OVERVIEW In Chapter 2, we defined the slope of a curve at a point as the limit of secant slopes. This limit, called a derivative, measures the rate at which a function changes, and it is one of the most important ideas in calculus. Derivatives are used to

Chapter

3 D IFFERENTIATION

OVERVIEW In Chapter 2, we defined the slope of a curve at a point as the limit of secant slopes. This limit, called a derivative, measures the rate at which a function changes, and it is one of the most important ideas in calculus. Derivatives are used to calculate velocity and acceleration, to estimate the rate of spread of a disease, to set levels of production so as to maximize efficiency, to find the best dimensions of a cylindrical can, to find the age of a prehistoric artifact, and for many other applications. In this chapter, we develop tech- niques to calculate derivatives easily and learn how to use derivatives to approximate com- plicated functions.

3.1 The Derivative as a Function

At the end of Chapter 2, we defined the slope of a curve y= ƒsxd at the point where

x=x 0 to be

H ISTORICAL E SSAY

ƒsx 0 + hd - ƒsx 0 d

lim .

The Derivative

h :0

We called this limit, when it existed, the derivative of ƒ at x 0 . We now investigate the derivative as a function derived from ƒ by considering the limit at each point of the do- main of ƒ.

DEFINITION Derivative Function

The derivative of the function ƒ(x) with respect to the variable x is the function

ƒ¿ whose value at x is ƒsx + hd - ƒsxd

ƒ¿sxd = lim ,

h :0

provided the limit exists.

Chapter 3: Differentiation

We use the notation ƒ(x) rather than simply ƒ in the definition to emphasize the inde- pendent variable x, which we are differentiating with respect to. The domain of ƒ¿ is the set

Secant slope is

f of points in the domain of ƒ for which the limit exists, and the domain may be the same or

Q (z, f (z))

smaller than the domain of ƒ. If ƒ¿ exists at a particular x, we say that ƒ is differentiable (has a derivative) at x. If ƒ¿ exists at every point in the domain of ƒ, we call ƒ differen- tiable .

P (x, f (x))

f If we write z=x+h , then h=z-x and h approaches 0 if and only if z approaches x . Therefore, an equivalent definition of the derivative is as follows (see Figure 3.1).

xz Derivative of f at x is

Alternative Formula for the Derivative

f f'

h →0

ƒszd - ƒsxd

ƒ¿sxd = lim .

z :x

z-x

z lim →x z

FIGURE 3.1 The way we write the difference quotient for the derivative of a

Calculating Derivatives from the Definition

function ƒ depends on how we label the points involved.

The process of calculating a derivative is called differentiation. To emphasize the idea that differentiation is an operation performed on a function y= ƒsxd , we use the notation

d ƒsxd dx

as another way to denote the derivative ƒ¿sxd . Examples 2 and 3 of Section 2.7 illustrate the differentiation process for the functions y = mx + b and y= 1 >x. Example 2 shows

that

d smx + bd = m . dx

For instance,

dx a 2 x- 4 b= 2 .

In Example 3, we see that

d 1 1 dx a xb=- 2 . x

Here are two more examples.

EXAMPLE 1 Applying the Definition Differentiate ƒsxd = x

x- 1 .

Solution

Here we have ƒsxd = x

x- 1

3.1 The Derivative as a Function

and sx + hd

ƒsx + hd =

, so

sx + hd - 1 ƒsx + hd - ƒsxd

ƒ¿sxd = lim

h :0

h x+h - x

x+h- 1 x- 1 = lim

h :0

= lim 1 # sx + hdsx - 1d - xsx + h - 1d a - c = ad - cb

h :0 h sx + h - 1dsx - 1d

b d bd

= lim 1 #

-h

h :0 h sx + h - 1dsx - 1d

= lim -1

h :0 sx + h - 1dsx - 1d

sx - 1d 2

EXAMPLE 2 Derivative of the Square Root Function (a) Find the derivative of y = 1x for x7 0.

(b) Find the tangent line to the curve y = 1x at x= 4.

You will often need to know the Solution derivative of 1x for x7 0:

(a) We use the equivalent form to calculate ƒ¿ :

d 1x = 1 .

ƒszd - ƒsxd

dx

2 1x

ƒ¿sxd = lim

z-x = lim 1z - 1x

z :x

z :x

z-x

= lim 1z - 1x

z :x A 1z - 1x BA 1z + 1x B

= lim 1 = 1 .

z :x 1z + 1x

2 1x

y 1 4 (b) The slope of the curve at x= 4 is ƒ¿s4d = 1 =1 .

The tangent is the line through the point (4, 2) with slope 1 >4 (Figure 3.2):

y=2+ 1 sx - 4d

FIGURE 3.2 The curve y = 1x and its

y= tangent at (4, 2). The tangent’s slope is 1 x+ 1.

found by evaluating the derivative at x= 4 (Example 2).

We consider the derivative of y = 1x when x= 0 in Example 6.

Chapter 3: Differentiation

Notations

There are many ways to denote the derivative of a function y= ƒsxd , where the independ- ent variable is x and the dependent variable is y. Some common alternative notations for the derivative are

dy

ƒ¿sxd = y¿ =

= d ƒsxd = Dsƒdsxd = D

x ƒsxd .

dx

dx

dx

The symbols d >dx and D indicate the operation of differentiation and are called differentiation operators . We read dy >dx as “the derivative of y with respect to x,” and

d ƒ >dx and ( d >dx )ƒ(x) as “the derivative of ƒ with respect to x.” The “prime” notations y¿ and ƒ¿ come from notations that Newton used for derivatives. The d >dx notations are simi- lar to those used by Leibniz. The symbol dy >dx should not be regarded as a ratio (until we introduce the idea of “differentials” in Section 3.8).

Be careful not to confuse the notation D(ƒ) as meaning the domain of the function ƒ instead of the derivative function ƒ¿ . The distinction should be clear from the context. To indicate the value of a derivative at a specified number x=a , we use the notation

df

dy

ƒ¿sad =

= d ƒsxd ` .

dx ` x=a dx ` x=a dx

x=a

For instance, in Example 2b we could write

dx 1x `

ƒ¿s4d = d 1 1

=1 . x= 4 2 1x x= 4 2 24 4

To evaluate an expression, we sometimes use the right bracket ] in place of the vertical bar ƒ.

Graphing the Derivative

We can often make a reasonable plot of the derivative of y= ƒsxd by estimating the slopes on the graph of ƒ. That is, we plot the points sx, ƒ¿sxdd in the xy-plane and connect them with a smooth curve, which represents y= ƒ¿sxd .

EXAMPLE 3 Graphing a Derivative Graph the derivative of the function y= ƒsxd in Figure 3.3a.

Solution

We sketch the tangents to the graph of ƒ at frequent intervals and use their slopes to estimate the values of ƒ¿sxd at these points. We plot the corresponding sx, ƒ¿sxdd pairs and connect them with a smooth curve as sketched in Figure 3.3b.

