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