differentiation: basic concepts - lamar university calculus/hoffman... · differentiation 3. the...

Chapter� 1. The Derivative:

Slope and Rates

� 2. Techniques ofDifferentiation

� 3. The Product andQuotient Rules

� 4. Marginal Analysis:Approximation byIncrements

� 5. The Chain Rule

� 6. The SecondDerivative

� 7. ImplicitDifferentiation andRelated Rates

� Chapter Summary andReview Problems

Differentiation:Basic Concepts

98 Chapter 2 Differentiation: Basic Concepts

Calculus is the mathematics of change, and the primary tool for studying rates ofchange is a procedure called differentiation. In this section, we describe this proce-dure and show how it can be used in rate problems and to find the slope of a tangentline to a curve.

As an illustration of the ideas we shall explore, consider the motion of an objectfalling from a great height. In physics, it is shown that after t seconds, the object willhave fallen s(t) � 16t2 feet. Suppose we wish to compute the velocity of the objectafter, say, 2 seconds.

Unless the falling object has a speedometer, it is hard to simply “read” its veloc-ity, but we can measure the distance it falls between time t � 2 and time t � 2 � hand compute the average velocity over the time period (2, 2 � h) by the ratio

If the elapsed time of h seconds is small, we would expect the average velocity to bevery close to the instantaneous velocity at t � 2. Thus, it is reasonable to computethe instantaneous velocity vins by the limit

That is, after 2 seconds, the falling object is traveling at the rate of 64 feet per second.

The procedure we have just described is illustrated geometrically in Figure 2.1.Figure 2.1a shows the graph of the distance function s � 16t2, along with the pointsP(2, 64) and Q(2 � h, 16(2 � h)2). The line joining P and Q is called a secant lineof the graph and has slope

msec �16(2 � h)2 � 64

(2 � h) � 2� 64 � 16h

vins � limhfi 0

vave � limhfi 0

(64 � 6h) � 64

�64h � 16h2

h� 64 � 16h

�16(2 � h)2 � 16(2)2

h�

16(4 � 4h � h2) � 16(4)

h

vave �distance traveled

elapsed time�

s(2 � h) � s(2)

h

A FALLING BODY PROBLEM

The Derivative:Slope and Rates

1

E x p l o r e !E x p l o r e !The graphing calculator can

simulate secant lines approach-

ing a tangent line. For a simple

example, store f(x) � x2 � 2

into Y1 of the equation editor,

selecting a bold graphing style.

In Y2 write L1*X � 2. Using

the stat edit menu, input the val-

ues �3.6, �2.4, �1.4, �.6, 0.

Graph in sequential mode, us-

ing a window of size [�4.7,

4.7]1 by [�2.2, 12.2]1. De-

scribe what you observe. What

is the limiting tangent line?

As indicated in Figure 2.1b, when we take smaller and smaller h, the correspond-ing secant lines PQ tend toward the position of what we intuitively think of as the tan-gent line at P. This suggests that we compute the slope of the tangent line mtan by find-ing the limiting value of the slopes of the approximating secant lines PQ; that is,

Thus, the slope of the tangent line to the graph of s(t) � 16t2 at the point where t � 2 is exactly the same as the instantaneous rate of change of s with respect to twhen t � 2.

The procedure illustrated for the falling body function s(t) � 16t2 applies to a vari-ety of other functions f(x). In particular, the average rate of change in f(x) over theinterval (x, x � h) is given by the ratio

f(x � h) � f(x)

h

RATES OF CHANGE AND SLOPE

mtan � limhfi 0

msec � limhfi 0

(64 � 16h) � 64

FIGURE 2.1 The graph of s � 16t2.

s

t2 2 + h

P

QSecant lines

Tangent line

s

t2 2 + h

s = 16t2

P(2, 64)

Q(2 + h, 16(2 + h)2)

Chapter 2 � Section 1 The Derivative: Slope and Rates 99

(a) The secant line through P(2,64) and Q(2 � h, 16(2 � h)2).

(b) As h→0 the secant line PQtends toward the tangentline at P.

which can be interpreted geometrically as the slope of the secant line through thepoints P(x, f(x)) and Q(x � h, f(x � h)) (Figure 2.2a). To find the instantaneous rateof change of f(x) at x, we compute the limit

which also gives the slope of the tangent line to the graph of f(x) at the point P(x, f(x)), as indicated in Figure 2.2b.

