ln |x|

(f ) lim

x’’∞ e’x

3. If possible, use some algebraic manipulations, plus l™Hôpital™s Rule, to

evaluate each of the following limits. In each case, check carefully that the

hypotheses of l™Hôpital™s Rule apply.

lim x 3 e’x

(a)

x’+∞

lim x · sin[1/x]

(b)

x’+∞

146 CHAPTER 5 Indeterminate Forms

lim ln[x/(x + 1)] · (x + 1)

(c)

x’+∞

lim ln x · e’x

(d)

x’+∞

lim e2x · x 2

(e)

x’’∞

lim x · e1/x

(f )

x’0

4. Evaluate each of the following improper integrals. In each case, be sure to

write the integral as an appropriate limit.

1

x ’3/4 dx

(a)

0

3

(x ’ 3)’4/3 dx

(b)

1

2 1

(c) dx

(x + 1)1/3

’2

6 x

(d) dx

(x ’ 1)(x + 2)

’4

x+4

8

(e) dx

(x ’ 8)1/3

4

3 sin x

(f ) dx

x2

0

5. Evaluate each of the following improper integrals. In each case, be sure to

write the integral as an appropriate limit.

∞

e’3x dx

(a)

1

∞

x 2 e’x dx

(b)

2

∞

(c) x ln x dx

0

∞ dx

(d)

1 + x2

1

∞ dx

(e)

x

1

’1 dx

(f )

x2 + x

’∞

CHAPTER 6

Transcendental

Functions

6.0 Introductory Remarks

There are two types of functions: polynomial and transcendental. A polynomial

of degree k is a function of the form p(x) = a0 + a1 x + a2 x 2 + · · · + ak x k .

Such a polynomial has precisely k roots, and there are algorithms that enable us to

solve for those roots. For most purposes, polynomials are the most accessible and

easy-to-understand functions. But there are other functions that are important in

mathematics and physics. These are the transcendental functions. Among this more

sophisticated type of functions are sine, cosine, the other trigonometric functions,

and also the logarithm and the exponential. The present chapter is devoted to the

study of transcendental functions.

6.1 Logarithm Basics

A convenient, intuitive way to think about the logarithm function is as the inverse

to the exponentiation function. Proceeding intuitively, let us consider the function

f (x) = 3x .

To operate with this f, we choose an x and take 3 to the power x. For example,

f (4) = 34 = 3 · 3 · 3 · 3 = 81

147

Copyright 2003 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.

148 CHAPTER 6 Transcendental Functions

1

f (’2) = 3’2 =

9

f (0) = 30 = 1.

The inverse of the function f is the function g which assigns to x the power to

which you need to raise 3 to obtain x. For instance,

g(9) = 2 because f (2) = 9

g(1/27) = ’3 because f (’3) = 1/27

g(1) = 0 because f (0) = 1.

We usually call the function g the “logarithm to the base 3” and we write g(x) =

log3 x. Logarithms to other bases are de¬ned similarly.

While this approach to logarithms has heuristic appeal, it has many drawbacks:

we do not really know what 3x means when x is not a rational number; we have

no way to determine the derivative of f or of g; we have no way to determine the

integral of f or of g. Because of these dif¬culties, we are going to use an entirely

Y

new method for studying logarithms. It turns out to be equivalent to the intuitive

FL

method described above, and leads rapidly to the calculus results that we need.

6.1.1 A NEW APPROACH TO LOGARITHMS

AM

When you studied logarithms in the past you learned the formula

log(x · y) = log x + log y;

TE

this says that logs convert multiplication to addition. It turns out that this property

alone uniquely determines the logarithm function.

Let (x) be a differentiable function with domain the positive real numbers

and whose derivative function (x) is continuous. Assume that satis¬es the

multiplicative law

(x · y) = (x) + (y) (—)

for all positive x and y. Then it must be that (1) = 0 and there is a constant C

such that

C

(x) = .

x

In other words

xC

(x) = dt.

t

1

A function (x) that satis¬es these properties is called a logarithm function.

The particular logarithm function which satis¬es (1) = 1 is called the natural

149

CHAPTER 6 Transcendental Functions

logarithm: In other words,

x 1

natural logarithm = ln x = dt.

t

1

For 0 < x < 1 the value of ln x is the negative of the actual area between the

graph and the x-axis. This is so because the limits of integration, x and 1, occur in

x

reverse order: ln x = 1 (1/t) dt with x < 1.

Fig. 6.1

Notice the following simple properties of ln x which can be determined from

looking at Fig. 6.1:

(i) When x > 1, ln x > 0 (after all, ln x is an area).

(ii) When x = 1, ln x = 0.

(iii) When 0 < x < 1, ln x < 0

1

x 1 1

dt = ’

since dt < 0 .

t t

1 x

(iv) If 0 < x1 < x2 then ln x1 < ln x2 .

We already know that the logarithm satis¬es the multiplicative property. By

applying this property repeatedly, we obtain that: If x > 0 and n is any integer then

ln(x n ) = n · ln x.

A companion result is the division rule: If a and b are positive numbers then

a

= ln a ’ ln b.

ln

b

EXAMPLE 6.1

Simplify the expression

a 3 · b2

A = ln .

c’4 · d

150 CHAPTER 6 Transcendental Functions

SOLUTION

We can write A in simpler terms by using the multiplicative and quotient

properties:

A = ln(a 3 · b2 ) ’ ln(c’4 · d)

= [ln a 3 + ln(b2 )] ’ [ln(c’4 ) + ln d]

= [3 ln a + 2 · ln b] ’ [(’4) · ln c + ln d]

= 3 ln a + 2 · ln b + 4 · ln c ’ ln d.

