Calculus and Economics

Microeconomics with Calculus: Tutorial #1

Calculus and Economics

Daniel S. ChristiansenALBION COLLEGE

[email protected] 14, 2002

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c©Copyright 1999–2002 Daniel S. Christiansen

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http://www.albion.edu/econ/christiansen/micwcalc/tut0.pdf

Table of Contents

1 Introduction 4

2 Using Calculus to Learn Economics 7

3 The Notion of a Derivative 103.1 The Mathematics of a Derivative . . . . . . . . . . . . 103.2 The Geometry of a Derivative . . . . . . . . . . . . . . 133.3 The Economics of a Derivative . . . . . . . . . . . . . 183.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 22

4 Derivatives of a Few Simple Functions 244.1 Power Functions . . . . . . . . . . . . . . . . . . . . . 244.2 Exponential and Logarithmic Functions . . . . . . . . 274.3 Section Quiz . . . . . . . . . . . . . . . . . . . . . . . 29

5 Rules of Differentiation 315.1 Sums, Differences, and Constants . . . . . . . . . . . . 325.2 Product Rule . . . . . . . . . . . . . . . . . . . . . . . 33

Table of Contents (cont.) 3

5.3 Quotient Rule . . . . . . . . . . . . . . . . . . . . . . . 355.4 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . 365.5 Inverses . . . . . . . . . . . . . . . . . . . . . . . . . . 405.6 Section Quiz . . . . . . . . . . . . . . . . . . . . . . . 42

6 Optimization 446.1 Maximization and Minimization . . . . . . . . . . . . . 456.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 51

7 A Few Words about Integration 52

8 Chapter Quiz 56

Answers to Quizzes 60

Solutions to Exercises 61

References 68

Section 1: Introduction 4

1 IntroductionThe idea of this project is to present an intermediate-level coursein microeconomic theory with the help of some simple notions fromcalculus. There are two distinguishing features. The first is that eachchapter comes in two versions — one designed for the printed page(the paper version) and one intended to be read on the computerscreen (the web version). The paper version comes in the form of atraditional textbook. The web version presents the same material buthas interactive features, for example quizzes that can be corrected onthe spot. Furthermore, all of the material is hyperlinked, so readingdoes not have to be done in a linear manner from beginning to end.This is useful in leaving and then returning to figures, equations,solutions to exercises, and other points of reference. It is also usefulin linking to other documents or web sites.

The other distinguishing feature is the way calculus is used to de-velop microeconomic theory. We presume a minimal background incalculus — just one course. You will surely benefit from having takenmore than one calculus course, and we eventually use some concepts


and techniques from more advanced courses. But when we do so, wedevelop what we need with the assumption that you have not seen itbefore. What is truly unique about this approach is the order in whichthe economics is presented. Since you already know about calculuswith one variable, we first present that portion of microeconomic the-ory that can be handled with this set of mathematical tools. It turnsout that we can get a lot of mileage out of one-variable calculus andthat we can cover almost all of the basic topics from principles of mi-croeconomics. At a later point we introduce partial derivatives andconstrained optimization and take up the topic of consumer theory.It is with consumer theory that almost every other intermediate mi-croeconomics textbook begins. But this is not a good place to startusing calculus with the background we have assumed.

Thus the prerequisites for following this text are one course inprinciples of microeconomics and one course in calculus.

If you should require further background in microeconomics at theprinciples level, almost any principles of economics textbook shouldprovide what you need. The book by Mankiw [4], for example, isa popular current textbook that provides a good grounding in basic


economics.1

The best background in mathematics is the most rigorous calculuscourse you can find — the kind a serious mathematician would like tooffer you. If you are looking to make up deficiencies in your trainingor simply looking for review, this advice is not likely to be timely. Inthis case any calculus textbook should be able to serve as referencematerial for the mathematics used here. One possibility is the bookby Hughes-Hallett [2]. Another good source of help is Donald P.Story’s e-calculus web site [6], an interactive tutorial for a first coursein calculus.2 Still another possibility is the old but classic CalculusMade Simple [7] written by Silvanus P. Thompson in 1910.3

1In particular, see Mankiw’s Supply and Demand I: How Markets Work, Supplyand Demand II: Markets and Welfare, and Firm Behavior and the Organizationof Industry.

2The e-calculus web site inspired this project, and Story’s exerquiz and web

packages, together with the mathematical typesetting system LATEX, made it pos-sible.

3Thompson’s book has the long and suggestive subtitle, Being a Very SimplestIntroduction to those Beautiful Methods of Reckoning Which are Generally Calledby the Terrifying Names of the Differential Calculus and the Integral Calculus.

http://www.math.uakron.edu/~dpstory/e-calculus.html

Section 2: Using Calculus to Learn Economics 7

There are many good textbooks on microeconomic theory at theintermediate level, among them those by Landsburg [3] and Varian [9].Neither of these employs calculus except in the appendices. Twobooks that use calculus are Binger and Hoffman [1] and Nicholson [5].You need to be aware that in each of these textbooks the material ispresented in an order different from that of this book.