What can we learn from the graph of y= ƒ¿sxd ? At a glance we can see

1. where the rate of change of ƒ is positive, negative, or zero;

2. the rough size of the growth rate at any x and its size in relation to the size of ƒ(x);

3. where the rate of change itself is increasing or decreasing. Here’s another example.

EXAMPLE 4 Concentration of Blood Sugar On April 23, 1988, the human-powered airplane Daedalus flew a record-breaking 119 km

from Crete to the island of Santorini in the Aegean Sea, southeast of mainland Greece. Dur-

3.1 The Derivative as a Function

Slope 0

A Slope –1

Slope – 4 

4 2 y-units/x-unit

(a) Slope

–2 B' Vertical coordinate –1

(b)

FIGURE 3.3 We made the graph of y= ƒ¿sxd in (b) by plotting slopes from the graph of y= ƒsxd in (a). The vertical coordinate of B¿ is the slope at B and so on. The graph of ƒ¿ is a visual record of how the slope of ƒ changes with x.

ing the 6-hour endurance tests before the flight, researchers monitored the prospective pilots’ blood-sugar concentrations. The concentration graph for one of the athlete-pilots is shown in Figure 3.4a, where the concentration in milligrams deciliter is plotted against time in hours. >

The graph consists of line segments connecting data points. The constant slope of each segment gives an estimate of the derivative of the concentration between measure- ments. We calculated the slope of each segment from the coordinate grid and plotted the derivative as a step function in Figure 3.4b. To make the plot for the first hour, for in- stance, we observed that the concentration increased from about 79 mg dL to 93 mg dL. > > The net increase was ¢y = 93 - 79 = 14 mg >dL. Dividing this by ¢t = 1 hour gave the rate of change as

¢y

= 14 = 14 mg

1 >dL per hour.

¢t

Notice that we can make no estimate of the concentration’s rate of change at times t=

1, 2, Á , 5 , where the graph we have drawn for the concentration has a corner and no slope. The derivative step function is not defined at these times.

Copyright © 2005 Pearson Education, Inc., publishing as Pearson AŁ ıyr™©Å‘n£«–

Chapter 3: Differentiation

Athens Ae

/dL TURKEY GREECE

gean Sea

SANTORINI

Concentration, mg RHODES

Sea of Crete

Time (h) (a)

Mediterranean

Heraklion

km y' Daedalus's flight path on April 23, 1988

FIGURE 3.4 (a) Graph of the sugar concentration in the blood of a Daedalus pilot

5 during a 6-hour preflight endurance test. (b) The derivative of the pilot’s blood-sugar

concentration shows how rapidly the concentration rose and fell during various portions

0 1 2 3 4 5 6 of the test.

Differentiable on an Interval; One-Sided Derivatives

Rate of change of concentration, Time (h)

A function y= ƒsxd is differentiable on an open interval (finite or infinite) if it has a de-

(b)

rivative at each point of the interval. It is differentiable on a closed interval [a, b] if it is differentiable on the interior (a, b) and if the limits

ƒsa + hd - ƒsad

lim + Right-hand derivative at a

h :0

h ƒsb + hd - ƒsbd

lim - Left-hand derivative at b

h :0

exist at the endpoints (Figure 3.5).

Right-hand and left-hand derivatives may be defined at any point of a function’s do- main. The usual relation between one-sided and two-sided limits holds for these derivatives. Because of Theorem 6, Section 2.4, a function has a derivative at a point if and only if it

lim f

h h →0 has left-hand and right-hand derivatives there, and these one-sided derivatives are equal.

f EXAMPLE 5

y=

ƒxƒ Is Not Differentiable at the Origin

h lim

Show that the function y= ƒxƒ is differentiable on s - q , 0d and s0, q d but has no deriva-

tive at x= 0.

Solution

To the right of the origin,

dx sƒxƒd= dx sxd = xd = 1. smx + bd = m , a b dx dx ƒxƒ=x s1

h 0 h 0 To the left,

FIGURE 3.5 Derivatives at endpoints are

# xd = - 1

one-sided limits. dx sƒxƒd= dx s - xd = dx ƒxƒ=-x s-1

3.1 The Derivative as a Function

(Figure 3.6). There can be no derivative at the origin because the one-sided derivatives dif-

x fer there: ƒ0+hƒ-ƒ0ƒ

ƒhƒ

Right-hand derivative of ƒ x ƒ at zero = lim

h = lim h :0 + h

y' y'

h :0 +

= lim h

h :0 + h 0 ƒ h ƒ = h when h 7 0.

y'

= lim + 1=1

right-hand derivative

h :0

left-hand derivative

ƒ0+hƒ-ƒ0ƒ ƒhƒ Left-hand derivative of ƒ x ƒ at zero = lim - = lim

h h :0 - h FIGURE 3.6 The function y= ƒxƒ is = lim not differentiable at the origin where -h -

h :0

h ƒ h ƒ = - h when h 6 0. the graph has a “corner.”

h :0

= lim - - 1 = -1.

h :0

EXAMPLE 6

y = 1x Is Not Differentiable at x=0

In Example 2 we found that for x7 0,

d 1x = 1 .

dx

2 1x

We apply the definition to examine if the derivative exists at x= 0:

20 + h - 20

lim + = lim =q.

h :0

h h :0 + 1h

Since the (right-hand) limit is not finite, there is no derivative at x= 0. Since the slopes of the secant lines joining the origin to the points A h , 1h B on a graph of y = 1x ap-

proach q, the graph has a vertical tangent at the origin.

When Does a Function Not Have a Derivative at a Point?

A function has a derivative at a point x 0 if the slopes of the secant lines through Psx 0 , ƒsx 0 dd and a nearby point Q on the graph approach a limit as Q approaches P. When- ever the secants fail to take up a limiting position or become vertical as Q approaches P, the derivative does not exist. Thus differentiability is a “smoothness” condition on the graph of ƒ. A function whose graph is otherwise smooth will fail to have a derivative at a point for several reasons, such as at points where the graph has

1. a corner, where the one-sided

2. a cusp, where the slope of PQ

derivatives differ.

approaches q from one side and -q from the other.

Chapter 3: Differentiation

3. a vertical tangent, where the slope of PQ approaches q from both sides or approaches -q from both sides (here, -q ).

4. a discontinuity.

Differentiable Functions Are Continuous

A function is continuous at every point where it has a derivative.

THEOREM 1

Differentiability Implies Continuity

If ƒ has a derivative at x=c , then ƒ is continuous at x = c .

Proof Given that ƒ¿scd exists, we must show that lim x:c ƒsxd = ƒscd , or equivalently,

that If lim h:0 ƒsc + hd = ƒscd . hZ 0, then ƒsc + hd = ƒscd + sƒsc + hd - ƒscdd

ƒsc + hd - ƒscd

= ƒscd +

3.1 The Derivative as a Function

Now take limits as h : 0. By Theorem 1 of Section 2.2,

ƒsc + hd - ƒscd

lim ƒsc + hd = lim ƒscd + lim # lim h

h h :0

= ƒscd + ƒ¿scd # 0 = ƒscd + 0 = ƒscd .

