SOLUTION

We calculate the limit (—), with the role of c played by 2:

f (2 + h) ’ f (2) [(2 + h)2 + (2 + h)] ’ [22 + 2]

= lim

lim

h h

h’0 h’0

[(4 + 4h + h2 ) + (2 + h)] ’ [6]

= lim

h

h’0

5h + h2

= lim

h

h’0

= lim 5 + h

h’0

= 5.

We see that the required limit (—) exists, and that it equals 5. Thus the function

f (x) = x 2 + x is differentiable at x = 2, and the value of the derivative is 5.

Math Note: When the derivative of a function f exists at a point c, then we denote

the derivative either by f (c) or by (d/dx)f (c) = (df /dx)(c). In some contexts

(e.g., physics) the notation f™(c) is used. In the last example, we calculated that

f (2) = 5.

The importance of the derivative is two-fold: it can be interpreted as rate of

change and it can be interpreted as the slope. Let us now consider both of these

ideas.

Suppose that •(t) represents the position (in inches or feet or some other standard

unit) of a moving body at time t. At time 0 the body is at •(0), at time 3 the body is

at •(3), and so forth. Imagine that we want to determine the instantaneous velocity

of the body at time t = c. What could this mean? One reasonable interpretation

is that we can calculate the average velocity over a small interval at c, and let the

67

CHAPTER 2 Foundations of Calculus

length of that interval shrink to zero to determine the instantaneous velocity. To

carry out this program, imagine a short interval [c, c + h]. The average velocity of

the moving body over that interval is

•(c + h) ’ •(c)

vav ≡ .

h

This is a familiar expression (see (—)).As we let h ’ 0, we know that this expression

tends to the derivative of • at c. On the other hand, it is reasonable to declare this

limit to be the instantaneous velocity. We have discovered the following important

rule:

Let • be a differentiable function on an interval (a, b). Suppose that •(t )

represents the position of a moving body. Let c ∈ (a, b). Then

• (c ) = instantaneous velocity of the moving body at c.

Now let us consider slope. Look at the graph of the function y = f (x) in Fig. 2.6.

We wish to determine the “slope” of the graph at the point x = c. This is the same

as determining the slope of the tangent line to the graph of f at x = c, where the

tangent line is the line that best approximates the graph at that point. See Fig. 2.7.

What could this mean? After all, it takes two points to determine the slope of a line,

yet we are only given the point (c, f (c)) on the graph. One reasonable interpretation

of the slope at (c, f (c)) is that it is the limit of the slopes of secant lines determined

Fig. 2.6

Fig. 2.7

68 CHAPTER 2 Foundations of Calculus

by (c, f (c)) and nearby points (c + h, f (c + h)). See Fig. 2.8. Let us calculate this

limit:

f (c + h) ’ f (c) f (c + h) ’ f (c)

= lim

lim .

(c + h) ’ c h

h’0 h’0

Fig. 2.8

We know that this last limit (the same as (—)) is the derivative of f at c. We have

Y

learned the following:

FL

Let f be a differentiable function on an interval (a, b). Let c ∈ (a, b). Then

the slope of the tangent line to the graph of f at c is f (c ).

AM

EXAMPLE 2.11

Calculate the instantaneous velocity at time t = 5 of an automobile whose

position at time t seconds is given by g(t) = t 3 + 4t 2 + 10 feet.

TE

SOLUTION

We know that the required instantaneous velocity is g (5). We calculate

g(5 + h) ’ g(5)

g (5) = lim

h

h’0

[(5 + h)3 + 4(5 + h)2 + 10] ’ [53 + 4 · 52 + 10]

= lim

h

h’0

((125 + 75h + 15h2 + h3 ) + 4 · (25 + 10h + h2 ) + 10)

= lim

h

h’0

(125 + 100 + 10)

’

h

115h + 19h2 + h3

= lim

h

h’0

= lim 115 + 19h + h2

h’0

= 115.

69

CHAPTER 2 Foundations of Calculus

We conclude that the instantaneous velocity of the moving body at time t = 5

is g (5) = 115 ft/sec.

Math Note: Since position (or distance) is measured in feet, and time in seconds,

then we measure velocity in feet per second.

EXAMPLE 2.12

Calculate the slope of the tangent line to the graph of y = f (x) = x 3 ’ 3x

at x = ’2. Write the equation of the tangent line. Draw a ¬gure illustrating

these ideas.

SOLUTION

We know that the desired slope is equal to f (’2). We calculate

f (’2 + h) ’ f (’2)

f (’2) = lim

h

h’0

[(’2 + h)3 ’ 3(’2 + h)] ’ [(’2)3 ’ 3(’2)]

= lim

h

h’0

[(’8 + 12h ’ 6h2 + h3 ) + (6 ’ 3h)] + [2]

= lim

h

h’0

3 ’ 6h2 + 9h

h

= lim

h

h’0

= lim h2 ’ 6h + 9

h’0

= 9.

We conclude that the slope of the tangent line to the graph of y = x 3 ’ 3x at

x = ’2 is 9. The tangent line passes through (’2, f (’2)) = (’2, ’2) and

has slope 9. Thus it has equation

y ’ (’2) = 9(x ’ (’2)).

The graph of the function and the tangent line are exhibited in Fig. 2.9.

You Try It: Calculate the tangent line to the graph of f (x) = 4x 2 ’ 5x + 2 at the

point where x = 2.

EXAMPLE 2.13

A rubber balloon is losing air steadily. At time t minutes the balloon contains

75 ’ 10t 2 + t cubic inches of air. What is the rate of loss of air in the balloon

at time t = 1?

70 CHAPTER 2 Foundations of Calculus

x

Fig. 2.9

SOLUTION

Let ψ(t) = 75 ’ 10t 2 + t. Of course the rate of loss of air is given by ψ (1).

We therefore calculate

ψ(1 + h) ’ ψ(1)

ψ (1) = lim

h

h’0

[75 ’ 10(1 + h)2 + (1 + h)] ’ [75 ’ 10 · 12 + 1]

= lim

h

h’0

[75 ’ (10 + 20h + 10h2 ) + (1 + h)] ’ [66]

= lim

h

h’0

’19h ’ 10h2

= lim

h

h’0

= lim ’19 ’ 10h

h’0

= ’19.