In this chapter we review the concepts from calculus that areneeded to read the chapters that make up the first part of this project,Decisions and Markets: Economics with One Variable.

2 Using Calculus to Learn EconomicsIt is a premise of this book that a calculus approach to the studyof microeconomics pays large dividends. There are several relatedreasons why this is so.

First, the use of mathematics helps develop problem-solving skills.It imposes a rigor that mandates keeping track of costs and benefits,and it provides a framework for determining which variables and pa-


rameters are important. We apply calculus and other mathematics toa wide variety of problems. Our view is that you can’t learn economicswithout solving problems.

Second, the calculus approach helps in learning to think clearly.You will be forced to translate weak verbal arguments into precise,consistent statements. Making a statement formal clarifies what isbeing said. When you are called upon to verify what you think isobvious, you may discover that it is not obvious or perhaps not eventrue.

Third, the use of calculus unifies the material by focusing on thecommon economic structure of problems. When we strip the specificdetails away, many problems look surprisingly alike and have com-mon solutions. More generally, this is an argument for the power ofabstraction. Calculus makes it easier, not harder, to learn economics.There are, in fact, only a small number of basic ideas in intermediatemicroeconomics; much of what we do is working out the details forparticular situations.

Fourth, calculus helps in becoming literate in the language of mod-ern economics. For better or worse, mathematics is the language of


economics. If you want to read the economic literature or get anadvanced degree in economics, you will have to deal with the mathe-matics. The best business schools demand mathematical maturity oftheir students; and if you want to undertake a serious study of finance,for example, you will find that calculus is only the beginning.

Fifth, this approach provides the opportunity to apply mathemat-ics to social science problems. Whereas many of the examples in afirst course in calculus come from physics, there are also examplesfrom economics that illustrate most of the same points. This showsthe power of calculus and aids in its understanding.

Sixth, this approach allows us to take the subject to a more ad-vanced point and thus gives a better appreciation of the power of eco-nomic theory. Economics is ultimately a policy science, and theoryplays an essential role. Hal Varian [8] has some enlightening answersto the question, “What use is economic theory?” Among other things,he suggests that theory is useful as a substitute for data that we needbut do not have; and he explains how theory can generate useful in-sights in explaining economic phenomena, even when the theory isonly approximately true. We will see examples of these insights as we

Section 3: The Notion of a Derivative 10

proceed.

3 The Notion of a DerivativeThe concept of a derivative plays a central role in this book. Wereview this notion from the perspective of mathematics, geometry,and economics.

3.1 The Mathematics of a Derivative

Consider an independent variable x and a dependent variable y, sothat y is a function of x. We choose to write this relationship asy = y(x). We have to be a bit careful because we are using theletter y in two different ways; it stands both for the value that y takesand for the function itself.

We discuss what is meant by the derivative of the function y at aparticular point x1. For another point x2 close to x1, let ∆x = x2−x1

be the change in x and let ∆y = y(x2)− y(x1) be the correspondingchange in y. Then the derivative of y with respect to x at the


point x1 is defined to be the limit of the ratio of the change in y tothe change in x, i.e.,

lim∆x→0

y(x1 + ∆x)− y(x1)∆x

.

If y is a linear function, y = a + bx, then the ratio of the changein y to the corresponding change in x is always the same and is equalto b. This is true regardless of the size of the change in x or thebeginning point x1. Thus the derivative of the function y = a + bx isthe constant b.

In principle, the limiting operation above could be used to calcu-late the derivative of an arbitrary function.4 For our purposes this isnot necessary; instead we presume that this has been done to gener-ate the derivatives of a few simple functions as given in Section 4 andsome rules of differentiation as given in Section 5. It is important toobserve that the derivative is typically not a constant (as is the case

4There are some technical issues related to whether or not the limit, and hencethe derivative, exists. We put these aside.


for a linear function) but depends on the value of x at which we arecalculating it.

We need notation for the derivative of y with respect to x. Thetwo most common ways of writing it are either as dy/dx or as y′. Weoften want to indicate where the derivative is being evaluated. Thederivative of the function y at the point x1 may be denoted either as

dy

dx

∣∣∣∣x=x1

or y′(x1).

We use both the dy/dx and y′(x) notation, freely switching from oneto the other.

Since the derivative of y with respect to x is a function of x, we canalso work with its derivative, the second derivative of y with respectto x. This is written as d2y/dx2 or as y′′. As the second derivative isalso a function of x, we can write its value at the point x1 in either ofthe following ways:

d2y

dx2

∣∣∣∣x=x1

or y′′(x1).