Here is an example from economics illustrating the relationship between rate ofchange and slope.

The graph shown in Figure 2.3 gives the relationship between the percentage of unem-ployment U and the corresponding percentage of inflation I. Use the graph to estimate

FIGURE 2.2 Secant lines approximating a tangent line.

y

x

P

TangentSecants

(x, f (x))

y

x

PQ(x + h, f (x + h))

Tangent line

Secant line

(x, f (x))

limhfi 0

f(x � h) � f(x)

h


EXAMPLE 1 .1EXAMPLE 1 .1

(a) The graph of f (x) with a se-cant line through points P (x,f (x)) and Q(x � h, f(x � h )).

(b) As h→0 the secant lines tendtoward the tangent line at P.

E x p l o r e !E x p l o r e !Store f(x) � x2 into Y1 of the

equation editor and graph using

the window [�4.7, 4.7]1 by

[�2.2, 12.2]1. Access the tan-

gent line option of the DRAW key

(second PRGM) and use the right

arrow to trace the cursor to the

point (2, 4) on your graph. Press

ENTER and observe what hap-

pens. Where does this line in-

tercept the y axis?

the rate at which I changes with respect to U when the level of unemployment is 3%and again when it is 10%.

SolutionFrom the figure, we estimate the slope of the tangent line at the point (3, 15), corre-sponding to U � 3, to be approximately �14. That is, when unemployment is 3%,inflation I is decreasing at the rate of 14 percentage points for each percentage pointincrease in unemployment U.

At the point (10, �5), the slope of the tangent line is approximately �0.4, whichmeans that when there is 10% unemployment, inflation is decreasing at the rate ofonly 0.4 percentage point for each percentage point increase in unemployment.

FIGURE 2.3 Inflation as a function of unemployment.Source: Adapted from Robert Eisner, The Misunderstood Economy: What Counts and How to Count It,Boston, MA: Harvard Business School Press, 1994, page 173.

Slope = = –140.5–7

Slope = = –0.4–12.5

Infl

atio

n, %

50

40

30

20

10

0

–10

Unemployment, %

0.0 2.5 5.0 7.5 10.0 12.5 15.0

–7

0.5

Tangent lines –12.5


The expression

that appears in both slope and rate of change computation is called a difference quo-tient of the function f. Specifically, in both applications, we compute the limit of adifference quotient as h approaches 0. To unify the study of these and other similarapplications, we introduce the following terminology and notation.

The advantage of the derivative notation is that the observations made earlier inthis section about slope and rates of change can be summarized in the following com-pact form.

In the first of the following two examples, we find the equation of a tangent line.Then in the second, we consider a business application involving rates.

First compute the derivative of f(x) � x3, and then use it to find the slope of the tan-gent line to the curve y � x3 at the point where x � �1. What is the equation of thetangent line at this point?

SolutionAccording to the definition of the derivative

Slope as a Derivative � The slope of the tangent line to the curvey � f(x) at the point (c, f (c)) is given by mtan � f �(c).

Instantaneous Rate of Change as a Derivative � Thequantity f(x) changes at the rate f�(c) with respect to x when x � c.

f(x � h) � f(x)

h

THE DERIVATIVE


The Derivative of a Function � The derivative of the functionf(x) with respect to x is the function f �(x) (read as “f prime of x”) given by

and the process of computing the derivative is called differentiation. We saythat f(x) is differentiable at c if f �(c) exists (that is, if the limit of the differ-ence quotient exists when x � c).

f�(x) � limhfi 0

f(x � h) � f(x)

h


Thus, the slope of the tangent line to the curve y � x3 at the point where x ��1 is f �(�1) � 3(�1)2 � 3 (Figure 2.4). To find an equation for the tangent line,we also need the y coordinate of the point of tangency; namely, y � (�1)3 � �1.Therefore, the tangent line passes through the point (�1, �1) with slope 3. By apply-ing the point-slope formula, we get

or

A manufacturer estimates that when x units of a certain commodity are produced andsold, the revenue derived will be R(x) � 0.5x2 � 3x � 2 thousand dollars. At whatrate is the revenue changing with respect to the level of production x when 3 unitsare being produced? Is the revenue increasing or decreasing at this time?