In conclusion, the rate of air loss in the balloon at time t = 1 is ψ (1) =

’19 ft 3 /sec. Observe that the negative sign in this answer indicates that the

change is negative, i.e., that the quantity is decreasing.

You Try It: The amount of water in a leaky tank is given by W (t) = 50 ’ 5t 2 + t

gallons. What is the rate of leakage of the water at time t = 2?

Math Note: We have noted that the derivative may be used to describe a rate of

change and also to denote the slope of the tangent line to a graph. These are really

two different manifestations of the same thing, for a slope is the rate of change of

rise with respect to run (see Section 1.4 on the slope of a line).

71

CHAPTER 2 Foundations of Calculus

2.5 Rules for Calculating Derivatives

Calculus is a powerful tool, for much of the physical world that we wish to analyze

is best understood in terms of rates of change. It becomes even more powerful when

we can ¬nd some simple rules that enable us to calculate derivatives quickly and

easily. This section is devoted to that topic.

I Derivative of a Sum [The Sum Rule]: We calculate the derivative of a sum

(or difference) by

(f (x) ± g(x)) = f (x) ± g (x).

In our many examples, we have used this fact implicitly. We are now just

enunciating it formally.

II Derivative of a Product [The Product Rule]: We calculate the derivative

of a product by

[f (x) · g(x)] = f (x) · g(x) + f (x) · g (x).

We urge the reader to test this formula on functions that we have worked with

before. It has a surprising form. Note in particular that it is not the case that

[f (x) · g(x)] = f (x) · g (x).

III Derivative of a Quotient [The Quotient Rule]: We calculate the derivative

of a quotient by

g(x) · f (x) ’ f (x) · g (x)

f (x)

= .

g 2 (x)

g(x)

In fact one can derive this new formula by applying the product formula to

g(x) · [f (x)/g(x)]. We leave the details for the interested reader.

IV Derivative of a Composition [The Chain Rule]: We calculate the

derivative of a composition by

[f —¦ g(x)] = f (g(x)) · g (x).

To make optimum use of these four new formulas, we need a library of functions

to which to apply them.

A Derivatives of Powers of x: If f (x) = x k then f (x) = k · x k’1 , where

k ∈ {0, 1, 2, . . .}.

Math Note: If you glance back at the examples we have done, you will notice that

we have already calculated that the derivative of x is 1, the derivative of x 2 is 2x,

72 CHAPTER 2 Foundations of Calculus

and the derivative of x 3 is 3x 2 . The rule just enunciated is a generalization of these

facts, and is established in just the same way.

B Derivatives of Trigonometric Functions: The rules for differentiating sine

and cosine are simple and elegant:

d

sin x = cos x.

1.

dx

d

cos x = ’ sin x.

2.

dx

We can ¬nd the derivatives of the other trigonometric functions by using these two

facts together with the quotient rule from above:

cos x(d/dx) sin x ’ sin x(d/dx) cos x

d sin x

d

3. tan x = =

(cos x)2

dx cos x

dx

(cos x)2 + (sin x)2 1

= = = (sec x)2 .

(cos x)2 (cos x)2

Similarly we have

d

cot x = ’(csc x)2 .

4.

dx

d

sec x = sec x tan x.

5.

dx

d

csc x = ’ csc x cot x.

6.

dx

C Derivatives of ln x and ex : We conclude our library of derivatives of basic

functions with

dx

e = ex

dx

and

1

d

ln x = .

dx x

We may apply the Chain Rule to obtain the following particularly useful general-

ization of this logarithmic derivative:

d • (x)

ln •(x) = .

dx •(x)

Now it is time to learn to differentiate the functions that we will commonly

encounter in our work. We do so by applying the rules for sums, products, quotients,

and compositions to the formulas for the derivatives of the elementary functions.

73

CHAPTER 2 Foundations of Calculus

Practice is the essential tool in mastery of these ideas. Be sure to do all the You Try

It problems in this section.

EXAMPLE 2.14

Calculate the derivative

d

[(sin x + x) · (x 3 ’ ln x)].

dx

SOLUTION

We know that (d/dx) sin x = cos x, (d/dx)x = 1, (d/dx)x 3 = 3x 2 , and

(d/dx) ln x = (1/x). Therefore, by the addition rule,

d d d

(sin x + x) = sin x + x = cos x + 1

dx dx dx

and

1

d3 d3 d

(x ’ ln x) = x’ ln x = 3x 2 ’ .

dx dx dx x

Now we may conclude the calculation by applying the product rule:

d

(sin x + x) · (x 3 ’ ln x)

dx

d d3

= (sin x + x) · (x 3 ’ ln x) + (sin x + x) · (x ’ ln x)

dx dx

1

= (cos x + 1) · (x 3 ’ ln x) + (sin x + x) · 3x 2 ’

x

1

= (4x 3 ’ 1) + x 3 cos x + 3x 2 sin x ’ sin x ’ (ln x cos x + ln x).

x

EXAMPLE 2.15

Calculate the derivative

ex + x sin x

d

.

tan x

dx

SOLUTION

We know that (d/dx)ex = ex , (d/dx)x = 1, (d/dx) sin x = cos x, and

(d/dx) tan x = sec2 x. By the product rule,

d d d

[x · sin x] = x · sin x + x · sin x = 1 · sin x + x · cos x.

dx dx dx

74 CHAPTER 2 Foundations of Calculus

Therefore, by the quotient rule,

ex + x sin x tan x · (d/dx)(ex + x sin x) ’ (ex + x sin x)(d/dx) tan x

d

=

(tan x)2

tan x

dx

tan x · (ex + sin x + x cos x) ’ (ex + x sin x) · (sec x)2

=

(tan x)2

ex tan x + tan x sin x + x sin x ’ ex sec2 x ’ x sin x sec2 x

= .

tan2 x

d x

sin x · cos x ’

You Try It: Calculate the derivative .

ex + ln x

dx

EXAMPLE 2.16

Calculate the derivative

d

(sin(x 3 ’ x 2 )).

dx

SOLUTION

This is the composition of functions, so we must apply the Chain Rule. It is

essential to recognize what function will play the role of f and what function

will play the role of g.