3.2 The Geometry of a Derivative

We have seen that, for a linear function y = a + bx, the derivative isequal to the slope b. If both a and b are positive, then Figure 1 showsthe function y on the left and the derivative y′ or dy/dx on the right.The first thing to note is that to do this right, two separate graphsare required. This is because the units of the function and the unitsof the derivative are not the same. This is evidenced by the mostcommon example used in calculus courses: distance y, measured infeet, depends on time x, measured in seconds. The derivative, velocity,is measured in feet per second.5 In general the units of dy/dx are thoseof y/x (and not those of y). Since the derivative is everywhere equalto b, a positive constant, its graph appears on the right as a horizontalline. The second derivative (not shown) is constant at zero.

5One could, of course, plot distance and time on the same axis, placing 30 feetat the same point as 30 feet per second. But this leads to nonsensical comparisonsbetween distance and velocity: is 20 feet per second more or less than 30 feet?To complicate the matter, the answer changes when we consider speed in feet perminute.


x

y

y

a

x

y′

y′b

Figure 1: (Left) A linear function y = a + bx. (Right) The derivative dy/dx

or y′ of the same function, which is the constant b.

The geometric interpretation is based on identifying the derivativeof a function with its slope. If the function is linear, it is clear what wemean by its slope. But what is meant by the slope of any arbitraryfunction or curve? Given a point on its graph and another pointnearby, we can take the ratio of the change in y to the change in x.If we let the two points become closer and closer together, then inthe limit we obtain the slope at the point. Geometrically, this is the


x

y

y

x

y′

y′

Figure 2: (Left) A function y(x) with positive and increasing derivative.

(Right) The derivative y′(x) of the same function.

tangent to the function. This is the same limiting operation we usedin finding the derivative mathematically.

For various shapes of the function y = y(x), we plot both thefunction and its derivative. This has already been done for a linearfunction in Figure 1.

Students often confuse the words “positive” and “increasing.” Thelinear function y on the left of Figure 1 is positive (i.e., it lies above


the x axis) and increasing (i.e., it slopes up). Its slope, the derivative,is plotted on the right. The slope is positive, but it is not increasing.(The slope as a function of x is constant!)

Consider next the function y(x) shown on the left of Figure 2. As xincreases the slope gets larger. Thus y′, plotted on the right, mustincrease as x increases. Here we can say that the function y is positiveand increasing. The derivative is also positive and increasing. If thederivative is linear, as we have drawn it, then the second derivative(not shown) is a positive constant.

Now examine the function y(x) shown on the left of Figure 3.As x increases the slope gets smaller. The slope, plotted on the right,must thus decrease as x increases. Again the function y is positiveand increasing. But its derivative is positive and decreasing. If thederivative is linear, as we have drawn it, then the second derivative(not shown) is a negative constant.

Finally look at the function y(x) shown on the left of Figure 4.For the values of x shown, the function is always positive. But itfirst increases and then decreases. Thus the derivative, shown onthe right, is positive at first but after a point becomes negative. It is


x

y

y

x

y′

y′

Figure 3: (Left) A function y(x) with positive but decreasing derivative.


always decreasing. If it is a linear function, then the second derivativemust be a negative constant.

The graph on the left of Figure 2 is an example of a convex function(one for which d2y/dx2 is positive or zero) and those on the left ofFigures 3 and 4 are concave functions (ones for which d2y/dx2 isnegative or zero).


x

y

y

x

y′

0

y′

Figure 4: (Left) A function y(x) with derivative first positive and then negative.


3.3 The Economics of a Derivative

In economics a derivative typically represents a marginal concept. Ifbenefits (B) and costs (C) depend on the level of an activity (x), asin Chapter 2, then the derivative of B with respect to x representsmarginal benefit and the derivative of C with respect to x representsmarginal cost. If revenue depends on the quantity of a good sold, thenthe derivative of revenue with respect to quantity represents marginal


revenue. If production depends on labor input, then the derivativeof production with respect to labor input represents the marginalproduct of labor. If resource cost depends on labor input, then thederivative of resource cost with respect to labor input represents themarginal resource cost of labor. We will see many such examplesthroughout the book.

We examine one of these concepts, marginal cost (MC), morecarefully. In a first course in economics, such an expression is usuallydefined as MC = ∆C/∆x and represents the cost of producing onemore (or the last) unit. In the formula it is understood that x takes ononly the discrete values 1, 2, 3, etc. This is precisely the point of viewthat must be taken, at least in principle, when a good is indivisible.Supertankers and dams, for example, come only in whole units.

Seeing the formula for marginal cost in the previous paragraph,students sometime wonder why they must divide by the change in x.The reason is to obtain the extra cost per unit of x. If $10 of extracost is incurred in producing two more bushels of output, then themarginal cost is $5 per bushel.

