restriction of tan x to the interval (’π/2, π/2).

Observe that the tangent function is unde¬ned at the endpoints of this interval.

Since

d

Tan x = sec2 x

dx

we see that Tan x is increasing, hence it is one-to-one (Fig. 6.19). Also Tan takes

arbitrarily large positive values when x is near to, but less than, π/2. And Tan takes

negative values that are arbitrarily large in absolute value when x is near to, but

greater than, ’π/2. Therefore Tan takes all real values. Since Tan : (’π/2, π/2) ’

(’∞, ∞) is one-to-one and onto, the inverse function Tan’1 : (’∞, ∞) ’

(’π/2, π/2) exists. The graph of this inverse function is shown in Fig. 6.20. It

is obtained by the usual procedure of re¬‚ecting in the line y = x.

EXAMPLE 6.38

Calculate

√ √

Tan’1 1, Tan’1 1/ 3, Tan’1 (’ 3).

186 CHAPTER 6 Transcendental Functions

y

y = Tan x

x

Fig. 6.19

y

_1

y = Tan x

x

Fig. 6.20

SOLUTION

We have

π

Tan’1 1 = ,

4

√ π

Tan’1 1/ 3 = ,

6

√ π

Tan’1 (’ 3) = ’ .

3

187

CHAPTER 6 Transcendental Functions

As with the ¬rst two trigonometric functions, we note that the tangent func-

√ √

tion takes each of the values 1, 1/ 3, ’ 3 at many different points of its

domain. But Tan x takes each of these values at just one point of its domain.

The derivative of our new function may be calculated in the usual way. The

result is

1

d

Tan’1 t = .

1 + t2

dt

Next we calculate some derivatives:

EXAMPLE 6.39

Calculate the following derivatives:

d d d

Tan’1 x Tan’1 (x 3 ) Tan’1 (ex )

, √, .

dx dx dx

x= 2

x=1 x=0

SOLUTION

We have

1 1

d

Tan’1 x = =,

1 + x2 2

dx x=1 x=1

1 2

d

Tan’1 (x 3 ) = · 3x 2 =,

√ √

1 + (x 3 )2 3

dx x= 2 x= 2

1 1

d

Tan’1 (ex ) = · ex =.

1 + (e x )2 2

dx x=0 x=0

You Try It: Calculate (d/dx)Tan’1 [ln x + x 3 ] and (d/dx) ln[Tan’1 x].

6.6.4 INTEGRALS IN WHICH INVERSE

TRIGONOMETRIC FUNCTIONS ARISE

Our differentiation formulas for inverse trigonometric functions can be written in

reverse, as antidifferentiation formulas. We have

du

= Sin’1 u + C;

√

1 ’ u2

du

= ’Cos’1 u + C;

√

1 ’ u2

du

du = Tan’1 u + C.

1+u 2

The important lesson here is that, while the integrands involve only polynomials

and roots, the antiderivatives involve inverse trigonometric functions.

188 CHAPTER 6 Transcendental Functions

EXAMPLE 6.40

Evaluate the integral

sin x

dx.

1 + cos2 x

SOLUTION

For clarity we set •(x) = cos x, • (x) = ’ sin x. The integral becomes

• (x) dx

’ .

1 + • 2 (x)

By what we have just learned about Tan’1 , this last integral is equal to

’Tan’1 •(x) + C.

Resubstituting •(x) = cos x yields that

sin x

dx = ’Tan’1 (cos x) + C.

Y

1 + cos 2x

FL

x/(1 + x 4 ) dx.

You Try It: Calculate

EXAMPLE 6.41

AM

Calculate the integral

3x 2

√ dx.

TE

1 ’ x6

SOLUTION

For clarity we set •(x) = x 3 , • (x) = 3x 2 . The integral then becomes

• (x) dx

.

1 ’ • 2 (x)

We know that this last integral equals

Sin’1 •(x) + C.

Resubstituting the formula for • gives a ¬nal answer of

3x 2

dx = Sin’1 (x 3 ) + C.

√

1 ’ x6

You Try It: Evaluate the integral

x dx

√ .

1 ’ x4

189

CHAPTER 6 Transcendental Functions

6.6.5 OTHER INVERSE TRIGONOMETRIC FUNCTIONS

The most important inverse trigonometric functions are Sin’1 , Cos’1 , and Tan’1 .

We say just a few words about the other three.

De¬ne Cot x to be the restriction of the cotangent function to the interval (0, π)

(Fig. 6.21). Then Cot is decreasing on that interval and takes on all real values.