Similar arguments with one-sided limits show that if ƒ has a derivative from one side (right or left) at x=c then ƒ is continuous from that side at x=c .

Theorem 1 on page 154 says that if a function has a discontinuity at a point (for in- stance, a jump discontinuity), then it cannot be differentiable there. The greatest integer function fails to be differentiable at every integer y= :x; = int x x=n (Example 4,

Section 2.6). CAUTION The converse of Theorem 1 is false. A function need not have a derivative at a

point where it is continuous, as we saw in Example 5.

The Intermediate Value Property of Derivatives

Not every function can be some function’s derivative, as we see from the following theorem.

THEOREM 2

If a and b are any two points in an interval on which ƒ is differentiable, then ƒ¿ FIGURE 3.7 The unit step

takes on every value between ƒ¿sad and ƒ¿sbd .

function does not have the Intermediate Value Property and cannot be the derivative of a function on the real line.

Theorem 2 (which we will not prove) says that a function cannot be a derivative on an in- terval unless it has the Intermediate Value Property there. For example, the unit step func- tion in Figure 3.7 cannot be the derivative of any real-valued function on the real line. In Chapter 5 we will see that every continuous function is a derivative of some function.

In Section 4.4, we invoke Theorem 2 to analyze what happens at a point on the graph of a twice-differentiable function where it changes its “bending” behavior.

3.1 The Derivative as a Function

EXERCISES 3.1

Finding Derivative Functions and Values

B Using the definition, calculate the derivatives of the functions in Exer-

4. k szd = 1-z ; k¿s -1d, k¿s1d, k¿ A 22

2z

cises 1–6. Then find the values of the derivatives as specified.

5. psud = 2 3u ; p¿s1d, p¿s3d, p¿s2 >3d

1. ƒsxd = 4 - x 2 ; ƒ¿s -3d, ƒ¿s0d, ƒ¿s1d

6. r ssd = 2 2s + 1 ; r¿s0d, r¿s1d, r¿s1 >2d

2. Fsxd = sx - 1d 2 + 1; F¿s -1d, F¿s0d, F¿s2d In Exercises 7–12, find the indicated derivatives.

st

1 ; dy

3. g std = 2 g¿s -1d, g¿s2d, g¿ A 23 B 7. if y = 2x 3 8. dr

ds if r = +1

dx

Chapter 3: Differentiation

9. ds t

dt if s = 2t + 1 Match the functions graphed in Exercises 27–30 with the derivatives

Graphs

10. dy graphed in the accompanying figures (a) – (d). dt if y = t - 1 t

Slopes and Tangent Lines

In Exercises 13–16, differentiate the functions and find the slope of

the tangent line at the given value of the independent variable. (b)

(a)

13. ƒsxd = x + 9 x , x = -3

(d) In Exercises 17–18, differentiate the functions. Then find an equation

(c)

of the tangent line at the indicated point on the graph of the function.

27. y

28. y

17. y= ƒsxd =

In Exercises 19–22, find the values of the derivatives. x 0 0

Using the Alternative Formula for Derivatives

31. a. The graph in the accompanying figure is made of line seg- Use the formula

ments joined end to end. At which points of the interval

[ - 4, 6] is ƒ¿ not defined? Give reasons for your answer. ƒ¿sxd = lim

ƒszd - ƒsxd

z :x

z-x

to find the derivative of the functions in Exercises 23–26.

23. ƒsxd = 1 y x+ 2

24. ƒsxd =

1 x sx - 1d 2

25. g sxd = x x- 1

26. g sxd = 1 + 1x

3.1 The Derivative as a Function

b. Graph the derivative of ƒ.

The graph should show a step function.

32. Recovering a function from its derivative

a. Use the following information to graph the function ƒ over

the closed interval [ - 2, 5] .

i) The graph of ƒ is made of closed line segments joined

end to end.

ii) The graph starts at the point s - 2, 3d .

iii) The derivative of ƒ is the step function in the figure

shown here.

Time (days)

b. During what days does the population seem to be increasing –2

0 1 3 5 fastest? Slowest?

One-Sided Derivatives

Compare the right-hand and left-hand derivatives to show that the functions in Exercises 35–38 are not differentiable at the point P.

b. Repeat part (a) assuming that the graph starts at s - 2, 0d

36. y instead of s - 2, 3d .

35. y

33. Growth in the economy The graph in the accompanying figure 2 y shows the average annual percentage change y= ƒstd in the U.S.

gross national product (GNP) for the years 1983–1988. Graph y 2 dy >dt (where defined). (Source: Statistical Abstracts of the United

P (1, 2) States , 110th Edition, U.S. Department of Commerce, p. 427.)

34. Fruit flies (Continuation of Example 3, Section 2.1.) Popula-

tions starting out in closed environments grow slowly at first, when there are relatively few members, then more rapidly as the number of reproducing individuals increases and resources are still abundant, then slowly again as the population reaches the

Differentiability and Continuity on an Interval

carrying capacity of the environment. Each figure in Exercises 39–44 shows the graph of a function over a closed interval D. At what domain points does the function appear to be

a. Use the graphical technique of Example 3 to graph the derivative of the fruit fly population introduced in Section 2.1. a. differentiable?

The graph of the population is reproduced here.

b. continuous but not differentiable?

c. neither continuous nor differentiable?

Chapter 3: Differentiation

Give reasons for your answers.

51. Tangent to a parabola Does the parabola y= 2x 2 - 13x + 5

39. 40. have a tangent whose slope is -1? If so, find an equation for the y

line and the point of tangency. If not, why not? y

D : –3 ⱕ xⱕ 2 y

52. Tangent to y ⴝ 1x Does any tangent to the curve y = 1x

2 2 D : –2 ⱕ xⱕ 3 cross the x-axis at x=- 1? If so, find an equation for the line and the point of tangency. If not, why not?

1 1 53. Greatest integer in x Does any function differentiable on

0 1 2 3 s-q, qd have y= int x , the greatest integer in x (see Figure –1

2.55), as its derivative? Give reasons for your answer.

54. Derivative of y ⴝƒxƒ Graph the derivative of ƒsxd = ƒxƒ. Then –2

graph y=s ƒ x ƒ - 0d >sx - 0d = ƒ x ƒ >x. What can you conclude?

55. Derivative of ⴚƒ Does knowing that a function ƒ(x) is differen-

41. 42. tiable at x=x 0 tell you anything about the differentiability of the y

function -ƒ at x=x 0 ? Give reasons for your answer. y

56. Derivative of multiples Does knowing that a function g (t) is

D : –3 ⱕ xⱕ 3 D : –2 ⱕ xⱕ 3 differentiable at t= 7 tell you anything about the differentiability

3 of the function 3g at t= 7? Give reasons for your answer.

1 2 57. Limit of a quotient Suppose that functions g(t) and h(t) are

defined for all values of t and Can gs 0d = hs0d = 0 . –3 –2 –1 0

1 lim t:0 sg stdd >shstdd exist? If it does exist, must it equal zero? –2

Give reasons for your answers.