We will assume that the variable x is continuous, i.e., that it can


take on any real value (or at least any nonnegative value) and not just1, 2, 3, etc. This is appropriate when we want to measure, in pounds,the amount of cheese produced in a day. Although commercial scalesmay not be capable of measuring cheese to the one-millionth of apound, we are not misled by presuming that we could do so. Evenwhen we deal with goods such as automobiles that are less homoge-neous than cheese, we can still make sense of requiring a productionof 61.5 automobiles per day if we interpret that to mean 123 automo-biles over a two day production run. The advantage of working witha continuous variable is that it greatly simplifies the mathematics.

For a continuous variable x the appropriate definition of marginalcost is as the derivative of the cost function: MC = dC/dx or MC =C ′(x). An advantage of using the C ′(x) notation is that it emphasizesthat marginal cost is itself a function of x.

In a continuous setting we often speak loosely of marginal costrepresenting the extra cost incurred by increasing x by one unit orthe cost saved by decreasing it by one unit. We should rememberthat, strictly speaking, the kinds of increments we are dealing withare infinitesimal ones. But the units are still those of rates: dollars


per bushel, dollars per pound, dollars per automobile, etc.We reexamine the graphs in Figures 1-4 from a total cost and

marginal cost perspective. We consider whether the figure on the leftmight represent a cost function. For concreteness, let us suppose thatwe are examining the cost of producing cereal. Then x is measuredin boxes of cereal per month and y is measured in dollars per month.Marginal cost, pictured on the right, is then measured in dollars perbox of cereal.

The graph on the left of Figure 1 represents a linear cost functionwhere the cost goes up by the same amount for each extra box ofcereal produced. This is another way of saying that marginal costis constant. In these examples there is some fixed cost as well. Thegraph on the left of Figure 2 represents a cost function where themarginal cost gets larger as more boxes are produced. The graph onthe left of Figure 3 shows a situation where marginal cost falls as moreboxes are produced.

The graph on the left of Figure 4 is not a good candidate for acost function. The problem is that, after a point, cost goes down asmore is produced. While it might be appropriate for the marginal


cost of producing cereal to fall in some situations, it is not reasonableto think that total cost would go down as more boxes of cereal areproduced. This would mean that marginal cost, the cost of producingan extra box of cereal, is negative.

3.4 Exercises

Exercise 3.1. If the amount of taxes paid (T ) depends on income (x),how would you use calculus notation to describe the marginal tax rate?If taxes and income are both measured in dollars per year, what arethe units of the marginal tax rate?

Exercise 3.2. Consider a function y(x) as shown in Figure 5. Drawa graph to show the derivative y′(x). Which of the words “increas-ing,” “decreasing,” “positive,” and “negative” apply to the function y?Which of these terms apply to the derivative y′? Which of these termsapply to the second derivative y′′?


x

y

y

Figure 5: See Exercises 3.2 and 3.3.

Exercise 3.3. If the function plotted in Figure 5 is a cost function,what can you say about the shape of marginal cost?

Exercise 3.4. If the derivative y′ of a function y is a straight line,can you be sure that the function y is also a straight line? Explain.

Exercise 3.5. Geometrically, the first derivative of a function at apoint measures its slope. What does the second derivative measure?

Section 4: Derivatives of a Few Simple Functions 24

4 Derivatives of a Few Simple FunctionsWe can accomplish most of our goals by working with derivatives ofonly a few functions. The process of taking a derivative is calleddifferentiation. There are three kinds of functions that we needto be able to differentiate: power functions such as x2, exponentialfunctions such as ex, and logarithmic functions such as lnx.

4.1 Power Functions

A power function is a variable raised to a constant power: y = xa

where a is a constant. From the definition of a derivative, it can beshown that the derivative of a power function is given by

dxa

dx= axa−1. (1)

The formula given in equation (1) covers more functions than youmight initially think. First, the formula works when a is a positive


integer. Thus

dx2

dx= 2x,

dx3

dx= 3x2, and

dx10

dx= 10x9.

For the special case a = 1, the derivative of x is the constant 1(since x0 = 1).

The formula is also valid when a is a negative integer. You shouldmake sure you understand what is meant by a variable x raised toa negative power: it is the same as one divided by x raised to thecorresponding positive power. Thus

x−2 =1x2

x−3 =1x3

and x−10 =1

x10.

If we apply the formula from equation (1) to the three functions abovewe find thatdx−2

dx= −2x−3,

dx−3

dx= −3x−4, and

dx−10

dx= −10x−11.

Given the meaning of negative exponents, we can write these as

d(1/x2

)dx

= − 2x3

,d

(1/x3

)dx

= − 3x4

, andd

(1/x10

)dx

= − 10x11

.


The case a = −1 is worth special attention:

dx−1

dx= −x−2.

This is the same as writing

d (1/x)dx

= − 1x2

.

