Bear H.S. Understanding Calculus

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UnderstandingCalculus

-3

-2

-1

We calculate

a" as follows:

d X

_ax

= lim

dX~O

aXa

d X

= lim

dX~O

d X

= lim a

- _

dX~O

d X

lim .

dX~O

It can be shown that this limit exists for any a > 0, and we let

d X

-1

= lim .

~x~o

With this notation we can write

2 3

Figure 7.1

d x x

-a

=kaa . (7.1)

The derivative

any exponential function (i.e., any a) is always proportional to the function

value.

Now we do some experimenting to see how the proportionality constant

depends on

a. First let a =2 and calculate

(2~X

- 1)/

for small values

d X

1)/

0.1

.0718

.718

.01

.00695

.695

.001

.000693

.693

.0001

.000693

.693

As far as our eight-place calculator can tell

d X

= lim

.693.

dX~O

we do the same calculations (Problem 7.1) for a = 3, we get the approximate limit

d X

= lim

1.099.

dX~O

Llx

It is clear that the difference quotient (a

- 1)/

Llx

gets bigger as a increases, for any fixed

Llx.

Therefore, k

increases as a increases. It is therefore plausible that there is some number

Chapter7 •

Exponential

Functions

and

Logarithms

e between 2 and 3 such that k, = 1, and experimentation shows (Problem 7.2) that e is

approximately equal to 2.718. Now for this number e,

_ex=e

(7.2)

and the elegance of (7.2) is the justification for introducing the number e. The chain rule then

gives the formulas

~e!(X)

e!(x)

['(x),

d u u

-e

=e-.

dx dx

As examples, we have

d 2 2

_eX

.2x;

_e-

(- 3);

d 1

-e~=e~--;

d·

SlO X

= e

SlO X

cosx.

The graph of y =e' is shown in Figure 7.2, along with its tangent line at (0, 1).

Figure 7.2

The function e' is strictly increasing, is not bounded above, and approaches zero as

-00.

Every positive number y is

for some unique number x. This number x is

called the natural logarithm of y and is denoted log y. In other words, y = e' and x = log y

mean the same thing. The functions

and log x are inverse functions;

is defined for all x

and takes as values all positive numbers; log x is defined for all positive numbers and takes as

values all numbers. For all

x > 0, elog X =x, and for all x, log

=x. The graphs of y =e'

and y =

logx

are shown together in Figure 7.3.

Notice that the graphs are symmetric about the line

y = x. This is true of every pair of

inverse functions and simply reflects the fact that (a, b) is on the graph of one function if and

only if

(b, a) is on the graph of its inverse.

Understanding

Calculus

Figure 7.3

Ingeneral, thenotation I

-I

(x) is usedforthefunction

inverse

to I (x), so1(1

-1

(x)) =

1-1

(x)) =x. Noticethattheexponent notation is not usedconsistently here: I (x)2 means

I(x)

I(x),

but

I-I

(x) nevermeans

1/I(x).

Thelogarithm of a numberis an exponent; thatis, f, = logN meanse

= N. Therefore,

the rulesfor logarithms arejust the algebraic rules for exponents. The log of a productis the

sumof the logs since

ell

· e

l 2

ell

+l2.

That is,

log(ab) = loga +Iogs.

Similarly,

log

= k logN

sinceif e

=N, e

and the log of N

is k timesthe log of N.

Tofindthe

derivative

of logx, we let y =logx, so e

=x, and then differentiate:

=x,

ydy

_ 1

= - .

The generalchainrule formula is

d 1 du

-logu

--.

For example,

d 2x

-d

10g(x

+ 1) =

-2--1

;

x x +

d sinx

-logcosx

---;

cosx

-log(1

+ eX) =

--.

1 +

Chapter 7 • Exponential Functions and Logarithms

Many times a logarithm formula can be simplified before it is differentiated; for example,

d d

-log[(x

+ 1)(x

3)] =

-[log(x

+ 1) + log(x

3)]

1 3x

---+-_.

- x + 1 x

dId

-log

2 =

-log(5x

1)-2

(5x + 1)

-(-2)

log(5x + 1)

=(-2)oSx+l;

d d

-log2

-x

log 2 = log 2.

We chose e as the primary base for an exponential function because k; = 1; that is,

because

. 1. Now let's go back to

for general a. Since log

=x log a and

x x

u = e

10g

U for all u, we can write

From this we get

d d

_ax

= eX

loga

-(x

loga)

loga.

Hence, the constant k

= log a. Earlier we computed the approximate value k

.693, so

from

= k

and

= 2

log 2 we conclude that log 2

.693.

x x

Functions that involve the variable in both the base and the exponent do not fall under

either of the rules

d d

=nx

;

_ax

loga.

dx dx

To differentiate such a function, we first take its logarithm. For example, let

y = (x

+ 2)3x.