Notice that, if x is the variable, then x 3 ’ x 2 is applied ¬rst and sin

applied next. So it must be that g(x) = x 3 ’ x 2 and f (s) = sin s. Notice that

(d/ds)f (s) = cos s and (d/dx)g(x) = 3x 2 ’ 2x. Then

sin(x 3 ’ x 2 ) = f —¦ g(x)

and

d d

(sin(x 3 ’ x 2 )) = (f —¦ g(x))

dx dx

df d

= (g(x)) · g(x)

ds dx

= cos(g(x)) · (3x 2 ’ 2x)

= [cos(x 3 ’ x 2 )] · (3x 2 ’ 2x).

EXAMPLE 2.17

Calculate the derivative

x2

d

ln .

x’2

dx

75

CHAPTER 2 Foundations of Calculus

SOLUTION

Let

x2

h(x) = ln .

x’2

Then

h = f —¦ g,

where f (s) = ln s and g(x) = x 2 /(x ’ 2). So (d/ds)f (s) = 1/s and

(d/dx)g(x) = (x ’ 2) · 2x ’ x 2 · 1/(x ’ 2)2 = (x 2 ’ 4x)/(x ’ 2)2 . As a

result,

d d

h(x) = (f —¦ g)(x)

dx dx

df d

= (g(x)) · g(x)

ds dx

x 2 ’ 4x

1

= ·

g(x) (x ’ 2)2

x 2 ’ 4x

1

=2 ·

x /(x ’ 2) (x ’ 2)2

x’4

= .

x(x ’ 2)

You Try It: Perform the differentiation in the last example by ¬rst applying a rule

of logarithms to simplify the function to be differentiated.

You Try It: Calculate the derivative of tan(ex ’ x).

EXAMPLE 2.18

2

Calculate the tangent line to the graph of f (x) = x · ex at the point (1, e).

SOLUTION

The slope of the tangent line will be the derivative. Now

2 2 2 2

f (x) = [x] · ex + x · ex = ex + x · 2x · ex .

In the last derivative we have of course used the Chain Rule. Thus f (1) =

e + 2e = 3e. Therefore the equation of the tangent line is

(y ’ e) = 3e(x ’ 1).

You Try It: Calculate the equation of the tangent line to the graph of g(x) =

cos((x 2 ’ 2)/ln x) at the point (2, cos[2/ ln 2]).

76 CHAPTER 2 Foundations of Calculus

Math Note: Calculate (d/dx)(x 2 /x) using the quotient rule. Of course x 2 /x = x,

and you may calculate the derivative directly. Observe that the two answers are

the same. The calculation con¬rms the validity of the quotient rule by way of an

example. Use a similar example to con¬rm the validity of the product rule.

2.5.1 THE DERIVATIVE OF AN INVERSE

An important formula in the calculus relates the derivative of the inverse of a

function to the derivative of the function itself. The formula is

1

[f ’1 ] (t) = ()

.

f (f ’1 (t))

We encourage you to apply the Chain Rule to the formula f (f ’1 (x)) = x to obtain

a formal derivation of the formula ( ).

EXAMPLE 2.19

Calculate the derivative of g(t) = t 1/3 .

SOLUTION

Set f (s) = s 3 and apply formula ( ). Then f (s) = 3s 2 and f ’1 (t) = t 1/3 .

With s = f ’1 (t) we then have

1 1 1 1

[f ’1 ] (t) = = · t ’2/3 .

= =

f (f ’1 (t)) 3 · [t 1/3 ]2

3s 2 3

Formula ( ) may be applied to obtain some interesting new derivatives to add

to our library. We record some of them here:

1

d

arcsin x = √

I.

1 ’ x2

dx

1

d

arccos x = ’ √

II.

1 ’ x2

dx

1

d

arctan x =

III.

1 + x2

dx

√

You Try√ Calculate the derivative of f (x) =

It: x. Calculate the derivative of

g(x) = x for any k ∈ {2, 3, 4, . . . }.

k

2.6 The Derivative as a Rate of Change

If f (t) represents the position of a moving body, or the amount of a changing quan-

tity, at time t, then the derivative f (t) (equivalently, (d/dt)f (t)) denotes the rate

of change of position (also called velocity) or the rate of change of the quantity.

77

CHAPTER 2 Foundations of Calculus

When f (t) represents velocity, then sometimes we calculate another derivative”

(f ) (t)”and this quantity denotes the rate of change of velocity, or acceleration.

In specialized applications, even more derivatives are sometimes used. For

example, sometimes the derivative of the acceleration is called jerk and sometimes

the derivative of jerk is called surge.

EXAMPLE 2.20

The position of a body moving along a linear track is given by p(t) = 3t 2 ’

5t + 7 feet. Calculate the velocity and the acceleration at time t = 3 seconds.

SOLUTION

The velocity is given by

p (t) = 6t ’ 5.

At time t = 3 we therefore ¬nd that the velocity is p (3) = 18 ’ 5 = 13 ft/sec.

The acceleration is given by the second derivative:

p (t) = (p ) (t) = (6t ’ 5) = 6.

The acceleration at time t = 3 is therefore 6 ft/sec2 .

Math Note: As previously noted, velocity is measured in feet per second (or

ft/sec). Acceleration is the rate of change of velocity with respect to time; therefore

acceleration is measured in “feet per second per second” (or ft/sec2 ).

EXAMPLE 2.21

A massive ball is dropped from a tower. It is known that a falling body

descends (near the surface of the earth) with an acceleration of about 32

ft /sec. From this information one can determine that the equation for the

position of the ball at time t is

p(t) = ’16t 2 + v0 t + h0 ft.

Here v0 is the initial velocity and h0 is the initial height of the ball in feet.1

Also t is time measured in seconds. If the ball hits the earth after 5 seconds,

then determine the height from which the ball is dropped.

SOLUTION

Observe that the velocity is

v(t) = p (t) = ’32t + v0 .

Obviously the initial velocity of a falling body is 0. Thus

0 = v(0) = ’32 · 0 + v0 .

1 We shall say more about this equation, and this technique, in Section 3.4.

78 CHAPTER 2 Foundations of Calculus

It follows that v0 = 0, thus con¬rming our intuition that the initial velocity

is 0. Thus

p(t) = ’16t 2 + h0 .