The same formula works, in fact, for any real number a. Forexample, it works for a = 1

2 , a = 13 , and a = − 1

2 . The meaning ofxm/n, when m and n are integers, is the nth root of xm. Thus x1/2

is the same as√

x, x1/3 is the same as 3√

x, and x−1/2 is the same as1/√

x.Once again we can use equation (1) to calculate derivatives:

dx1/2

dx=

12x−1/2 or

d√

x

dx=

12√

x

anddx1/3

dx=

13x−2/3 or

d 3√

x

dx=

13 3√

x2.


Similarly,

dx−1/2

dx= −1

2x−3/2 or

d (1/√

x)dx

= − 12 2√

x3.

4.2 Exponential and Logarithmic Functions

In a first course in calculus, you discussed the constant e = 2.718 . . ..Two functions are built from from this fundamental mathematicalconstant. The first is the exponential function y = ex. Generalizationsof this simple function are important in economics because they allowfor constant proportional rates of growth. Thus they appear naturallywhen dealing with interest rates and other growth rates over time.

The derivative of ex is especially easy to remember as it is a func-tion whose derivative is the function you start with:

dex

dx= ex. (2)


x

y

y = ln x

y = ex

Figure 6: The exponential function y = ex and the logarithmic function y =

ln x. The two functions are inverses of one another. (As such each is the

reflection of the other around the 45◦ line, the dotted line in the graph.)

The second function is the logarithm to the base e of the variable x,written as y = ln x. The derivative of the logarithmic function y = ln xis

d lnx

dx=

1x

. (3)


The two functions ex and lnx, plotted in Figure 6, are closelyrelated as they are inverses of one another. This means, for example,that if y = ln x, then x = ey. Geometrically it means that one functionis transformed into the other by rotating about the 45 degree line.

Note that ex is equal to one when x = 0 and is well-defined fornegative values of x, where it is less than one but greater than zero.The function ln x is zero when x = 1 and is negative when x is betweenzero and one. For zero and negative real numbers, however, lnx isnot defined. Note also that the exponential function is convex whilethe logarithmic function is concave.

4.3 Section Quiz

Click on the “Begin Quiz” button to begin, choose your answers, clickon the “End Quiz” button to see how many questions you answeredcorrectly, and click on the “Correct” button to see the correct answers.

1. If y = x4 then dy/dx is equal to


4x4 4x3 3x4 3x3

2. If y = x−4 then dy/dx is equal to

−4x−3 −4x−4 −4x−5 −4x

3. If y = x3/2 then dy/dx is equal to32x1/2 3

2x−1/2 32x2/3 3

2x−2/3

4. If y = ex then dy/dx is equal to

x e ex lnx

5. If y = ln x then dy/dx is equal to

ex x 1/x√

x

6. If y = x−6 then dy/dx is equal to

−6x−7 −6x−5 −x−7 −x−5

7. If y =√

x then dy/dx is equal to√

x√

x/2 2√

x 1/ (2√

x)

8. If y = x.8 then dy/dx is equal to

.8x.2 .8x−.2 .8x−.8 .2x.8

Section 5: Rules of Differentiation 31

9. If y = x−1.6 then dy/dx is equal to

1.6x−2.6 1.6x−.6 −1.6x−2.6 −1.6x−.6

10. If y = 1/x then dy/dx is equal to

1/x 1/x2 −1/x −1/x2

5 Rules of DifferentiationNew functions can be made out of the simple functions introduced inSection 4 by combining them in obvious ways. For example, we mightadd or multiply two functions together. We need to discuss how todifferentiate these new functions.


5.1 Sums, Differences, and Constants

Suppose we have two functions, y1(x) and y2(x). We can constructnew functions by adding the two together and by subtracting one fromthe other. The derivative of the sum of the two functions is the sum ofthe derivatives. Thus the derivative of y1(x) + y2(x) is y′1(x) + y′2(x).For example, the derivative of x2 + x3 is 2x + 3x2. Likewise, thederivative of the difference of the two functions is the difference of thederivatives. Thus the derivative of y1(x)− y2(x) is y′1(x)− y′2(x). Forexample, the derivative of x2 − x3 is 2x− 3x2.

Constant functions are particularly easy to work with. The deriva-tive of a constant is zero. But the derivative of a constant times afunction is the constant times the derivative of the function. Thusif we have a function y = y(x) and a constant c, the function c hasderivative zero but the function cy(x) has derivative cy′(x). From therule for sums given above, the derivative of y(x)+c is just y′(x). Thusthe derivative of the function 5 is zero, the derivative of 5x2 is 10x,and the derivative of 5x2 + 5 is 10x.

Example 5.1. We calculate the first and second derivatives of the


function y(x) = x3 + 3x2 − 6x + 2. The first derivative is y′(x) =3x2 + 6x − 6. The second derivative is the derivative of the firstderivative. Thus y′′(x) = 6x + 6.

Example 5.2. We calculate the first and second derivatives of thefunction y(x) = x−1+lnx. The first derivative is y′(x) = −x−2+x−1,and the second derivative is y′′(x) = 2x−3 − x−2.