SolutionFirst, since x represents the number of units produced, we must have x � 0. The dif-ference quotient of R(x) is

Thus, the derivative of R(x) is

and since

R�(3) � (3) � 3 � 6

it follows that revenue is changing at the rate of $6,000 per unit with respect to thelevel of production when 3 units are being produced.

R�(x) � limhfi 0

R(x � h) � R(x)

h� lim

hfi 0 (x � 0.5h � 3) � x � 3

�xh � 0.5h2 � 3h

h� x � 0.5h � 3

R(x � h) � R(x)

h�

[0.5(x2 � 2xh � h2) � 3(x � h) � 2] � [0.5x2 � 3x � 2]

h

y � 3x � 2

y � (�1) � 3[x � (�1)]

� 3x2

� limhfi 0

(x3 � 3x2h � 3xh2 � h3) � x3

h� lim

hfi 0 (3x2 � 3xh � h2)

f�(x) � lim hfi 0

f(x � h) � f(x)

h� lim

hfi 0

(x � h)3 � x3

h


y

y = x3

x(–1, –1)

FIGURE 2.4 The graph of y � x3.


R (thousands of dollars)

x(units

produced)3

FIGURE 2.5 The graph of R (x) �0.5x2 � 3x � 2, for x � 0.

Since R�(3) � 6 is positive, the tangent line at the point on the graph of the rev-enue function where x � 3 must be sloped upward. This observation suggests thatrevenue is increasing when x � 3, as confirmed by the graph of R(x) shown in Figure2.5.

The derivative f�(x) of y � f(x) is sometimes written as (read as “dee y, dee x”),

and in this notation, the value of the derivative at x � c (that is, f �(c)) is written as

For example, if y � x2, then

and the value of this derivative at x � �3 is

The notation for derivative suggests slope, , and can also be thought of as

“the rate of change of y with respect to x.” Sometimes it is convenient to condense a statement such as

“when y � x2, then ”

by writing simply

which reads, “the derivative of x2 with respect to x is 2x.”

The following example illustrates how the different notational forms for the deriv-ative can be used.

First compute the derivative of f(x) � , then use it to(a) Find the equation of the tangent line to the curve y � at the point where

x � 4.�x

�x

d

dx(x2) � 2x

dy

dx� 2x

�y

�x

dy

dx

dy

dx�x��3� 2x�x��3

� 2(�3) � �6

dy

dx� 2x

dy

dx�x�c

dy

dxDERIVATIVE NOTATION



E x p l o r e !E x p l o r e !Many graphing calculators

have a special utility for com-

puting derivatives numerically,

called the numerical derivative

(nDeriv). It can be accessed via

the MATH key. This derivative

can also be accessed through the

CALC (second TRACE) key, espe-

cially if a graphical presentation

is desired. For instance, store

f(x) � into Y1 of the equa-

tion editor and display its graph

using a decimal window. Use

the option of the CALC key

and observe the numerical de-

rivative value at x � 1.

dy

dx

�x

(b) Find the rate at which y � is changing with respect to x when x � 1.

SolutionThe derivative of y � with respect to x is given by

(a) When x � 4, the corresponding y coordinate on the graph of f(x) � is y �

� 2, so the point of tangency is P(4, 2). Since f �(x) � , the slope of the

tangent line to the graph of f(x) at the point P(4, 2) is given by

and by substituting into the point-slope formula, we find that the equation of thetangent line at P is

or

(b) The rate of change of y � when x � 1 is

If a function f(x) is differentiable at the point P(x0, f(x0)), then the graph of y � f(x)has a nonvertical tangent line at P and at all points “near” P. Intuitively, this suggeststhat a function must be continuous at any point where it is differentiable, since a graphcannot have a “hole” or “gap” at any point where a well-defined tangent can be drawn.

DIFFERENTIABILITYAND CONTINUITY

dy

dx�x�1�

1

2�1�

1

2

�x

y �1

4x � 1

y � 2 �1

4(x � 4)

f�(4) �1

2�4�

1

4

1

2�x�4

�x

� limhfi 0

1

�x � h � �x�

1

2�x

� limhfi 0

x � h � x

h(�x � h � �x)� lim

hfi 0

h

h(�x � h � �x)

� limhfi 0

(�x � h � �x)(�x � h � �x)

h(�x � h � �x)

d

dx ��x� � lim

hfi 0 f(x � h) � f(x)

h� lim

hfi 0 �x � h � �x

h

�x

�x


The converse, however, is not true; that is, a continuous function need not beeverywhere differentiable. For instance, consider the three graphs shown in Figure2.6. The first has a gap at x � 0, and certainly has no tangent there. The graphs inFigures 2.6b and 2.6c are continuous at x � 0. In both cases, however, there are sharppoints at (0, 0) (a “corner” in Figure 2.6b and a “cusp” in Figure 2.6c), which pre-vent the construction of a well-defined tangent line there.