–2 –1 0 1 2 3 58. a. Let ƒ(x) be a function satisfying ƒ ƒsxd ƒ … x 2 for -1 … x … 1 . Show that ƒ is differentiable at x= 0 and find ƒ¿s0d .

43. 44. b. Show that

D : –3 ⱕ xⱕ 3 x sin 1 x, xZ 0

D : –1 ⱕ xⱕ 2 4 ƒsxd = L

x= 0

1 2 is differentiable at x= 0 and find ƒ¿s0d . T 59. Graph y= 1 > A 2 1x B in a window that has 0…x…2. Then, on

the same screen, graph

Theory and Examples

1, - 0.5, - 0.1 . Explain what In Exercises 45–48,

for h=

1, 0.5, 0.1 . Then try h=-

is going on.

a.

T Find the derivative 60. of the given function Graph y= ƒ¿sxd

3x 2 in a window that has -2 … x … 2, 0 … y … 3 .

y= ƒsxd .

Then, on the same screen, graph

b. Graph y= ƒsxd and y= ƒ¿sxd side by side using separate sets of

coordinate axes, and answer the following questions. sx + hd -x 3

y=

c.

For what values of x, if any, is h ƒ¿ positive? Zero? Negative?

d. Over what intervals of x-values, if any, does the function

2, - 1, - 0.2 . Explain what is y= ƒsxd increase as x increases? Decrease as x increases? How

for h=

2, 1, 0.2 . Then try h=-

going on.

is this related to what you found in part (c)? (We will say more

61. Weierstrass’s nowhere differentiable continuous function

about this relationship in Chapter 4.) The sum of the first eight terms of the Weierstrass function q

45. y=-x 2 46. y=- g 1 >x ƒ(x) = n= 0 s2 >3d n cos s9 n pxd is

47. y=x 3 >3

48. y=x 4 >4

g sxd = cos spxd + s2 >3d 1 cos s9pxd + s2 >3d 2 cos s9 2 pxd

49. 3 + s2 >3d 3 cos s9 Does the curve 3 y=x ever have a negative slope? If so, where? pxd + Á + s 2 >3d 7 cos s9 7 pxd . Give reasons for your answer.

Graph this sum. Zoom in several times. How wiggly and bumpy

50. Does the curve y= 2 1x have any horizontal tangents? If so, is this graph? Specify a viewing window in which the displayed where? Give reasons for your answer.

portion of the graph is smooth.

3.1 The Derivative as a Function

159

COMPUTER EXPLORATIONS

f. Graph the formula obtained in part (c). What does it mean when Use a CAS to perform the following steps for the functions in Exer-

its values are negative? Zero? Positive? Does this make sense cises 62–67.

with your plot from part (a)? Give reasons for your answer. Plot y= ƒsxd to see that function’s global behavior.

62. a. 3 ƒsxd = x +x 2 - x, x 0 =1

>3

63. ƒsxd = x 1 >3 +x 2 , x 0 =1

b. Define the difference quotient q at a general point x, with general

step size h.

64. 4x ,

=2

65. ƒsxd = x- ƒsxd = 1 2 x 0 2 , x 0 =-1

c. Take the limit as h:0. What formula does this give?

x +1

3x +1

67. ƒsxd = x d. 2 Substitute the value x=x 0 and plot the function y= ƒsxd >2 cos x , x 0 =p >4 together with its tangent line at that point.

66. ƒsxd = sin 2x, x 0 =p

e. Substitute various values for x larger and smaller than x 0 into the

formula obtained in part (c). Do the numbers make sense with your picture?

3.2 Differentiation Rules

3.2 Differentiation Rules

This section introduces a few rules that allow us to differentiate a great variety of func- tions. By proving these rules here, we can differentiate functions without having to apply the definition of the derivative each time.

Powers, Multiples, Sums, and Differences

The first rule of differentiation is that the derivative of every constant function is zero.

RULE 1

Derivative of a Constant Function

If ƒ has the constant value ƒsxd = c , then

dx scd = 0 .

dx

EXAMPLE 1

If ƒ has the constant value ƒsxd = 8 , then

df

c (x, c)

= d s8d = 0 .

dx

dx

Similarly,

dx a- 2b dx a23b = 0.

h d p =0

and

FIGURE 3.8 The rule sd >dxdscd = 0 is Proof of Rule 1 We apply the definition of derivative to ƒsxd = c , the function whose another way to say that the values of

outputs have the constant value c (Figure 3.8). At every value of x, we find that constant functions never change and that the slope of a horizontal line is zero at

ƒ¿sxd = lim = lim c-c = lim 0=0. every point.

ƒsx + hd - ƒsxd

h :0

h h :0 h h :0

Chapter 3: Differentiation

The second rule tells how to differentiate x n if n is a positive integer.

RULE 2

Power Rule for Positive Integers

If n is a positive integer, then

d x n = nx n- 1 dx .

To apply the Power Rule, we subtract 1 from the original exponent (n) and multiply the result by n.

EXAMPLE 2 Interpreting Rule 2 ƒ

1 2x

3x 2 4x 3 Á

H ISTORICAL B IOGRAPHY

First Proof of Rule 2 The formula

Richard Courant

(1888–1972) n- +z 2 x + Á + zx 2 +x n- 1 d can be verified by multiplying out the right-hand side. Then from the alternative form for

z n -x n = sz - xdsz n- 1 n-

the definition of the derivative,

ƒszd - ƒsxd

ƒ¿sxd = lim z-x z = lim

n -x n

z :x

z :x z-x

= lim sz n- 1 +z n- 2 x + Á + zx n- 2 +x n- 1 d

z :x

= nx n- 1

Second Proof of Rule 2 If then ƒsxd = x n , ƒsx + hd = sx + hd n . Since n is a positive integer, we can expand sx + hd n by the Binomial Theorem to get

ƒsx + hd - ƒsxd

sx + hd n -x

h = lim h :0

ƒ¿sxd = lim

h :0

n + nx n- 1 nsn - h+ 1d x n- 2 h 2 + Á + nxh n- 1 +h n cx n

2 d-x

= lim

h :0

nx n- 1 nsn - h+ 1d x n- 2 h 2 + Á + nxh n- 2 1 +h n

= lim

h :0

h = lim n- 1 nsn - + 1d n- h + Á + nxh n- cnx 2 x 2

2 +h n- h :0 1 d

= nx n- 1 The third rule says that when a differentiable function is multiplied by a constant, its

derivative is multiplied by the same constant.

3.2 Differentiation Rules

RULE 3

Constant Multiple Rule

If u is a differentiable function of x, and c is a constant, then

d scud = c du .

dx

dx

In particular, if n is a positive integer, then

d scx

n d = cnx n- 1 .

dx

EXAMPLE 3

6x

3 (1, 3) Slope

(a) The derivative formula

2 d s3x 2 d=3 # 2x = 6x

dx

2 says that if we rescale the graph of y=x 2 by multiplying each y-coordinate by 3,

then we multiply the slope at each point by 3 (Figure 3.9).