As economists we are interested in functions such as those we haveconstructed here. For example, we care about sums because we addthe actions of various participants together to get the market result.We work with differences when we construct profit as revenue minuscost. We face constant functions when we deal with fixed or overheadcost. And we see a constant times a variable when we constructrevenue as a price times quantity in cases where price is given.

5.2 Product Rule

The product rule applies to calculating the derivative of a functionobtained by multiplying two functions together when neither one is


necessarily a constant. The result is that the derivative of the productof two functions is the first times the derivative of the second plus thesecond times the derivative of the first. Thus if we have two functionsy1(x) and y2(x), we calculate the derivative of the product function,y(x) = y1(x) · y2(x), as follows:

y′(x) = y1(x) · y′2(x) + y2(x) · y′1(x)

or, using the dy/dx notation,

dy

dx= y1 ·

dy2

dx+ y2 ·

dy1

dx.

A legendary calculus teacher teaches this as the hi-ho rule because“the derivative of hi-ho equals hi-dee-ho plus ho-dee-hi.” (Herehi is the function y1(x), ho is the function y2(x), and dee refers totaking the derivative.)

Example 5.3. We calculate the derivative of the function y(x) = xex.The product rule is appropriate because the two functions x and ex aremultiplied together. Since the derivative of ex is ex and the derivativeof x is 1, we have that y′(x) = xex + ex.


Example 5.4. As another example we calculate the derivative ofy(x) = x2 ·x3. Using the product rule, y′(x) = x2 ·3x2 +x3 ·2x = 5x4.(Note that y(x) = x5 and, from our results in Section 4.1, its derivativemust be 5x4.)

Product functions will be of great interest to us because expen-diture is the product of price and quantity; and, in some situations,price is not constant.

5.3 Quotient Rule

The quotient rule applies when one function is divided by another.Suppose we have two functions y1(x) and y2(x). We consider the“quotient” function y(x) = y1(x)/y2(x). (We presume that we arenot dividing by zero.) The derivative of the quotient is given by

y′(x) =y2(x) · y′1(x)− y1(x) · y′2(x)

[y2(x)]2.


Example 5.5. Let y1(x) = ln x and y2(x) = x. We consider thequotient function y(x) = (lnx)/x. Then

y′(x) =x · (1/x)− lnx

x2=

1− lnx

x2.

Example 5.6. We calculate the derivative of the function y(x) =x3/x2 (which is equal to x and should have derivative 1):

y′(x) =x2 · 3x2 − x3 · 2x

x4= 1.

Average cost, which is total cost divided by quantity, is a quotient.We will want to differentiate this function early in Chapter 3.

5.4 Chain Rule

The chain rule applies when we have two functions with a connection:the independent variable of the first function is the dependent variableof the second.

Suppose we have two functions y = y(w) and w = w(x). Since ydepends on w and w depends on x, ultimately y depends on x. We


write this composite function as

y(x) = y(w(x)). (4)

We start with x, first apply the function w, and then apply the func-tion y to the result. We have to be a bit careful here because weare using the same letter y to denote the first function of w and thecomposite function of x. It should be clear that when we differentiatewith respect to w we are referring to the former function and whenwe differentiate with respect to x we are referring to the latter. Donot confuse the composite function with a product function.

The chain rule says that the derivative of the function in equa-tion (4) is equal to the product of the derivative of y with respect to wand the derivative of w with respect to x. We can write this using ourdy/dx notation as

dy

dx=

dy

dw· dw

dx(5)

and using our y′(x) notation as

y′(x) = y′(w(x)) · w′(x). (6)


Equation (5) has the advantage of being easy to remember. It appearsas if the dw terms cancel out. Equation (6) makes it very clear whereeach function is being evaluated.

Example 5.7. Consider the function y(x) = (1 + x2)10. This can bewritten as y = w10 with w = 1 + x2. Thus

dy

dx=

dw10

dw· d(1 + x2)

dx= 10w9 · 2x = 20x(1 + x2)9.

It is much easier to solve the problem this way than to expand theoriginal function and then differentiate. Notice that we must substi-tute for w so that the answer is in terms of x alone.

Example 5.8. Consider the function ex2. This can be written as

y = ew with w = x2. Thus

dy

dx=

dew

dw· dx2

dx= ew · 2x = 2xex2

.


Example 5.9. Consider the function y = 1/ lnx. Here y = 1/w withw = ln x. Then

dy

dx=

d(1/w)dw

· d lnx

dx= − 1

w2· 1x

= − 1x ln2 x

.

Applied to Example 5.7 the chain rule says that the derivative ofsomething to the 10th power is 10 times the something to the 9thpower times the derivative of the something. Applied to Example 5.8it says that the derivative of e raised to the something is e raisedto the something times the derivative of the something. Applied toExample 5.9 it says that the derivative of something to the minusone is minus something to the minus two times the derivative of thesomething. The most common mistake students make with the chainrule is forgetting to multiply by “the derivative of the something.”