Then

log y = 3x log(x

+2),

1 dy 2 2x

- - = 310g(x +2) +3x .

+ 2

dy [ 6x

]

- = Y 310g(x

+2) +

-2-

[

6x2 ]

= (x

+ 2)3x 310g(x

+ 2) +

-2--

PROBLEMS

7.1 Approximate k

by evaluating

(3~X

1)/

for

= .1, .01, .001, .0001.

7.2 Approximate

k, by calculating

[(2.718)~X

1]/

for

= .1, .01, .001, .0001.

Understanding Calculus

Find

7.3 y = e

7.4 y =e

x 2

7.5 y = 1/e

x 2

7.6 y =

(ev'X

+ 1)2

7.7

y=~

7.8 y =3

7.9 y = 2

x 2

7.10 y =x

+ x cosx

7.11 y =

cosx +

sinx

7.12 y =

log x

7.13

y = e

x 2

log(1 + x

)

7.14 y = log 3x

7.15

y = log(x

7.16 y = 10g[xe

]

7.17 y =log[(x + 1)(x + 2)]

7.18 y =log(x

2/3

)

7.19 y = XX

Hint: Find t log y first.

7.20 y = (2x +

1)X

7.21 y = (sin x + cos

r)"

7.22 The number y

bacteria at time t in any given colony is given by y = Ae

for some

constants A and k. If

y = 1000 at t = 0 and y = 3000 at t = 2, what are A and k? What

y when t = 4?

7.23 Show that if y = Ae

and y = 2A when t = to,then y will double in any time interval

to;

that is, for all t ,

Aek(t+to)

= 2.

7.24 If

Y is the number of bacteria in a culture, and y = 500 at t = 0, so y =

soo-«

(t is hours),

and

y doubles every hour, what is k?

7.25 If an amount A of money is invested at 6%, compounded annually, then after n years the

investment is worth A

(1.06)n. How long does it take to double your money at 6%?

7.26 A common rule of thumb says that money doubles at interest rate

after n years, where

n =

1f-.

For example, at 7% money doubles in 10 years. Explain, keeping in mind that

log 2

.70 and for small x (e.g., .05, .07, etc.), log(1 +x)

7.27 Show that the only functions y such that *= y are the functions y = Ae

•

Hint:

Assume

y is such a function; let u =

ye-

and show that

7.28 The hyperbolic cosine and hyperbolic sine are defined as follows:

e' + e-

e' - e-

cosh x = 2

,sinh

x = 2 .

Show that

(a)

t cosh x = sinh x and t sinh x = cosh x;

(b) coslr' x - sinlr' X = 1.

7.29 (a) Graph the curve y = e!x,

-1

(b) Find the point

P = (x, e

) on the curve y =

such that the tangent line at P passes

through the origin. Does your answer look right if

a =

at P intersects the x-axis.

7.30 (a) Find the point(s) on the graph of y =

xe'

where the tangent line is horizontal.

(b) Graph the curve for

-5

Inverse

Functions

A function I is said to be one-to-one on an interval I provided I does not take the same value

for different values of x in I. For the continuous functions we consider here, a one-to-one

function is either strictly increasing on I (I

(X2)

> I (Xl) if X2 > Xl), or strictly decreasing

(X2)

< I (Xl) if X2 >

Xl)'

If I is a one-to-one function, then for each function value y

there is a unique X such that

I(x)

= y. This mapping - from y to X - is denoted

I-I

(read

I -inverse), so that X =

1-1

(y) and y = I (x) mean the same thing. In the last chapter we

saw that log

X and e" are inverse functions, so that if I (x) = e", 1-1(x) = log x. For any

one-to-one function

I(/-I(x))

= X and

I-I(/(x))

= x.

Now consider the inverses of the power functions x", If n =

1,3,5,

...

, then x" is

strictly increasing for all x, and if n = 2, 4, 6,

...

, then x

is strictly increasing for X

use

to denote the inverse of x

so if I (x) = x", then

I-I

(x) = x

, and x

is the nth root

x. If n is odd, then x

is defined for all x, but if n is even, x

is defined only for x

Thus, x 4 denotes the positive square root of the positive number x.

We let y = x

and calculate

Since yn = x, we differentiate and get

n-l

dy _ 1

--,

(

)n-I

n X

1 1

dy 1

nyn-I

Now write

yn-I

in terms of x. Since y =

x~,

=----

1-1

= - x

•

The rule for differentiating x

is the same as that for differentiating x" or x

-n

: multiply

by the exponent, and subtract one from the exponent to get the new exponent. The chain

Understanding

Calculus

rule extends the exponent rule to fractional exponents

(x;:-

= (x

)m) and negative exponents

(x-;:- =

(x;)-I):

d m m !!!._I d m m

_!!!_I

-Xii

-x

n ;

-x

n =

--x

Indeed, the exponent rule holds for

any exponent:

= rx

•

Here are some examples:

d I 1 2

-x'3

-x-'3·

3 '

d 4 4 I

-x

-x-so

5 '

d 7 7 4

-x'3

-x'3·

d 5 5 7

-x-

--X-i·

.!!.....x~

J2x~-I.