Therefore the inverse

Cot’1 : (’∞, ∞) ’ (0, π)

y

y = Cot x

F x

Fig. 6.21

is well de¬ned. Look at Fig. 6.22 for the graph. It can be shown that

1

d

Cot’1 x = ’ .

1 + x2

dx

De¬ne Sec x to be the function sec x restricted to the set [0, π/2) ∪ (π/2, π]

(Fig. 6.23). Then Sec x is one-to-one. For these values of the variable x, the cosine

function takes all values in the interval [’1, 1] except for 0. Passing to the recip-

rocal, we see that secant takes all values greater than or equal to 1 and all values

less than or equal to ’1. The inverse function is

Sec’1 : (’∞, ’1] ∪ [1, ∞) ’ [0, π/2) ∪ (π/2, π]

190 CHAPTER 6 Transcendental Functions

y

F

_1

y = Cot x

x

Fig. 6.22

y

y = Sec x

1

F x

Fig. 6.23

(Fig. 6.24). It can be shown that

1

d

Sec’1 x = |x| > 1.

√ ,

|x| · x 2 ’ 1

dx

The function Csc x is de¬ned to be the restriction of Csc x to the set [’π/2, 0) ∪

(0, π/2]. The graph is exhibited in Fig. 6.25. Then Csc x is one-to-one. For these

values of the x variable, the sine function takes on all values in the interval [’1, 1]

except for 0. Therefore Csc takes on all values greater than or equal to 1 and all

values less than or equal to ’1; Csc’1 therefore has domain (’∞, ’1] ∪ [1, ∞)

and takes values in [’1, 0) ∪ (0, 1] (Fig. 6.26).

191

CHAPTER 6 Transcendental Functions

y

F

_1

y = Sec x

x

1

Fig. 6.24

y

_ F/2 F/2 x

Fig. 6.25

It is possible to show that

1

d

Csc’1 x = ’ |x| > 1.

√ ,

|x| · x 2’1

dx

√ √

’1 ’1

You Try It: What is Sec (’2/ 3)? What is Csc (’ 2)?

192 CHAPTER 6 Transcendental Functions

y

F/2

x

_ F/2

Fig. 6.26

Summary of Key Facts About the Inverse Trigonometric Functions

π π

Sin x = sin x, ’ ¤ x ¤ ; Cos x = cos x, 0 ¤ x ¤ π;

2 2

π π

Tan x = tan x, ’ < x < ; Cot x = cot x, 0 < x < π;

2 2

Sec x = sec x, x ∈ [0, π/2) ∪ (π/2, π]; Csc x = csc x, x ∈ [’π/2, 0) ∪ (0, π/2].

1 1

d d

Sin’1 x = √ Cos’1 x = ’ √

, ’1 < x < 1; , ’1 < x < 1;

1 ’ x2 1 ’ x2

dx dx

1 1

d d

Tan’1 x = Cot’1 x = ’

, ’∞ < x < ∞; , ’∞ < x < ∞;

1 + x2 1 + x2

dx dx

1 1

d d

Sec’1 x = Csc’1 x = ’

, |x| > 1; , |x| > 1;

√ √

|x| · x 2 ’ 1 |x| · x 2 ’ 1

dx dx

du du

= Sin’1 u + C; = ’Cos’1 u + C;

√ √

1 ’ u2 1 ’ u2

du du

du = Tan’1 u + C; du = ’Cot’1 u + C;

1+u 1+u

2 2

du du

= Sec’1 u + C; = ’Csc’1 u + C.

√ √

|u| · u 2’1 |u| · u 2’1

You Try It: What is the derivative of Sec’1 x 2 ?

193

CHAPTER 6 Transcendental Functions

6.6.6 AN EXAMPLE INVOLVING INVERSE

TRIGONOMETRIC FUNCTIONS

EXAMPLE 6.42

Hypatia is viewing a ten-foot-long tapestry that is hung lengthwise on a

wall. The bottom end of the tapestry is two feet above her eye level. At what

distance should she stand from the tapestry in order to obtain the most

favorable view?

SOLUTION

For the purposes of this problem, view A is considered more favorable than

view B if it provides a greater sweep for the eyes. In other words, form the

triangle with vertices (i) the eye of the viewer, (ii) the top of the tapestry, and

(iii) the bottom of the tapestry (Fig. 6.27). Angle ± is the angle at the eye of

the viewer. We want the viewer to choose her position so that the angle ± at the

eye of the viewer is maximized.