We are interested in composite functions in economics so that wecan attribute revenues and costs to their ultimate sources. If, forexample, revenue depends on output and output depends on laborinput, then revenue ultimately depends on labor input. We will needto be able to differentiate such functions.


5.5 Inverses

The function y = y(x) is a rule that leads from any value of inde-pendent variable x to a unique value for the dependent variable y.The inverse function x = x(y) is a rule that goes the other direc-tion: for a given value of y it provides the unique value of x such thaty = y(x). We will restrict ourselves to working with cases where theinverse function is well-defined.6

The inverse function is quite different from the reciprocal of thefunction y(x), which is 1/y(x).7

As long as we avoid dividing by zero, the derivative of the inversefunction is one divided by the derivative of the original function:

dx

dy=

1dy/dx

or x′(y) =1

y′(x(y)). (7)

6We have to be careful because there may be more than one value of x associ-ated with each value of y. If y = x2, for example, then x = ±√y; i.e., there aretwo values of x associated with a given y. Thus x = x(y) is not well defined overall real numbers. If we look only at positive values of x, then there is no problem.

7Note that x(y(x)) = x; this involves composition of two function. Alterna-tively, y(x) times 1/y(x) equals one; this involves the product of two functions.


Again the first set of notation is the most suggestive: it appears thatthe dx and dy symbols can be manipulated as algebraic quantities.The second set of notation is much more precise about where deriva-tives are evaluated.

Example 5.10. If y = 100− 2x the inverse function is x = 50− 12y.

Differentiating the inverse function directly, we find that dx/dy =− 1

2 . Alternatively, using equation (7) and dy/dx = −2, we find thatdx/dy = − 1

2 .

Example 5.11. If we consider only positive values of x and y, thenthe function y(x) = x2 has inverse x(y) =

√y. Using equation (7)

dx

dy=

1dy/dx

=12x

=1

2√

y.

Notice that we substituted for x so that our answer was left in termsof y, the variable on which dx/dy depends.

Example 5.12. Recall from Section 4.2 that the logarithmic andexponential functions are inverses of one another so that if y = lnx


then x = ey. Hencedx

dy=

1dy/dx

=1

(1/x)= x = ey.

This is the same result we gave for the derivative of the exponentialfunction in Section 4.2.

In economics we frequently need to invert demand functions: in-stead of the quantity that can be sold at various prices, we will oftenbe interested in the price at which various quantities can be sold.

5.6 Section Quiz

Click on the “Begin Quiz” button to begin, choose your answers, clickon the “End Quiz” button to see how many questions you answeredcorrectly, and click on the “Correct” button to see the correct answers.

1. If y = x− x2 then dy/dx is equal to

1− x 1− 2x 2− x 2− 2x


2. If y = x lnx then dy/dx is equal to

1 + lnx x + lnx x +1x

1 +1x

3. If y = ex/x then dy/dx is equal toxex − ex

e2x

ex − xex

e2x

ex − xex

x2

xex − ex

x2

4. If y =√

x/x then dy/dx is equal to

− 32x−3/2 − 1

2x−1/2 − 32x−1/2 − 1

2x−3/2

5. If y = ln x2 then dy/dx is equal to

2/x2 2/x 1/x2 1/x

6. If y = (1 + 2x)5 then dy/dx is equal to

5(1 + x)4 5(1 + 2x)4 10(1 + 2x)4 10(1 + x)4

7. If y =√

10 + x2 then dy/dx is equal tox√

10 + x2

2√10 + x2

2x√10 + x2

x

2√

10 + x2

8. What is the second derivative of x−3/2?

Section 6: Optimization 44

− 35x5/2

154x7/2

3x5/2

34x1/2

9. If y = w2 and w = 3x+4, the chain rule can be used to calculatethe derivative of

9x2 + 4 x2(3x + 4) 3x2 + 4 (3x + 4)2

10. If y = 3x + 5, calculate dx/dy.

1/3 3 3x 3y

6 OptimizationOptimization is one of the basic analytical techniques employed inmicroeconomics. We will get lots of mileage out of the idea that thedecisions of economic agents (households, firms, social planners, etc.)are the result of these parties trying to do the best they can subjectto the constraints they face.


6.1 Maximization and Minimization

In this chapter we limit ourselves to a discussion of how to maximize orminimize a function y(x) of one variable when there are no constraints,i.e., when x may be chosen freely.8 Except for a few complications,the idea is to find values of x for which the derivative of y with respectto x is zero.