Now consider the inverses of the trigonometric functions. The functions sin x and cos x

are periodic (sin x =sin(x +21l')) and so are not one-to-one. However, the functions are one-

to-one on certain intervals. For example, sin

x is one-to-one (strictly increasing) on

I]'

and

cosx

is one-to-one (strictly decreasing) on [0, 1l']. We define sin"! x and cos"! x to be

the inverses

the functions on these restricted domains. This is

just

what we did when we

defined

to be the inverse of x

restricted to [0,

(0).

So sin

-1

for

-1

1, is the

unique

y between - I and I such that sin y = x. Similarly, cos"! x is the unique number y

between 0 and n such that sin y = x.

Now we calculate the derivatives. Let y =

sin-

x, so

siny

= x,

cosy

= 1,

(8.1)

= costsin "

A similar process shows that if y = cos" x, then

dy d

-I

- =

-cos

x = (8.2)

sintcosr!

To simplify the expressions

costsin"!

and sintcos"!

x),

we draw the right triangle (Figure

8.1) in which the angles 8

sin-

x and

cos-

x are displayed. From the figure we can

read the identities

costsin"

=cos s =

.Jf=X2,

sin(cos-

= sin

.Jf=X2.

(8.3)

Chapter8 •

Inverse

Functions

Figure 8.1

Now rewriting (8.1) and (8.2) using (8.3), we have

d .

-I

-SIn

x = ,

.JT=X2

-I

-cos

x = - .

.JT=X2

Here are some chain rule examples with

sin-I

x and

COS-I

..!!-

sin-

(3x)

= 3 ;

- (3x)2

-I

2 1

-COS

x = - . (2x);

.Jf=X4

d .

sin-

e' = ;

dx JI - e

d 1 1 1

cos-

(logx)

= - -

- (log x)? x

Notice that all these functions have limited domains:

sin-

3x is defined if - 1

cos" x

is defined if - 1

sin-

e' is defined if -

< x

The function tan x is strictly increasing for - I < x < I and takes on all real values

on this interval. Therefore, tan

-I

x is defined for all x and takes values in

I).

To find

tan-

x we let y =

tan:"

x, so

tany = x,

sec

= I,

- = cos?y.

From Figure 8.2 we see that cos y = J 1

l+x

-tan

x=--.

dx 1 + x

(8.4)

Figure 8.2

y =tan-

Understanding Calculus

variant

(8.4)

thatis

important

for ourlater

work

integration

d 1

-1

X 1

--tan

dx a a a

PROBLEMS

Find

8.1 y = 1 +x:J

3 1

8.2 y

=X-

+5x

7 7

8.3 y = 3xs -

2x-:S

8.4 y =

x,jX

- 3x

2,jX

8.5 y = (x + 1) 3

8.6 y =

sin(x~

+ 1)

8.7 y = (1 + x

)

8.8 y = (cos' X + 1)

8.9 y =

tanu

: +1)

8.10 y = (secx +

1)-j

Find the exact values;that is, sin" f =

(not .785).

8.11

sin-

8.12 cos" 2

-I

8.13 tan ,J3

8.14 cos" 2

8.15

sin-

( -

v;)

8.16 cos" (

-D

8.17 cos"

(-1)

8.18 sin-I (1)

8.19

sin-l(sinn)

8.20 cos

-I

(cos ( -

))

Find

8.21 y = sin-

8.22 y = sin"! 3x

8.23

y = sin"!(2x - 1)

8.24

y =

COS-I

,jX

8.25 y =cos-

(eX)

8.26 y =cos" (x + 1)

8.27 y = tan'" 3x

8.28 y = tan" (x + 2)

8.29 y = tan:" (log

(8.5)

Chapter 8 • Inverse Functions

(

SinX)

8.30 y =

tan-

cosx

8.31 (a) Verify that t

tarr

-t

tan" x

(b) Draw a right triangle which shows that tan-I x +tan-I

= I'

• d In d 1 //I I

8.32 Use the chain rule to show that d; (x

(x'") ii = ; x n - .

8.33 Show that the identity costsin "

-JI=X2followsfromtheidentitycosu

= )1 - sin

8.34 (a) Verify: -dd !

tan-Ict+~)

~+(l+b)

•

x a a a x

(b) Find y if

;1.;

25+(~-2)2'

8.35 Use Figures 8.1 and 8.2 to write the following as algebraic functions of x:

(a) tantsin" x)

(b) costtan" x)

(d) tantsec" Jf+X2)