(b) A useful special case

1 (1, 1) Slope

The derivative of the negative of a differentiable function u is the negative of the func-

tion’s derivative. Rule 3 with c=- 1 gives

d d s-1

d du

s - ud = ud = - 1 # sud = - .

0 1 2 dx

dx

dx

dx

FIGURE 3.9 The graphs of y=x 2 and

Proof of Rule 3

y= 3x 2 . Tripling the y-coordinates triples

the slope (Example 3). cu = lim Derivative definition

d cusx + hd - cusxd

dx

h :0

h with ƒsxd = cusxd

usx + hd - usxd

= c lim Limit property

h :0

du

=c u is differentiable .

dx

The next rule says that the derivative of the sum of two differentiable functions is the sum of their derivatives.

Denoting Functions by u and Y The functions we are working with

when we need a differentiation formula are likely to be denoted by letters like ƒ

RULE 4

Derivative Sum Rule

and g. When we apply the formula, we do not want to find it using these same

If u and y are differentiable functions of x, then their sum u+y is differentiable letters in some other way. To guard

at every point where u and y are both differentiable. At such points, against this problem, we denote the

functions in differentiation rules by

d su + yd = du + dy .

letters like u and y that are not likely to

dx

dx

dx

be already in use.

Chapter 3: Differentiation

EXAMPLE 4 Derivative of a Sum y=x 4 + 12x

dy

dx sx d+ dx s12xd

dx

= 4x 3 + 12

Proof of Rule 4 We apply the definition of derivative to ƒsxd = usxd + ysxd :

d [usx + hd + ysx + hd] - [usxd + ysxd] [usxd + ysxd] = lim

dx

h :0

usx + hd - usxd

= lim c +

ysx + hd - ysxd

h :0

usx + hd - usxd

ysx + hd - ysxd = lim = du + + lim dy .

h dx dx Combining the Sum Rule with the Constant Multiple Rule gives the Difference Rule,

h :0

h h :0

which says that the derivative of a difference of differentiable functions is the difference of their derivatives.

d su - yd = d [u + s - 1dy] = du

dx + s -1d dx

dy = du - dy

dx dx The Sum Rule also extends to sums of more than two functions, as long as there are

dx

dx

only finitely many functions in the sum. If u 1 ,u 2 ,Á,u n are differentiable at x, then so is

u 1 +u 2 +Á+u n , and

dx su 1 +u 2 +Á+u n

d du d= 1 du 2 du n

dx

dx

dx

EXAMPLE 5 Derivative of a Polynomial

y=x 3 +4 x 2

3 - 5x + 1

dy

dx s1d

= d x d 3 + 4 2 a d x b- s5xd + d

dx

dx

dx 3 dx

= 3x 2 +4 # 2x - 5 + 0

3 = 3x 2 +8 x- 5

Notice that we can differentiate any polynomial term by term, the way we differenti- ated the polynomial in Example 5. All polynomials are differentiable everywhere.

Proof of the Sum Rule for Sums of More Than Two Functions We prove the statement

dx su 1 +u 2 +Á+u n d=

d du 1 du 2 du n

dx by mathematical induction (see Appendix 1). The statement is true for n= 2, as was just

dx

dx

proved. This is Step 1 of the induction proof.

3.2 Differentiation Rules

Step 2 is to show that if the statement is true for any positive integer n=k , where

kÚn 0 = 2, then it is also true for n=k+ 1. So suppose that

du k su 1 + dx 2 1 +u 2 +Á+u k d= +Á+ .

d du

du

dx (1)

dx

dx

Then

d (u

+u +Á+u k +u k+ )

dx

Call the function

Call this

defined by this sum u.

function y.

= d du k+ d+ 1 Rule 4 for su d +u 2 +Á+u k su + yd

dx 1 dx

dx

du 1 du 2 du k

du k+

dx

dx

dx

dx

Eq. (1)

With these steps verified, the mathematical induction principle now guarantees the Sum Rule for every integer nÚ 2.

EXAMPLE 6 Finding Horizontal Tangents

yy

4 2 Does the curve y=x 4 2 2 - 2x +2 have any horizontal tangents? If so, where?

Solution

The horizontal tangents, if any, occur where the slope dy >dx is zero. We have,

dy

= d - 2x

dx sx 4 2 + 2d = 4x - 4x.

dx

dy

1 Now solve the equation

= 0 for x:

dx

4x - 4x = 0

4xsx 2 - 1d = 0

x = 0, 1, - 1 .

FIGURE 3.10

The curve y=x 4 - 2x 2 +2 and its horizontal

0, 1 , and - 1. The corre- tangents (Example 6).

The curve y=x 4 - 2x 2 +2 has horizontal tangents at x=

sponding points on the curve are (0, 2), (1, 1) and s - 1, 1d . See Figure 3.10.

Products and Quotients

While the derivative of the sum of two functions is the sum of their derivatives, the deriva- tive of the product of two functions is not the product of their derivatives. For instance,

# xd = sx 2 d = 2x, while

dx sx

dx sxd dx sxd = 1

dx

The derivative of a product of two functions is the sum of two products, as we now explain.

RULE 5

Derivative Product Rule

If u and y are differentiable at x, then so is their product uy, and

d dy

du

dx suyd = u dx +y dx .

Chapter 3: Differentiation

Picturing the Product Rule

The derivative of the product uy is u times the derivative of y plus y times the deriva- If u(x) and y(x) are positive and

tive of u. In prime notation, suyd¿ = uy¿ + yu¿ . In function notation, increase when x increases, and if h 7 0 ,

d [ƒsxdg sxd] = ƒsxdg¿sxd + g sxdƒ¿sxd .

dx

yu

EXAMPLE 7 Using the Product Rule

y (x)

Find the derivative of

u (x)y(x)

y=

x ax + 1x b.

u (x) u

Solution

We apply the Product Rule with u= 1 and y=x >x 2 + s1 >xd:

dx dx , and is

+ 1x b d = x a2x - x 2 b + ax + 1x b a- x 2 b d 1 usx + hdysx + hd - usxdysxd

c 1 x ax 2 1 1 2 1 dx

d suyd = u then the total shaded area in the picture dy d +y du

dx

a 1 xb=- x 2 dx by

= usx + hd ¢y + ysx + hd

=2- 3 -1-

Example 3, Section 2.7.

¢u - ¢u¢y . Dividing both sides of this equation by

=1- 2 .

h gives

usx + hdysx + hd - usxdysxd

h Proof of Rule 5

= usx + hd ¢y ¢u h usx + hdysx + hd - usxdysxd + ysx + hd h d

suyd = lim

¢y

dx

h :0

- ¢u h . To change this fraction into an equivalent one that contains difference quotients for the de-

As h:0 + , rivatives of u and y, we subtract and add usx + hdysxd in the numerator:

# ¢y # ¢u dy :0

h dx =0,

usx + hdysx + hd - usx + hdysxd + usx + hdysxd - usxdysxd

d suyd = lim

leaving

dx

h :0

suyd = u dy dx dx +y du .

h h h :0 d

ysx + hd - ysxd

= lim cusx + hd + ysxd

usx + hd - usxd

dx

ysx + hd - ysxd

usx + hd - usxd

= lim usx + hd # lim + ysxd # lim .

h As h approaches zero, usx + hd approaches u(x) because u, being differentiable at x, is con-

h :0

h :0

h h :0

tinuous at x. The two fractions approach the values of dy >dx at x and at du >dx x . In short,

d suyd = u dy +y du .

dx

dx

dx

In the following example, we have only numerical values with which to work.