Now we also know that p(5) = 0; that is, at time t = 5 the ball is at height 0.

Thus

0 = p(5) = ’16 · 52 + h0 .

We may solve this equation for h0 to determine that h0 = 400.

We conclude that

p(t) = ’16t 2 + 400.

Furthermore, p(0) = 400, so the initial height of the ball is 400 feet.

You Try It: Suppose that a massive ball falls from a height of 600 feet. After how

many seconds will it strike the ground?

Y

FL

Exercises

AM

1. Calculate, if possible, each of these limits. Give reasons for each step of

your solution.

lim x · ex

(a)

TE

x’0

x2 ’ 1

(b) lim

x’1 x ’ 1

lim (x ’ 2) · cot(x ’ 2)

(c)

x’2

lim x · ln x

(d)

x’0

t 2 ’ 7t + 12

(e) lim

t ’3

t’3

s 2 ’ 3s ’ 4

(f) lim

s’4

s’4

ln x

(g) lim

x’1 x ’ 1

x2 ’ 9

(h) lim

x’’3 x + 3

79

CHAPTER 2 Foundations of Calculus

2. Determine whether the given function f is continuous at the given point c.

Give careful justi¬cations for your answers.

x’1

f (x) = c = ’1

(a)

x+1

x’1

f (x) = c=3

(b)

x+1

f (x) = x · sin(1/x) c=0

(c)

f (x) = x · ln x c=0

(d)

if x ¤ 1

x2

f (x) = c=1

(e)

if 1 < x

x

if x ¤ 1

x2

f (x) = c=1

(f)

2x if 1 < x

if x ¤ π

sin x

f (x) = c=π

(g)

(x ’ π ) if π < x

f (x) = eln x+x c=2

(h)

3. Use the de¬nition of derivative to calculate each of these derivatives.

f (2) when f (x) = x 2 + 4x

(a)

f (1) when f (x) = ’1/x 2

(b)

4. Calculate each of these derivatives. Justify each step of your calculation.

x

(a)

x2 + 1

d

sin(x 2 )

(b)

dx

d

t · tan(t 3 ’ t 2 )

(c)

dt

d x2 ’ 1

(d)

dx x 2 + 1

[x · ln(sin x)]

(e)

d s(s+2)

(f) e

ds

80 CHAPTER 2 Foundations of Calculus

d sin(x 2 )

(g) e

dx

[ln(ex + x)]

(h)

5. Imitate the example in the text to do each of these falling body problems.

(a) A ball is dropped from a height of 100 feet. How long will it take

that ball to hit the ground?

(b) Suppose that the ball from part (a) is thrown straight down with an

initial velocity of 10 feet per second. Then how long will it take the

ball to hit the ground?

(c) Suppose that the ball from part (a) is thrown straight up with an

initial velocity of 10 feet per second. Then how long will it take the

ball to hit the ground?

6. Use the Chain Rule to perform each of these differentiations:

d

(a) sin(ln(cos x))

dx

d sin(cos x)

(b) e

dx

d

ln(esin x + x)

(c)

dx

d

arcsin(x 2 + tan x)

(d)

dx

d

arccos(ln x ’ ex /5)

(e)

dx

d

arctan(x 2 + ex )

(f)

dx

7. If a car has position p(t) = 6t 2 ’ 5t + 20 feet, where t is measured in

seconds, then what is the velocity of that car at time t = 4? What is the

average velocity of that car from t = 2 to t = 8? What is the greatest

velocity over the time interval [5, 10]?

8. In each of these problems, use the formula for the derivative of an inverse

function to ¬nd [f ’1 ] (1).

f (0) = 1, f (0) = 3

(a)

f (3) = 1, f (3) = 8

(b)

f (2) = 1, f (2) = π 2

(c)

f (1) = 1, f (1) = 40

(d)

CHAPTER 3

Applications of

the Derivative

3.1 Graphing of Functions

We know that the value of the derivative of a function f at a point x represents the

slope of the tangent line to the graph of f at the point (x, f (x)). If that slope is

positive, then the tangent line rises as x increases from left to right, hence so does

the curve (we say that the function is increasing). If instead the slope of the tangent

line is negative, then the tangent line falls as x increases from left to right, hence

so does the curve (we say that the function is decreasing). We summarize:

On an interval where f > 0 the graph of f goes uphill.

On an interval where f < 0 the graph of f goes downhill.

See Fig. 3.1.

With some additional thought, we can also get useful information from the second

derivative. If f = (f ) > 0 at a point, then f is increasing. Hence the slope of the

tangent line is getting ever greater (the graph is concave up). The picture must be as

in Fig. 3.2(a) or 3.2(b). If instead f = (f ) < 0 at a point then f is decreasing.

Hence the slope of the tangent line is getting ever less (the graph is concave down).

The picture must be as in Fig. 3.3(a) or 3.3(b).

Using information about the ¬rst and second derivatives, we can render rather

accurate graphs of functions. We now illustrate with some examples.

EXAMPLE 3.1

Sketch the graph of f (x) = x 2 .

81

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

82 CHAPTER 3 Applications of the Derivative

Fig. 3.1

Fig. 3.2

Fig. 3.3

SOLUTION

Of course this is a simple and familiar function, and you know that its graph

is a parabola. But it is satisfying to see calculus con¬rm the shape of the graph.

Let us see how this works.

83

CHAPTER 3 Applications of the Derivative

First observe that f (x) = 2x. We see that f < 0 when x < 0 and f > 0

when x > 0. So the graph is decreasing on the negative real axis and the graph

is increasing on the positive real axis.

Next observe that f (x) = 2. Thus f > 0 at all points. Thus the graph is

concave up everywhere.

Finally note that the graph passes through the origin.

We summarize this information in the graph shown in Fig. 3.4.

Fig. 3.4

EXAMPLE 3.2

Sketch the graph of f (x) = x 3 .

SOLUTION

First observe that f (x) = 3x 2 . Thus f ≥ 0 everywhere. The function is

always increasing.

Second observe that f (x) = 6x. Thus f (x) < 0 when x < 0 and

f (x) > 0 when x > 0. So the graph is concave down on the negative real

axis and concave up on the positive real axis.

Finally note that the graph passes through the origin.