In general, the functions you encounter in this text will be differentiable at almostall points. In particular, polynomials are everywhere differentiable and rational func-tions are differentiable wherever they are defined.

In Problems 1 through 8, compute the derivative of the given function and find theslope of the line that is tangent to its graph for the specified value of the independentvariable.

1. f(x) � 5x � 3; x � 2 2. f(x) � x2 � 1; x � �1

3. f(x) � 2x2 � 3x � 5; x � 0 4. f(x) � x3 � 1; x � 2

5. g(t) � 6. f(x) � ; x � 2

7. f(x) � ; x � 9 8. h(u) � ; u � 4

In Problems 9 through 12, compute the derivative of the given function and find theequation of the line that is tangent to its graph for the specified value of x0.

9. f(x) � x2 � x � 1; x0 � 2 10. f(x) � x3 � x; x0 � �2

11. f(x) � 12. f(x) � ; x0 � 42�x3

x2 ; x0 �1

2

1

�u�x

1

x2

2

t; t �

1

2

FIGURE 2.6 Three functions that are not differentiable at (0, 0). (a) The graph has a gap at x � 0. (b) There is a sharp “corner” at (0, 0). (c) There is a “cusp” at (0, 0).

0

y = x2/3

x

y(c)

x

y

0

y = |x |

(b)

y =1x

0x

y(a)


P . R . O . B . L . E . M . S 2.1P . R . O . B . L . E . M . S 2.1

E x p l o r e !E x p l o r e !Store f(x) � abs(X) into Y1 of

the equation editor. The ab-

solute value function can be ob-

tained through the MATH key by

accessing the NUM menu. Use a

decimal window and compute

the numerical derivative at

x � 0. What do you observe and

how does this answer reconcile

with Figure 2.6(b)? The curve

is too sharp at the point (0, 0) to

possess a well-defined tangent

line there. Hence, the derivative

of f(x) � abs(x) does not exist at

x � 0. Note that the numerical

derivative must be used with

caution at cusps and unusual

points. Try computing the

numerical derivative of y � at

x � 0 and explaining how such

a result could occur numeri-

cally.

1

x

dy

dx

In Problems 13 through 16, find the rate of change where x � x0.

13. y � 3; x0 � 2 14. y � 6 � 2x; x0 � 3

15. y � x(1 � x); x0 � �1 16. y � ; x0 � �3

17. Suppose f(x) � x3.(a) Compute the slope of the secant line joining the points on the graph of f whose

x coordinates are x � 1 and x � 1.1.(b) Use calculus to compute the slope of the line that is tangent to the graph when

x � 1 and compare this slope with your answer in part (a).

18. Suppose f(x) � x2.(a) Compute the slope of the secant line joining the points on the graph of f whose

x coordinates are x � �2 and x � �1.9.(b) Use calculus to compute the slope of the line that is tangent to the graph when

x � �2 and compare this slope with your answer in part (a).

In Problems 19 and 20, sketch the graph of the function f(x). Determine the values ofx for which the derivative is zero. What happens to the graph at the corresponding

points?

19. f(x) � x3 � 3x2 20. f(x) � x3 � x2

21. In Example 4.5 of Chapter 1, we obtained the profit function P(x) � 400(15 � x)(x � 2) for the production of high-grade blank videocassettes. The graph of y �P(x) is the downward opening parabola shown in the accompanying figure.(a) Find P�(x).(b) Find where P�(x) � 0. This is where the graph of the profit function has a

horizontal tangent. What can be said about the profit at the corresponding valueof x?

22. Sketch the graph of the function y � x2 � 3x and use calculus to find its lowestpoint.

23. Sketch the graph of the function y � 1 � x2 and use calculus to find its highestpoint.

MAXIMIZATION OF PROFIT 24. A manufacturer can produce tape recorders at a cost of $20 apiece. It is estimatedthat if the tape recorders are sold for x dollars apiece, consumers will buy 120 � xof them each month. Use calculus to determine the price at which the manufac-turer’s profit will be the greatest.