EXAMPLE 8 Derivative from Numerical Values

Let y = uy be the product of the functions u and y. Find y¿s 2d if

us 2d = 3,

u¿s 2d = - 4,

ys 2d = 1,

and

y¿s 2d = 2 .

Solution

From the Product Rule, in the form

y¿ = suyd¿ = uy¿ + yu¿ ,

3.2 Differentiation Rules

we have

y¿s2d = us2dy¿s2d + ys2du¿s2d

= s3ds2d + s1ds - 4d = 6 - 4 = 2 . EXAMPLE 9 Differentiating a Product in Two Ways

Find the derivative of y = sx 2 + 1dsx 3 + 3d.

Solution

(a) From the Product Rule with u=x 2 +1 and y=x 3 + 3, we find

x 2 +1 x 3 = sx 2 2 dx 3 CA BA +3 BD + 1ds3x d + sx + 3ds2xd

= 3x 4 + 3x 2 + 2x 4 + 6x

= 5x 4 + 3x 2 + 6x.

(b) This particular product can be differentiated as well (perhaps better) by multiplying out the original expression for y and differentiating the resulting polynomial:

y = sx 2 + 1dsx 3 + 3d = x 5 +x 3 + 3x 2 +3

dy = 5x 4 + 3x 2 + 6x. dx

This is in agreement with our first calculation. Just as the derivative of the product of two differentiable functions is not the product of

their derivatives, the derivative of the quotient of two functions is not the quotient of their derivatives. What happens instead is the Quotient Rule.

RULE 6

Derivative Quotient Rule

If u and y are differentiable at x and if ysxd Z 0, then the quotient u >y is differ- entiable at x, and

d -u

y du dy

2 dx . yb= y

dx

dx

In function notation,

d ƒsxd

g sxd d=

g sxd ƒ¿sxd - ƒsxdg¿sxd

dx c .

g 2 sxd

EXAMPLE 10 Using the Quotient Rule Find the derivative of

t 2 -1 y= 2 . t +1

Chapter 3: Differentiation

Solution

We apply the Quotient Rule with u=t 2 -1 and y=t 2 + 1:

st 2 + 1d # 2t - st 2 - 1d # 2t

dy

d = ysdu u >dtd - usdy>dtd

2 2 a dt yb=

dt

y st 2 + 1d

2t 3 + 2t - 2t 3

+ 2t

st 2 + 1d 2 4t

= 2 . st + 1d 2

Proof of Rule 6

usx + hd

usxd

a y b = lim

ysx + hd

ysxd

dx

h :0

h ysxdusx + hd - usxdysx + hd

= lim

h :0

hysx + hdysxd

To change the last fraction into an equivalent one that contains the difference quotients for the derivatives of u and y, we subtract and add y(x)u(x) in the numerator. We then get

ysxdusx + hd - ysxdusxd + ysxdusxd - usxdysx + hd

dx y b = lim h :0

hysx + hdysxd

usx + hd - usxd

ysx + hd - ysxd

ysxd - usxd

= lim h h .

h :0

ysx + hdysxd

Taking the limit in the numerator and denominator now gives the Quotient Rule.

Negative Integer Powers of x

The Power Rule for negative integers is the same as the rule for positive integers.

RULE 7

Power Rule for Negative Integers

If n is a negative integer and xZ 0, then

d sx n d = nx n- 1 . dx

EXAMPLE 11

dx a xb=

(a) d d -1

sx d = s - 1dx =-1

dx

x 2 Agrees with Example 3, Section 2.7

dx a x 3 b=4 sx d = 4s - 3dx dx = - 12 x 4

d (b) 4 d -3

3.2 Differentiation Rules

Proof of Rule 7 The proof uses the Quotient Rule. If n is a negative integer, then n=-m , where m is a positive integer. Hence, x n =x -m =1 m >x , and

d n d= d sx 1 dx dx a

x m # d 1 -1 d # x m

dx A B dx A B m

sx m d 2 Quotient Rule with u= 1 and y=x

0 - mx m- y 1 =

Since d m7 0, sx m d = mx m- 1

x 2m

dx 4

= - mx -m - 1

= nx n- 1 .

Since -m=n 3

2 EXAMPLE 12 Tangent to a Curve Find an equation for the tangent to the curve

y=x+

at the point (1, 3) (Figure 3.11).

FIGURE 3.11 The tangent to the curve y=x+s 2 >xd at (1, 3) in Example 12.

Solution

The slope of the curve is

The curve has a third-quadrant portion not shown here. We see how to graph

= d d 1 1 sxd + 2 2 a x b = 1 + 2 a- 2 b=1- 2 . functions like this one in Chapter 4.

dy

dx

dx

dx

The slope at x= 1 is

dy

c1 - 2 d = 1 - 2 = -1.

dx ` x=

1 x x= 1

The line through (1, 3) with slope m=- 1 is y - 3 = s - 1dsx - 1d

Point-slope equation

y=-x+1+3 y=-x+4.

The choice of which rules to use in solving a differentiation problem can make a dif- ference in how much work you have to do. Here is an example.

EXAMPLE 13 Choosing Which Rule to Use Rather than using the Quotient Rule to find the derivative of

sx - 1dsx 2 - 2xd

y=

expand the numerator and divide by x 4 :

= x 3 - 3x 4 + 2x

sx - 1dsx 2 - 2xd

y=

=x -1 - 3x -2 + 2x -3 .

Chapter 3: Differentiation

Then use the Sum and Power Rules:

dy

dx =-x - 3s -2dx + 2s -3dx

=- 1 -6

Second- and Higher-Order Derivatives

If y= ƒsxd is a differentiable function, then its derivative ƒ¿sxd is also a function. If ƒ¿ is also differentiable, then we can differentiate ƒ¿ to get a new function of x denoted by ƒ– . So ƒ– = sƒ¿ d¿ . The function ƒ– is called the second derivative of ƒ because it is the deriv- ative of the first derivative. Notationally,

d dy

dy¿

ƒ–sxd =

dx b dx = y– = D 2 sƒdsxd = D x 2 ƒsxd . dx The symbol D 2 means the operation of differentiation is performed twice.

dx a =

If y=x 6 , then y¿ = 6x 5 and we have

dy¿

A 6x 5 B = 30x 4 dx . dx

y– =

Thus D 2 x 6 A 4 B = 30x .