We summarize our ¬ndings in the graph shown in Fig. 3.5.

You Try It: Use calculus to aid you in sketching the graph of f (x) = x 3 + x.

EXAMPLE 3.3

Sketch the graph of g(x) = x + sin x.

SOLUTION

We see that g (x) = 1 + cos x. Since ’1 ¤ cos x ¤ 1, it follows that

g (x) ≥ 0. Hence the graph of g is always increasing.

Now g (x) = ’ sin x. This function is positive sometimes and negative

sometimes. In fact

’ sin x is positive when kπ < x < (k + 1)π, k odd

84 CHAPTER 3 Applications of the Derivative

Fig. 3.5

and

’ sin x is negative when kπ < x < (k + 1)π, k even.

So the graph alternates being concave down and concave up. Of course it also

passes through the origin. We amalgamate all our information in the graph

shown in Fig. 3.6.

6

4

2

_F F 2F

_2

Fig. 3.6

EXAMPLE 3.4

Sketch the graph of h(x) = x/(x + 1 ).

85

CHAPTER 3 Applications of the Derivative

SOLUTION

First note that the function is unde¬ned at x = ’1.

We calculate that h (x) = 1/((x + 1)2 ). Thus the graph is everywhere

increasing (except at x = ’1).

We also calculate that h (x) = ’2/((x + 1)3 ). Hence h > 0 and the graph

is concave up when x < ’1. Likewise h < 0 and the graph is concave down

when x > ’1.

Finally, as x tends to ’1 from the left we notice that h tends to +∞ and as

x tends to ’1 from the right we see that h tends to ’∞.

Putting all this information together, we obtain the graph shown in Fig. 3.7.

Fig. 3.7

√

You Try It: Sketch the graph of the function k(x) = x · x + 1.

EXAMPLE 3.5

Sketch the graph of k(x) = x 3 + 3x 2 ’ 9x + 6.

SOLUTION

We notice that k (x) = 3x 2 + 6x ’ 9 = 3(x ’ 1)(x + 3). So the sign of k

changes at x = 1 and x = ’3. We conclude that

k is positive when x < ’3;

k is negative when ’3 < x < 1;

k is positive when x > 3.

Finally, k (x) = 6x + 6. Thus the graph is concave down when x < ’1 and

the graph is concave up when x > ’1.

Putting all this information together, and using the facts that k(x) ’ ’∞

when x ’ ’∞ and k(x) ’ +∞ when x ’ +∞, we obtain the graph shown

in Fig. 3.8.

86 CHAPTER 3 Applications of the Derivative

Fig. 3.8

3.2 Maximum/Minimum Problems

One of the great classical applications of the calculus is to determine the maxima

and minima of functions. Look at Fig. 3.9. It shows some (local) maxima and (local)

minima of the function f .

Fig. 3.9

Notice that a maximum has the characterizing property that it looks like a hump:

the function is increasing to the left of the hump and decreasing to the right of the

hump. The derivative at the hump is 0: the function neither increases nor decreases

87

CHAPTER 3 Applications of the Derivative

at a local maximum. This is sometimes called Fermat™s test. Also, we see that the

graph is concave down at a local maximum.

It is common to refer to the points where the derivative vanishes as critical points.

In some contexts, we will designate the endpoints of the domain of our function to

be critical points as well.

Now look at a local minimum. Notice that a minimum has the characterizing

property that it looks like a valley: the function is decreasing to the left of the valley

and increasing to the right of the valley. The derivative at the valley is 0: the function

neither increases nor decreases at a local minimum. This is another manifestation

of Fermat™s test. Also, we see that the graph is concave up at a local minimum.

Let us now apply these mathematical ideas to some concrete examples.

EXAMPLE 3.6

Find all local maxima and minima of the function k(x) = x 3 ’ 3x 2 ’ 24x + 5.

SOLUTION

We begin by calculating the ¬rst derivative:

k (x) = 3x 2 ’ 6x ’ 24 = 3(x + 2)(x ’ 4).

We notice that k vanishes only when x = ’2 or x = 4. These are the only

candidates for local maxima or minima. The second derivative is k (x) =

6x ’ 6. Now k (4) = 18 > 0, so x = 4 is the location of a local minimum.

Also k (’2) = ’18 < 0, so x = ’2 is the location of a local maximum.

A glance at the graph of this function, as depicted in Fig. 3.10, con¬rms our

calculations.

EXAMPLE 3.7

Find all local maxima and minima of the function g(x) = x + sin x.

SOLUTION

First we calculate that

g (x) = 1 + cos x.

Thus g vanishes at the points (2k + 1)π for k = . . . , ’2, ’1, 0, 1, 2, . . ..

Now g (x) = sin x. And g ((2k + 1)π ) = 0. Thus the second derivative test

is inconclusive. Let us instead look at the ¬rst derivative. We notice that it is

always ≥ 0. But, as we have already noticed, the ¬rst derivative changes sign at

a local maximum or minimum. We conclude that none of the points (2k + 1)π

is either a maximum nor a minimum. The graph in Fig. 3.11 con¬rms this

calculation.

You Try It: Find all local maxima and minima of the function g(x) = 2x 3 ’

15x 2 + 24x + 6.

88 CHAPTER 3 Applications of the Derivative

Y

FL

AM

Fig. 3.10

TE

6

4

2

_F F 2F

_2

Fig. 3.11

EXAMPLE 3.8

A box is to be made from a sheet of cardboard that measures 12 — 12 .

The construction will be achieved by cutting a square from each corner of

89

CHAPTER 3 Applications of the Derivative

the sheet and then folding up the sides (see Fig. 3.12). What is the box of

greatest volume that can be constructed in this fashion?

Fig. 3.12

SOLUTION

It is important in a problem of this kind to introduce a variable. Let x be the

side length of the squares that are to be cut from the sheet of cardboard. Then

the side length of the resulting box will be 12 ’ 2x (see Fig. 3.13). Also the

height of the box will be x. As a result, the volume of the box will be

V (x) = x · (12 ’ 2x) · (12 ’ 2x) = 144x ’ 48x 2 + 4x 3 .