10 ft

G

=

2 ft

O

x ft

Fig. 6.27

The ¬gure shows a mathematical model for the problem. The angle ± is the

angle θ less the angle ψ. Thus we have

± = θ ’ ψ = Cot’1 (x/12) ’ Cot’1 (x/2).

Notice that when the viewer is standing with her face against the wall then

θ = ψ = π/2 so that ± = 0. Also when the viewer is far from the tapestry then

θ ’ ± is quite small. So the maximum value for ± will occur for some ¬nite,

positive value of x. That value can be found by differentiating ± with respect

to x, setting the derivative equal to zero, and solving for x.

194 CHAPTER 6 Transcendental Functions

√

We leave it to you to perform the calculation and discover that 24 ft is the

optimal distance at which the viewer should stand.

You Try It: Redo the last example if the tapestry is 20 feet high and the bottom

of the tapestry is 6 inches above eye level.

Exercises

1. Simplify these logarithmic expressions.

a 2 · b’3

(a) ln 4

c ·d

log3 (a 2 b)

(b)

log2 (a 2 b)

(c) ln[e3x · z4 · w ’3 ]

√

(d) log10 [100w · 10]

2. Solve each of these equations for x.

(a) 3x · 5’x = 2x · e3

3x

= 10x · 102

(b)

5’x · 42x

(c) 23x · 34x · 45x = 6

5 2

= x ’x

(d)

2 ·3

23x · e5x

3. Calculate each of these derivatives.

d

ln[sin(x 2 )]

(a)

dx

x2

d

(b) ln

x’1

dx

d sin(ex )

(c) e

dx

d

(d) sin(ln x)

dx

4. Calculate each of these integrals.

e’x x 3 dx [Hint: Guess p(x) · e’x , p a polynomial.]

(a)

195

CHAPTER 6 Transcendental Functions

x 2 · ln2 x dx [Hint: Guess p(x) · ln2 x + q(x) ln x + r(x), p, q, r

(b)

polynomials.]

e ln x

(c) dx

x

1

2 ex

(d) dx

ex + 1

1

5. Use the technique of logarithmic differentiation to calculate the derivative

of each of the following functions.

x2 + 1

(a) x · 3

3

x ’x

sin x · (x 3 + x)

(b)

x 2 (x + 1)

(c) (x 2 + x 3 )4 · (x 2 + x)’3 · (x ’ 1)

x · cos x

(d)

ln x · ex

6. There are 5 grams of a certain radioactive substance present at noon on

January 10 and 3 grams present at noon on February 10. How much will be

present at noon on March 10?

7. A petri dish has 10,000 bacteria present at 10:00 a.m. and 15,000 present at

1:00 p.m. How many bacteria will there be at 2:00 p.m.?

8. A sum of $1000 is deposited on January 1, 2005 at 6% annual interest,

compounded continuously. All interest is re-invested. How much money

will be in the account on January 1, 2009?

9. Calculate these derivatives.

d

Sin’1 (x · ex )

(a)

dx

d x

Tan’1

(b)

x+1

dx

d

Tan’1 [ln(x 2 + x)]

(c)

dx

d

Sec’1 (tan x)

(d)

dx

10. Calculate each of these integrals.

2x

(a) x dx

1 + x4

196 CHAPTER 6 Transcendental Functions

3x 2

√

(b) dx

1 ’ x6

π/2 2 sin x cos x

(c) dx

1 ’ sin x 4

0

dx

(d)

5 + 2x 2

CHAPTER 7

Methods of

Integration

7.1 Integration by Parts

We learned in Section 4.5 that the integral of the sum of two functions is the sum

of the respective integrals. But what of the integral of a product? The following

reasoning is incorrect:

x 2 dx = x · x dx = x dx · x dx

because the left-hand side is x 3 /3 while the right-hand side is (x 2 /2) · (x 2 /2) =

x 4 /4.

The correct technique for handling the integral of a product is a bit more subtle,

and is called integration by parts. It is based on the product rule

(u · v) = u · v + u · v .

Integrating both sides of this equation, we have

(u · v) dx = u · v dx + u · v dx.

The Fundamental Theorem of Calculus tells us that the left-hand side is u · v. Thus

u·v = u · v dx + u · v dx

197

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198 CHAPTER 7 Methods of Integration

or

u · v dx = u · v ’ v · u dx.

It is traditional to abbreviate u (x) dx = du and v (x) dx = dv. Thus the

integration by parts formula becomes

u dv = uv ’ v du.

Let us now learn how to use this simple new formula.

EXAMPLE 7.1

Calculate

x · cos x dx.