If y– is differentiable, its derivative, y‡ = dy–

3 y >dx = d 3 >dx is the third derivative

How to Read the Symbols for

of y with respect to x. The names continue as you imagine, with

Derivatives

y¿ n “y prime”

=D y– n “y double prime” dx y n y

snd = d sn - 1d

dx

y “d squared y dx squared”

dx 2 denoting the nth derivative of y with respect to x for any positive integer n. y‡ “y triple prime”

We can interpret the second derivative as the rate of change of the slope of the tangent y snd “y super n”

to the graph of y= ƒsxd at each point. You will see in the next chapter that the second de-

d n y rivative reveals whether the graph bends upward or downward from the tangent line as we dx n “d to the n of y by dx to the n”

move off the point of tangency. In the next section, we interpret both the second and third

D n “D to the n”

derivatives in terms of motion along a straight line.

EXAMPLE 14 Finding Higher Derivatives The first four derivatives of y=x 3 - 3x 2 +2 are

First derivative:

y¿ = 3x 2 - 6x

Second derivative: y– = 6x - 6 Third derivative:

y‡ = 6

Fourth derivative: y s4d = 0.

The function has derivatives of all orders, the fifth and later derivatives all being zero.

3.2 Differentiation Rules

EXERCISES 3.2

Derivative Calculations

q 2 +3 q 4 -1

q 2 +3

In Exercises 1–12, find the first and second derivatives. 12q b a q 3 b sq - 1d 3 + sq + 1d 3

37. p= a 38. p=

1. y=-x 2 +3

2. y=x 2 +x+8

3. s= 5t 3 - 3t 5 4. w= 3z 7 - 7z 3 + 21z 2 Using Numerical Values

+ x 2 + 39. x y= Suppose u and y are functions of x that are differentiable at

3 2 4 and that

7. w= 3z -2 - 1 8. s=- 2t -1 + 4 us 0d = 5, z u¿s0d = -3, ys0d = -1, y¿s0d = 2.

Find the values of the following derivatives at x= 0.

11. r= 1 5 12 4 1 a. dx suyd

Suppose u and y are differentiable functions of x and that In Exercises 13–16, find y¿ (a) by applying the Product Rule and

(b) by multiplying the factors to produce a sum of simpler terms to 1d = 2, u¿s1d = 0, ys1d = 5, y¿s1d = -1.

us

differentiate. Find the values of the following derivatives at x= 1.

13. y=s 3-x 2 dsx 3 - x + 1d 14. y = sx - 1dsx 2 + x + 1d

a. d b. d suyd u a yb

x b ax - x+1b

Find the derivatives of the functions in Exercises 17–28.

Slopes and Tangents

17. y= 2x + 5

18. z= 2x + 1

41. a. Normal to a curve Find an equation for the line perpendicular

3x - 2 3

to the tangent to the curve y=x - 4x + 1 at the point (2, 1). x 2 -4

x 2 -1

t 2 b. Smallest slope 19. What is the smallest slope on the curve? At g sxd = 20. -1 x+ 0.5 ƒstd = 2 t what point on the curve does the curve have this slope? +t-2

21. y=s 1 - tds1 + t 2 d -1

22. w=s 2x - 7d -1

c. Tangents having specified slope Find equations for the

tangents to the curve at the points where the slope of the 1s - 1

sx + 5d

23. 24. u= ƒssd = 5x + 1

curve is 8.

1s + 1

2 1x

42. a. Horizontal tangents Find equations for the horizontal tan-

25. y=

1 + x - 4 1x

26. 1 gents to the curve y=x 3 - 3x - 2 . x Also find equations for r= 2 a + 2u b

2u

the lines that are perpendicular to these tangents at the points

sx + 1dsx + 2d

of tangency.

27. y=

2 2 28. sx y= - 1dsx + x + 1d sx - 1dsx - 2d

b. Smallest slope What is the smallest slope on the curve? At what point on the curve does the curve have this slope? Find

Find the derivatives of all orders of the functions in Exercises 29 and an equation for the line that is perpendicular to the curve’s

tangent at this point.

29. y= - 3 x 2 -x

43. Find the tangents to Newton’s serpentine (graphed here) at the ori-

30. y=

gin and the point (1, 2).

Find the first and second derivatives of the functions in Exercises 31–38. y

su - 1dsu 2 + u + 1d

sx 2 + xdsx 2 - x + 1d

36. w = sz + 1dsz - 1dsz + 1d

a 1 + 3z 2

Chapter 3: Differentiation

44. Find the tangent to the Witch of Agnesi (graphed here) at the point how to find the amount of medicine to which the body is most (2, 1).

sensitive.

51. Suppose that the function y in the Product Rule has a constant

8 value c. What does the Product Rule then say? What does this say x 2 about the Constant Multiple Rule?

2 52. The Reciprocal Rule

a.

The Reciprocal Rule says that at any point where the function

0 12 3 y (x) is differentiable and different from zero,

2 dx a yb=- 2 dx ax . + bx + c passes through the point (1, 2) and is tangent to the y line y=x at the origin. Find a, b, and c.

45. Quadratic tangent to identity function The curve y=

d 1 1 dy

Show that the Reciprocal Rule is a special case of the

46. Quadratics having a common tangent The curves y=

Quotient Rule.

b. Show that the Reciprocal Rule and the Product Rule together the point (1, 0). Find a, b, and c.

x 2 + ax + b and y = cx - x 2 have a common tangent line at

imply the Quotient Rule.

47. a. Find an equation for the line that is tangent to the curve

53. Generalizing the Product Rule The Product Rule gives the y=x 3 -x at the point s - 1, 0d .

formula

T b. Graph the curve and tangent line together. The tangent

intersects the curve at another point. Use Zoom and Trace to dy du

+y dx dx estimate the point’s coordinates.

dx suyd = u

T c. Confirm your estimates of the coordinates of the second for the derivative of the product uy of two differentiable functions intersection point by solving the equations for the curve and

of x.

tangent simultaneously (Solver key).

a. What is the analogous formula for the derivative of the

48. a. Find an equation for the line that is tangent to the curve product uyw of three differentiable functions of x ? y=x 3 - 6x 2 + 5x at the origin.

b. What is the formula for the derivative of the product u 1 u 2 u 3 u 4 T b. Graph the curve and tangent together. The tangent intersects

of four differentiable functions of x ? the curve at another point. Use Zoom and Trace to estimate

c. What is the formula for the derivative of a product the point’s coordinates.

u 1 u 2 u 3 Áu n of a finite number n of differentiable functions T c. Confirm your estimates of the coordinates of the second

of x ?

intersection point by solving the equations for the curve and

54. Rational Powers

tangent simultaneously (Solver key).

a. Find d 3 A >2 x B by writing x 3 >2 as x # x 1 >2

and using the Product

dx

Theory and Examples

Rule. Express your answer as a rational number times a

49. The general polynomial of degree n has the form rational power of x. Work parts (b) and (c) by a similar method.

Psxd = a n x n +a n- 1 x n- 1 +Á+a 2 x 2 +a 1 x+a 0

b. Find

sx 5 >2 d.

where Find a n Z0. P¿sxd .

dx

d.