Our job is to maximize this function V .

x

12 _ 2x

Fig. 3.13

Now V (x) = 144 ’ 96x + 12x 2 . We may solve the quadratic equation

144 ’ 96x + 12x 2 = 0

to ¬nd the critical points for this problem. Using the quadratic formula, we

¬nd that x = 2 and x = 6 are the critical points. Now V (x) = ’96 + 24x.

Since V (2) = ’48 < 0, we conclude that x = 2 is a local maximum for the

problem. In fact we can sketch a graph of V (x) using ideas from calculus and

see that x = 2 is a global maximum.

We conclude that if squares of side 2 are cut from the sheet of cardboard

then a box of maximum volume will result.

Observe in passing that if squares of side 6 are cut from the sheet then (there

will be no cardboard left!) the resulting box will have zero volume. This value

for x gives a minimum for the problem.

90 CHAPTER 3 Applications of the Derivative

EXAMPLE 3.9

A rectangular garden is to be constructed against the side of a garage. The

gardener has 100 feet of fencing, and will construct a three-sided fence;

the side of the garage will form the fourth side. What dimensions will give

the garden of greatest area?

SOLUTION

Look at Fig. 3.14. Let x denote the side of the garden that is perpendicular

to the side of the garage. Then the resulting garden has width x feet and length

100 ’ 2x feet. The area of the garden is

A(x) = x · (100 ’ 2x) = 100x ’ 2x 2 .

x

garage

100 _ 2x

x

Fig. 3.14

We calculate A (x) = 100 ’ 4x and ¬nd that the only critical point for the

problem is x = 25. Since A (x) = ’4 for all x, we determine that x = 25 is

a local maximum. By inspection, we see that the graph of A is a downward-

opening parabola. So x = 25 must also be the global maximum that we seek.

The optimal dimensions for the garden are

width = 25 ft. length = 50 ft.

You Try It: Find the right circular cylinder of greatest volume that can be

contained in a sphere of radius 1.

EXAMPLE 3.10

The sum of two positive numbers is 60. How can we choose them so as to

maximize their product?

SOLUTION

Let x be one of the two numbers. Then the other is 60 ’ x. Their product is

P (x) = x · (60 ’ x) = 60x ’ x 2 .

91

CHAPTER 3 Applications of the Derivative

Thus P is the quantity that we wish to maximize. Calculating the derivative,

we ¬nd that

P (x) = 60 ’ 2x.

Thus the only critical point for the problem is x = 30. Since P (x) ≡ ’2, we

¬nd that x = 30 is a local maximum. Since the graph of P is a downward-

opening parabola, we can in fact be sure that x = 30 is a global maximum.

We conclude that the two numbers that add to 60 and maximize the product

are 30 and 30.

You Try It: A rectangular box is to be placed in the ¬rst quadrant {(x, y) : x ≥

0, y ≥ 0} in such a way that one side lies on the positive x-axis and one side lies

on the positive y-axis. The box is to lie below the line y = ’2x + 5. Give the

dimensions of such a box having greatest possible area.

3.3 Related Rates

If a tree is growing in a forest, then both its height and its radius will be increasing.

These two growths will depend in turn on (i) the amount of sunlight that hits the

tree, (ii) the amount of nutrients in the soil, (iii) the proximity of other trees. We

may wish to study the relationship among these various parameters. For example,

if we know that the amount of sunlight and nutrients are increasing at a certain rate

then we may wish to know how that affects the rate of change of the radius. This

consideration gives rise to related rates problems.

EXAMPLE 3.11

A toy balloon is in the shape of a sphere. It is being in¬‚ated at the rate

of 20 in.3 /min. At the moment that the sphere has volume 64 cubic inches,

what is the rate of change of the radius?

SOLUTION

We know that volume and radius of a sphere are related by the formula

4π 3

V= (—)

r.

3

The free variable in this problem is time, so we differentiate equation (—) with

respect to time t. It is important that we keep the chain rule in mind as we do

so.1 The result is

4π

dV dr

= · 3r 2 · . (——)

3

dt dt

1 The point is that we are not differentiating with respect to r.

92 CHAPTER 3 Applications of the Derivative

Now we are given that dV /dt = 20. Our question is posed at the moment that

√

V = 64. But, according to (—), this means that r = 3 48/π. Substituting these

values into equation (——) yields

2

4π dr

20 = ·3 ·

3

48/π .

3 dt

Solving for dr/dt yields

5

dr

= 2/3 1/3 .

48 · π

dt

EXAMPLE 3.12

A 13-foot ladder leans against a wall (Fig. 3.15).The foot of the ladder begins

to slide away from the wall at the rate of 1 foot per minute. When the foot is

5 feet from the wall, at what rate is the top of the ladder falling?

Fig. 3.15

SOLUTION

Let h(t) be the height of the ladder at time t and b(t) be the distance of the base

of the ladder to the wall at time t. Then the Pythagorean theorem tells us that

h(t)2 + b(t)2 = 132 .

We may differentiate both sides of this equation with respect to the variable t

(which is time in minutes) to obtain

2 · h(t) · h (t) + 2 · b(t) · b (t) = 0.

Solving for h (t) yields

b(t) · b (t)

h (t) = ’ .

h(t)

93

CHAPTER 3 Applications of the Derivative

At the instant the problem is posed, b(t) = 5, h(t) = 12 (by the Pythagorean

theorem), and b (t) = 1. Substituting these values into the equation yields

5·1 5

h (t) = ’ = ’ ft/min.

12 12

Observe that the answer is negative, which is appropriate since the top of the

ladder is falling.

You Try It: Suppose that a square sheet of aluminum is placed in the hot sun.

It begins to expand very slowly so that its diagonal is increasing at the rate of

1 millimeter per minute. At the moment that the diagonal is 100 millimeters, at

what rate is the area increasing?

EXAMPLE 3.13

A sponge is in the shape of a right circular cone (Fig. 3.16). As it soaks up

water, it grows in size. At a certain moment, the height equals 6 inches, and

is increasing at the rate of 0.3 inches per second. At that same moment, the

radius is 4 inches, and is increasing at the rate of 0.2 inches per second.

How is the volume changing at that time?

Fig. 3.16

SOLUTION

We know that the volume V of a right circular cone is related to the height

h and the radius r by the formula

1

V = π r 2 h.