ANIMAL BEHAVIOR 25. Experiments indicate that when a flea jumps, its height (in meters) after t secondsis given by the function

H(t) � (4.4)t � (4.9)t2

�1 � x

dy

dx


P

x152

PROBLEM 21

Using calculus, determine the time at which the flea will be at the top of its jump.What is the maximum height reached by the flea?

RENEWABLE RESOURCES 26. The accompanying graph shows how the volume of lumber V in a tree varies withtime t (the age of the tree). Use the graph to estimate the rate at which V is chang-ing with respect to time when t � 30 years. What seems to be happening to therate of change of V as t increases without bound (that is, in the “long run”)?

27. (a) Find the derivative of the linear function f(x) � 3x � 2.(b) Find the equation of the tangent line to the graph of this function at the point

where x � �1.(c) Explain how the answers to parts (a) and (b) could have been obtained from

geometric considerations with no calculation whatsoever.

28. (a) Find the derivatives of the functions y � x2 and y � x2 � 3 and account geo-metrically for their similarity.

(b) Without further computation, find the derivative of the function y � x2 � 5.

29. (a) Find the derivative of the function y � x2 � 3x.(b) Find the derivatives of the functions y � x2 and y � 3x separately.(c) How is the derivative in part (a) related to those in part (b)?(d) In general, if f(x) � g(x) � h(x), what would you guess is the relationship

between the derivative of f and those of g and h?

30. (a) Compute the derivatives of the functions y � x2 and y � x3.(b) Examine your answers in part (a). Can you detect a pattern? What do you

think is the derivative of y � x4? How about the derivative of y � x27?

31. Explain why the graph of a function f(x) is rising over an interval a � x � b iff �(x) 0 throughout the interval. What can you say about the graph if f�(x) 0throughout the interval a � x � b?

In Problems 32 and 33, sketch the graph of a function f that has all of the givenproperties. You may need to refer to the result of Problem 31.

32. (a) f �(x) 0 when x 1 and when x 5(b) f �(x) 0 when 1 x 5(c) f �(1) � 0 and f �(5) � 0

33. (a) f �(x) 0 when x �2 and when �2 x 3(b) f �(x) 0 when x 3(c) f �(�2) � 0 and f �(3) � 0


Volume of lumber V (units)

302010

605040

10 20 30 40 50 60Time (years)

t0

PROBLEM 26 Graph showinghow the volume of lumber V in atree varies with time t.Source: Adapted from Robert H.Frank, Microeconomics andBehavior, 2nd ed., New York, NY:McGraw-Hill, Inc., 1994, page 623.

UNEMPLOYMENT 34. In economics, the graph in Figure 2.3 is called the Phillips curve, after A. W.Phillips, a New Zealander associated with the London School of Economics. Until Phillips published his ideas in the 1950s, many economists believed thatunemployment and inflation were linearly related. Read an article on the Phillipscurve (the source cited with Example 1.1 would be a good place to start) and writea paragraph on the nature of unemployment in the U.S. economy.

35. Find the slope of the line that is tangent to the graph of the function f(x) �

at the point where x � 3.85 by filling in the following chart.Record all calculations using five decimal places.

36. Find the x values at which the peaks and valleys of the graph of y � 2x3 �0.8x2 � 4 occur. Use four decimal places.

37. Show that f(x) � is not differentiable at x � 1.

If we had to use the limit definition every time we wanted to compute a derivative,it would be both tedious and difficult to use calculus in applications. Fortunately, thisis not necessary, and in this section and the next, we develop techniques that greatlysimplify the process of differentiation. We begin with a rule for the derivative of aconstant.

You can see this by considering the graph of a constant function f(x) � c, whichis a horizontal line (see Figure 2.7). Since the slope of such a line is 0 at all its points,it follows that f �(x) � 0. Here is a proof using the limit definition:

�x2 � 1�x � 1

�x �0.02 �0.01 �0.001 —0— 0.001 0.01 0.02

x � �x

f(x)

f(x � �x)

f(x � �x) � f(x)

�x

�x2 � 2x � �3x

Chapter 2 � Section 2 Techniques of Differentiation 109

Techniques ofDifferentiation

2

The Constant Rule � For any constant c,

That is, the derivative of a constant is zero.

d

dx(c) � 0

y

x

y = c

Slope 0

FIGURE 2.7 The graph of f (x) � c.

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