The last basic property of the logarithm is the reciprocal law: For any x > 0

we have

ln(1/x) = ’ ln x.

EXAMPLE 6.2

Express ln(1/7) in terms of ln 7. Express ln(9/5) in terms of ln 3 and ln 5.

SOLUTION

We calculate that

ln(1/7) = ’ ln 7,

ln(9/5) = ln 9 ’ ln 5 = ln 32 ’ ln 5 = 2 ln 3 ’ ln 5.

You Try It: Simplify ln(a 2 b’3 /c5 ).

6.1.2 THE LOGARITHM FUNCTION AND THE

DERIVATIVE

Now you will see why our new de¬nition of logarithm is so convenient. If we want

to differentiate the logarithm function we can apply the Fundamental Theorem of

Calculus:

x 1 1

d d

ln x = dt = .

dx dx t x

1

More generally,

1 du

d

ln u = .

dx u dx

EXAMPLE 6.3

Calculate

d d d d d

ln(4 + x), ln(x 3 ’ x), [(ln x)5 ], [(ln x) · (cot x)].

ln(cos x),

dx dx dx dx dx

151

CHAPTER 6 Transcendental Functions

SOLUTION

For the ¬rst problem, we let u = 4 + x and du/dx = 1. Therefore we have

1 1

d d

ln(4 + x) = · (4 + x) = .

4 + x dx 4+x

dx

Similarly,

3x 2 ’ 1

1

d d3

ln(x ’ x) = 3 · (x ’ x) = 3

3

x ’ x dx x ’x

dx

’ sin x

1

d d

ln(cos x) = · (cos x) =

cos x dx cos x

dx

5(ln x)4

41

d d

[(ln x) ] = 5(ln x) · (ln x) = 5(ln x) · =

5 4

dx dx x x

d d d

[(ln x) · (cot x)] = ln x · (cot x) + (ln x) · cot x

dx dx dx

1

= · cot x + (ln x) · (’ csc2 x).

x

You Try It: What is the derivative of the function ln(x 3 + x 2 ) ?

Now we examine the graph of y = ln x. Since

1

d

(ln x) =

(i) > 0,

dx x

d2 d1 1

(ln x) = =’

(ii) < 0,

x2

dx dx x

(iii) ln(1) = 0,

we know that ln x is an increasing, concave down function whose graph passes

through (1, 0). There are no relative maxima or minima (since the derivative is

never 0). Certainly ln .9 < 0; the formula ln(.9n ) = n ln .9 therefore tells us that

ln x is negative without bound as x ’ 0+ . Since ln x = ’ ln(1/x), we may also

conclude that ln x is positive without bound as x ’ +∞. A sketch of the graph of

y = ln x appears in Fig. 6.2.

We learned in the last paragraph that the function ln x takes negative values which

are arbitrarily large in absolute value when x is small and positive. In particular,

the negative y axis is a vertical asymptote. Since ln(1/x) = ’ ln x, we then ¬nd

that ln x takes arbitrarily large positive values when x is large and positive. The

graph exhibits these features.

Since we have only de¬ned the function ln x when x > 0, the graph is only

sketched in Fig. 6.2 to the right of the y-axis. However it certainly makes sense to

152 CHAPTER 6 Transcendental Functions

Fig. 6.2

discuss the function ln |x| when x = 0 (Fig. 6.3):

Fig. 6.3

If x = 0 then

1

d

(ln |x|) = .

dx x

In other words,

1

dx = ln |x| + C.

x

More generally, we have

1 du

d

ln |u| =

dx u dx

and

1 du

dx = ln |u| + C.

u dx

EXAMPLE 6.4

Calculate

4 1

dx, dx.

x+1 ’2 + 3x

153

CHAPTER 6 Transcendental Functions

SOLUTION

4 1

dx = 4 dx = 4 ln |x + 1| + C

x+1 x+1

1 1

dx = ln | ’ 2 + 3x| + C.

’2 + 3x 3

You Try It: Calculate the integral

cos x

dx.

2 + sin x

You Try It: Calculate the integral

e 1

dx.

x · [ln x]3/2

1

EXAMPLE 6.5

Evaluate the integral

cos x

dx.

3 sin x ’ 4

SOLUTION

For clarity we set •(x) = 3 sin x ’ 4, • (x) = 3(cos x). The integral then

has the form

1 1

• (x)

dx = ln |•(x)| + C.

3 3

•(x)

Resubstituting the expression for •(x) yields that

cos x 1

dx = ln |3 sin x ’ 4| + C.

3 sin x ’ 4 3

x2

You Try It: Evaluate dx.

1 ’ x3

EXAMPLE 6.6

Calculate

cot x dx.

SOLUTION

We rewrite the integral as

cos x

dx.

sin x

154 CHAPTER 6 Transcendental Functions

For clarity we take •(x) = sin x, • (x) = cos x. Then the integral becomes

• (x)

dx = ln |•(x)| + C.

•(x)

Resubstituting the expression for • yields the solution:

cot x dx = ln | sin x| + C.