3

Differentiating both sides with respect to the variable t (for time in seconds)

yields

1

dV dr dh

= π 2r h + r 2 .

3

dt dt dt

94 CHAPTER 3 Applications of the Derivative

Substituting the values for r, dr/dt, h, and dh/dt into the right-hand side yields

1 1 24π

dV

= π 2 · 4 · (0.2) · 6 + 42 · (0.3) = π [9.6 + 4.8] = .

3 3 5

dt

You Try It: In the heat of the sun, a sheet of aluminum in the shape of an equilateral

triangle is expanding so that its side length increases by 1 millimeter per hour. When

the side length is 100 millimeters, how is the area increasing?

3.4 Falling Bodies

It is known that, near the surface of the earth, a body falls with acceleration (due

to gravity) of about 32 ft/sec2 . If we let h(t) be the height of the object at time t

(measured in seconds), then our information is that

h (t) = ’32.

Observe the minus sign to indicate that height is decreasing.

Now we will do some organized guessing. What could h be? It is some function

whose derivative is the constant ’32. Our experience indicates that polynomials

decrease in degree when we differentiate them. That is, the degree goes from 5 to 4,

or from 3 to 2. Since, h is a polynomial of degree 0, we therefore determine

that h will be a polynomial of degree 1. A moment™s thought then suggests that

h (t) = ’32t. This works! If h (t) = ’32t then h (t) = [h (t)] = ’32. In fact

we can do a bit better. Since constants differentiate to zero, our best guess of what

the velocity should be is h (t) = ’32t + v0 , where v0 is an undetermined constant.

Now let us guess what form h(t) should have. We can learn from our experience

in the last paragraph. The “antiderivative” of ’32t (a polynomial of degree 1)

should be a polynomial of degree 2. After a little ¬ddling, we guess ’16t 2 . And this

works. The antiderivative of v0 (a polynomial of degree 0) should be a polynomial

of degree 1. After a little ¬ddling, we guess v0 t. And this works. Taking all this

information together, we ¬nd that the “antiderivative” of h (t) = ’32t + v0 is

h(t) = ’16t 2 + v0 t + h0 . (†)

Notice that we have once again thrown in an additive constant h0 . This does no

harm:

h (t) = [’16t 2 ] + [v0 t] + [h0 ] = ’32t + v0 ,

just as we wish. And, to repeat what we have already con¬rmed,

h (t) = [h (t)] = [’32t] + [v0 ] = ’32.

95

CHAPTER 3 Applications of the Derivative

We now have a general formula (namely (†)) for the position of a falling body at

time t. [Recall that we were ¬rst introduced to this formula in Section 2.6.] See

Fig. 3.17.

Fig. 3.17

Before doing some examples, we observe that a falling body will have initial

velocity 0. Thus

0 = h (0) = ’32 · 0 + v0 .

Hence, for a falling body, v0 = 0. In some problems we may give the body an

initial push, and then v0 will not be zero.

EXAMPLE 3.14

Suppose that a falling body hits the ground with velocity ’100 ft/sec. What

was the initial height of the body?

SOLUTION

With notation as developed above, we know that velocity is given by

h (t) = ’32t + 0.

We have taken v0 to be 0 because the body is a falling body; it had no initial

push. If T is the time at which the body hits the ground, then we know that

’100 = h (T ) = ’32 · T .

As a result, T = 25/8 sec.

96 CHAPTER 3 Applications of the Derivative

When the body hits the ground, its height is 0. Thus we know that

0 = h(T ) = h(25/8) = ’16 · (25/8)2 + h0 .

We may solve for h0 to obtain

625

h0 = .

4

Thus all the information about our falling body is given by

625

h(t) = ’16t 2 + .

4

At time t = 0 we have

625

h(0) = .

4

Thus the initial height of the falling body is 625/4 ft = 156.25 ft.

Notice that, in the process of solving the last example, and in the discussion

preceding it, we learned that h0 represents the initial height of the falling body and

v0 represents the initial velocity of the falling body. This information will be useful

in the other examples that we examine.

EXAMPLE 3.15

A body is thrown straight down with an initial velocity of 10 feet per second.

It strikes the ground in 12 seconds. What was the initial height?

SOLUTION

We know that v0 = ’10 and that h(12) = 0. This is the information that we

must exploit in solving the problem. Now

h(t) = ’16t 2 ’ 10t + h0 .

Thus

0 = h(12) = ’16 · 122 ’ 10 · 12 + h0 .

We may solve for h0 to obtain

h0 = 2424 ft.

The initial height is 2424 feet.

You Try It: A body is thrown straight up with initial velocity 5 feet per second

from a height of 40 feet. After how many seconds will it hit the ground? What will

be its maximum height?

97

CHAPTER 3 Applications of the Derivative

EXAMPLE 3.16

A body is launched straight up from height 100 feet with some initial velocity.

It hits the ground after 10 seconds. What was that initial velocity?

SOLUTION

We are given that h0 = 100. Thus

h(t) = ’16t 2 + v0 t + 100.

Our job is to ¬nd v0 . We also know that

0 = h(10) = ’16 · 102 + v0 · 10 + 100.

We solve this equation to ¬nd that v0 = 150 ft /sec.

You Try It: On a certain planet, bodies fall with an acceleration due to gravity

of 10 ft/sec2 . A certain body is thrown down with an initial velocity of 5 feet per

second, and hits the surface 12 seconds later. From what height was it launched?

Exercises

1. Sketch the graph of f (x) = x/[x 2 + 3], indicating all local maxima and

minima together with concavity properties.

2. What is the right circular cylinder of greatest volume that can be inscribed

upright in a right circular cone of radius 3 and height 6?

3. An air mattress (in the shape of a rectangular parallelepiped) is being in¬‚ated

in such a way that, at a given moment, its length is increasing by 1 inch

per minute, its width is decreasing by 0.5 inches per minute, and its height

is increasing by 0.3 inches per minute. At that moment its dimensions are

= 100 , w = 60 , and h = 15 . How is its volume changing at that time?

4. A certain body is thrown straight down at an initial velocity of 15 ft /sec. It

strikes the ground in 5 seconds. What is its initial height?

5. Because of viral infection, the shape of a certain cone-shaped cell is

changing. The height is increasing at the rate of 3 microns per minute.