50. The body’s reaction to medicine The reaction of the body to a

c. Find d sx 7 >2

dose of medicine can sometimes be represented by an equation of

dx

the form

d. What patterns do you see in your answers to parts (a), (b), and

R=M 2 C M

(c)? Rational powers are one of the topics in Section 3.6.

a 2 - 3b ,

55. Cylinder pressure If gas in a cylinder is maintained at a con- stant temperature T, the pressure P is related to the volume V by a where C is a positive constant and M is the amount of medicine

formula of the form

absorbed in the blood. If the reaction is a change in blood pres- sure, R is measured in millimeters of mercury. If the reaction is a

P= nRT - an 2 V - nb 2 , change in temperature, R is measured in degrees, and so on.

V Find dR >dM . This derivative, as a function of M, is called the

in which a, b, n, and R are constants. Find dP >dV . (See accompa- sensitivity of the body to the medicine. In Section 4.5, we will see

nying figure.)

3.2 Differentiation Rules

171

56. The best quantity to order One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is

km

hq Asqd = q + cm + 2 ,

where q is the quantity you order when things run low (shoes, ra- dios, brooms, or whatever the item might be); k is the cost of plac- ing an order (the same, no matter how often you order); c is the cost of one item (a constant); m is the number of items sold each week (a constant); and h is the weekly holding cost per item (a constant that takes into account things such as space, utilities, in-

surance, and security). Find dA and d >dq 2 A >dq 2 .

3.3 The Derivative as a Rate of Change

3.3 The Derivative as a Rate of Change

In Section 2.1, we initiated the study of average and instantaneous rates of change. In this section, we continue our investigations of applications in which derivatives are used to model the rates at which things change in the world around us. We revisit the study of mo- tion along a line and examine other applications.

It is natural to think of change as change with respect to time, but other variables can

be treated in the same way. For example, a physician may want to know how change in dosage affects the body’s response to a drug. An economist may want to study how the cost of producing steel varies with the number of tons produced.

Instantaneous Rates of Change

If we interpret the difference quotient sƒsx + hd - ƒsxdd >h as the average rate of change in ƒ over the interval from x to x+h , we can interpret its limit as h:0 as the rate at which ƒ is changing at the point x.

DEFINITION

Instantaneous Rate of Change

The instantaneous rate of change of ƒ with respect to x at x 0 is the derivative

ƒsx 0 + hd - ƒsx 0 d ƒ¿sx 0 d = lim ,

h :0

provided the limit exists.

Thus, instantaneous rates are limits of average rates. It is conventional to use the word instantaneous even when x does not represent time. The word is, however, frequently omitted. When we say rate of change, we mean instantaneous rate of change .

Chapter 3: Differentiation

EXAMPLE 1 How a Circle’s Area Changes with Its Diameter The area A of a circle is related to its diameter by the equation

A= 4 D 2 .

How fast does the area change with respect to the diameter when the diameter is 10 m?

Solution

The rate of change of the area with respect to the diameter is

dA = p # 2D = pD .

dD 4 2

When D=

10 m , the area is changing at rate sp

>2d10 = 5p m 2 >m.

Motion Along a Line: Displacement, Velocity, Speed,

Position at time t …

∆t

∆s

Acceleration, and Jerk

Suppose that an object is moving along a coordinate line (say an s-axis) so that we know its position s on that line as a function of time t:

FIGURE 3.12 The positions of a body moving along a coordinate line at time t

s= ƒstd .

and shortly later at time t + ¢t . The displacement of the object over the time interval from t to (Figure 3.12) t + ¢t is

¢s = ƒst + ¢td - ƒstd, and the average velocity of the object over that time interval is

displacement

¢s ƒst + ¢td - ƒstd

y ay =

travel time

¢t

¢t

To find the body’s velocity at the exact instant t, we take the limit of the average ve- locity over the interval from t to t + ¢t as ¢t shrinks to zero. This limit is the derivative of ƒ with respect to t.

DEFINITION

Velocity

Velocity (instantaneous velocity) is the derivative of position with respect to time. If a body’s position at time t is s= ƒstd , then the body’s velocity at time t is

ds

ƒst + ¢td - ƒstd

ystd =

= lim dt .

¢t:0

¢t

EXAMPLE 2 Finding the Velocity of a Race Car Figure 3.13 shows the time-to-distance graph of a 1996 Riley & Scott Mk III-Olds WSC

race car. The slope of the secant PQ is the average velocity for the 3-sec interval from

t= 2 to t= 5 sec ; in this case, it is about 100 ft sec or 68 mph. > The slope of the tangent at P is the speedometer reading at t= 2 sec , about 57 ft sec > or 39 mph. The acceleration for the period shown is a nearly constant

28.5 ft >sec 2 during

3.3 The Derivative as a Rate of Change

Secant slope is

average velocity for interval from Q

Distance (ft) 300

Tangent slope is speedometer

1 2 3 4 5 6 7 8 Elapsed time (sec)

FIGURE 3.13 The time-to-distance graph for Example 2. The slope of the tangent line at P is the instantaneous velocity at t = 2 sec.

each second, which is about 0.89g, where g is the acceleration due to gravity. The race car’s top speed is an estimated 190 mph. (Source: Road and Track, March 1997.)

Besides telling how fast an object is moving, its velocity tells the direction of motion. When the object is moving forward (s increasing), the velocity is positive; when the body is moving backward (s decreasing), the velocity is negative (Figure 3.14).

s increasing:

s decreasing:

positive slope so

negative slope so

moving backward FIGURE 3.14 For motion s= ƒstd along a straight line, y = ds /dt is

moving forward

positive when s increases and negative when s decreases.

If we drive to a friend’s house and back at 30 mph, say, the speedometer will show 30 on the way over but it will not show -30 on the way back, even though our distance from home is decreasing. The speedometer always shows speed, which is the absolute value of velocity. Speed measures the rate of progress regardless of direction.

Chapter 3: Differentiation

DEFINITION

Speed

Speed is the absolute value of velocity.

Speed =

ƒ ystd ƒ = ds `

dt `

EXAMPLE 3 Horizontal Motion Figure 3.15 shows the velocity y= ƒ¿std of a particle moving on a coordinate line. The

particle moves forward for the first 3 sec, moves backward for the next 2 sec, stands still for a second, and moves forward again. The particle achieves its greatest speed at time t= 4, while moving backward.

MOVES FORWARD

FORWARD AGAIN

f '(t)

Stands still

t (sec) 0 1 2 3 4 5 6 7

Greatest speed

MOVES BACKWARD

FIGURE 3.15 The velocity graph for Example 3.

H ISTORICAL B IOGRAPHY The rate at which a body’s velocity changes is the body’s acceleration. The accelera- Bernard Bolzano

tion measures how quickly the body picks up or loses speed.

A sudden change in acceleration is called a jerk. When a ride in a car or a bus is jerky, it is not that the accelerations involved are necessarily large but that the changes in accel- eration are abrupt.

3.3 The Derivative as a Rate of Change

DEFINITIONS

Acceleration, Jerk

Acceleration is the derivative of velocity with respect to time. If a body’s posi- tion at time t is s= ƒstd , then the body’s acceleration at time t is

astd = dy = d 2 s .

dt