For metabolic reasons, the volume remains constantly equal to 20 cubic

microns. At the moment that the radius is 5 microns, what is the rate of

change of the radius of the cell?

6. A silo is to hold 10,000 cubic feet of grain. The silo will be cylindrical

in shape and have a ¬‚at top. The ¬‚oor of the silo will be the earth. What

dimensions of the silo will use the least material for construction?

7. Sketch the graph of the function g(x) = x·sin x. Show maxima and minima.

98 CHAPTER 3 Applications of the Derivative

8. A body is launched straight down at a velocity of 5 ft /sec from height

400 feet. How long will it take this body to reach the ground?

9. Sketch the graph of the function h(x) = x/(x 2 ’ 1). Exhibit maxima,

minima, and concavity.

10. A punctured balloon, in the shape of a sphere, is losing air at the rate of

2 in.3 /sec. At the moment that the balloon has volume 36π cubic inches,

how is the radius changing?

11. A ten-pound stone and a twenty-pound stone are each dropped from height

100 feet at the same moment. Which will strike the ground ¬rst?

12. A man wants to determine how far below the surface of the earth is the

water in a well. How can he use the theory of falling bodies to do so?

13. A rectangle is to be placed in the ¬rst quadrant, with one side on the x-axis

and one side on the y-axis, so that the rectangle lies below the line 3x +5y =

15. What dimensions of the rectangle will give greatest area?

14. A rectangular box with square base is to be constructed to hold 100 cubic

inches. The material for the base and the top costs 10 cents per square

Y

inch and the material for the sides costs 20 cents per square inch. What

FL

dimensions will give the most economical box?

15. Sketch the graph of the function f (x) = [x 2 ’1]/[x 2 +1]. Exhibit maxima,

minima, and concavity.

AM

16. On the planet Zork, the acceleration due to gravity of a falling body near the

surface of the planet is 20 ft /sec. A body is dropped from height 100 feet.

How long will it take that body to hit the surface of Zork?

TE

CHAPTER 4

The Integral

4.0 Introduction

Many processes, both in mathematics and in nature, involve addition. You are famil-

iar with the discrete process of addition, in which you add ¬nitely many numbers

to obtain a sum or aggregate. But there are important instances in which we wish

to add in¬nitely many terms. One important example is in the calculation of area”

especially the area of an unusual (non-rectilinear) shape. A standard strategy is to

approximate the desired area by the sum of small, thin rectangular regions (whose

areas are easy to calculate). A second example is the calculation of work, in which

we think of the work performed over an interval or curve as the aggregate of small

increments of work performed over very short intervals. We need a mathematical

formalism for making such summation processes natural and comfortable. Thus we

will develop the concept of the integral.

4.1 Antiderivatives and Indefinite

Integrals

4.1.1 THE CONCEPT OF ANTIDERIVATIVE

Let f be a given function. We have already seen in the theory of falling bodies

(Section 3.4) that it can be useful to ¬nd a function F such that F = f . We call

99

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

100 CHAPTER 4 The Integral

such a function F an antiderivative of f . In fact we often want to ¬nd the most

general function F , or a family of functions, whose derivative equals f . We can

sometimes achieve this goal by a process of organized guessing.

Suppose that f (x) = cos x. If we want to guess an antiderivative, then we

are certainly not going to try a polynomial. For if we differentiate a polynomial

then we get another polynomial. So that will not do the job. For similar reasons

we are not going to guess a logarithm or an exponential. In fact the way that we

get a trigonometric function through differentiation is by differentiating another

trigonometric function. What trigonometric function, when differentiated, gives

cos x? There are only six functions to try, and a moment™s thought reveals that

F (x) = sin x does the trick. In fact an even better answer is F (x) = sin x + C. The

constant differentiates to 0, so F (x) = f (x) = cos x. We have seen in our study

of falling bodies that the additive constant gives us a certain amount of ¬‚exibility

in solving problems.

Now suppose that f (x) = x 2 . We have already noted that the way to get a

polynomial through differentiation is to differentiate another polynomial. Since

differentiation reduces the degree of the polynomial by 1, it is natural to guess that

the F we seek is a polynomial of degree 3. What about F (x) = x 3 ? We calculate

that F (x) = 3x 2 . That does not quite work. We seek x 2 for our derivative, but

we got 3x 2 . This result suggests adjusting our guess. We instead try F (x) = x 3 /3.

Then, indeed, F (x) = 3x 2 /3 = x 2 , as desired. We will write F (x) = x 3 /3 + C

for our antiderivative.

More generally, suppose that f (x) = ax 3 + bx 2 + cx + d. Using the reasoning

in the last paragraph, we may ¬nd fairly easily that F (x) = ax 4 /4 + bx 3 /3 +

cx 2 /2 + dx + e. Notice that, once again, we have thrown in an additive constant.

You Try It: Find a family of antiderivatives for the function f (x) = sin 2x ’

x 4 + ex .

4.1.2 THE INDEFINITE INTEGRAL

In practice, it is useful to have a compact notation for the antiderivative. What we

do, instead of saying that “the antiderivative of f (x) is F (x) + C,” is to write

f (x) dx = F (x) + C.

So, for example,

cos x dx = sin x + C

and

x4 x2

x + x dx = + +C

3

4 2

101

CHAPTER 4 The Integral

and

e2x

dx = + C.

2x

e

2

The symbol is called an integral sign (the symbol is in fact an elongated “S”)

and the symbol “dx” plays a traditional role to remind us what the variable is. We

call an expression like

f (x) dx

an inde¬nite integral. The name comes from the fact that later on we will have a

notion of “de¬nite integral” that speci¬es what value C will take”so it is more

de¬nite in the answer that it gives.

EXAMPLE 4.1

Calculate

sin(3x + 1) dx.

SOLUTION

We know that we must guess a trigonometric function. Running through

the choices, cosine seems like the best candidate. The derivative of cos x is

’ sin x. So we immediately see that ’ cos x is a better guess”its derivative

is sin x. But then we adjust our guess to F (x) = ’ cos(3x + 1) to take into

account the form of the argument. This almost works: we may calculate that

F (x) = 3 sin(3x + 1). We determine that we must adjust by a factor of 1/3.

Now we can record our ¬nal answer as

1