Suppose that we wish to find a value x∗, if such a value exists,that maximizes y(x). It cannot be the case that y′(x∗) > 0; if sothe function would be increasing, and a value of x higher than x∗

would give a higher value of y. Likewise, it cannot be the case thaty′(x∗) < 0; if so the function would be decreasing, and a value of xlower than x∗ would give a higher value of y. This means that if x∗

maximizes y(x), then y′(x∗) must equal zero.Similar reasoning suggests that if x∗ minimizes y(x), then y′(x∗)

must also equal zero.8In many economic contexts it does not make sense for the variable x to be

negative; in those cases there is an implicit constraint.


x

y

y

x∗ x

y

y

x∗

Figure 7: The point x∗ gives a maximum of y(x) on the left; x∗ gives a

minimum of y(x) on the right.

Points x∗ for which y′(x∗) = 0 are called stationary points.A necessary condition that x∗ provide a maximum (or a minimum) isthat x∗ be a stationary point.9 We call this the first-order conditionfor a maximum or a minimum. The points labeled x∗ in the twographs of Figure 7 are stationary points. The function on the left is

9This might not be the case if x∗ is a boundary point.


x

y

y

x∗

Figure 8: The function y(x) has an inflection point at x∗; y(x) is neither a

maximum nor minimum at this point.

maximized at this point, and the function on the right is minimized.From a operational point of view, stationary points provide can-

didates for points that optimize the given function.10 Unless we havemore information, we cannot be sure that a particular stationary pointis the one that solves our problem. For example, if our interest is in

10The only other conceivable candidates are points at which the function is notdifferentiable and boundary points resulting from the implicit constraints.


maximizing y(x) and we have found a stationary point x∗, that pointcould be associated with a maximum as in the graph on the left ofFigure 7. But it could also be associated with a minimum as in thegraph on the right of Figure 7 or an inflection point as in Figure 8.

It is also possible that a stationary point x∗ provides only a localmaximum (a maximization of y(x) only over points very close to x∗)but not a global maximum (a maximization over all values of x).We note that there may not be a unique x that maximizes y(x), orthere may be no solution at all.11

There is a way of using calculus to help determine whether a sta-tionary point is a local maximum, a local minimum, or neither. Thiscondition involves second derivatives as well as first derivatives andis called the second-order condition. There are three possibilities:(a) If y′(x∗) = 0 and y′′(x∗) < 0, then x∗ gives a local maximum.(b) If y′(x∗) = 0 and y′′(x∗) > 0, then x∗ gives a local minimum.(c) If y′(x∗) = 0 and y′′(x∗) = 0, then x∗ may provide a local maxi-mum, a local minimum, or neither. You should be able to tell from

11For example, there is no value of x that maximizes the function y = x.


the graph on the left of Figure 7 that the second derivative is negativeat x∗. The second-order condition thus says that the function is max-imized there. This illustrates possibility (a). At x∗ in the graph onthe right of Figure 7 the second derivative is positive and the functionis minimized there. This illustrates possibility (b). At x∗ in Figure 8the second derivative is zero, and we have neither a maximum nor aminimum. This is one illustration of possibility (c). Another is thefunction y = x4 which does have a minimum at x = 0 but whose firstand second derivatives are zero there.

We should be very careful to distinguish between necessary andsufficient conditions for a maximum. If x∗ maximizes y(x), theny′(x∗) = 0 and y′′(x∗) ≤ 0. This is a necessary condition for a maxi-mum. If y′(x∗) = 0 and y′′(x∗) < 0, then x∗ gives a local maximum.This is a sufficient condition.12

We should say a few words about how to use these conditions tofind the maximum value when a specific function is given. We need to

12Similar conditions apply for a minimum if we replace ≤ and < with ≥ and >,respectively.


first find the stationary points, i.e., points where the first derivative iszero. This involves solving one equation y′(x) = 0 in one unknown (x).If this is a linear or quadratic equation, then a solution can be easilyobtained. For example, in determining stationary points we may haveto solve the quadratic equation

ax2 + bx + c = 0.

One way to find such values of x is to use the quadratic formula,given by

x =−b±

√b2 − 4ac

2a. (8)

Note that there are two solutions, one associated with the plus signand one with the minus sign. We need to apply the second-ordercondition to each of these values to distinguish which might be themaximum and which the minimum.

Example 6.1. We find the value of x that maximizes the functiony(x) = 10 − x2. Since the first derivative is given by y′(x) = −2x,we find the stationary point by solving −2x = 0 or x∗ = 0. The


second derivative is y′′(x) = −2 < 0 for all x, so x∗ = 0 maximizesthe function.

Example 6.2. We determine a local maximum and a local minimumfor the function y(x) = x3−12x2+36x. The first derivative is y′(x) =3x2 − 24x + 36, so the stationary points are solutions to the equation3x2 − 24x + 36 = 0 or, after simplifying, x2 − 8x + 12 = 0. This is aquadratic equation with solution given by Equation (8):

x =8±

√64− 482

=8± 4

2= 4± 2.

Thus there are two stationary points, x = 2 and x = 6. To determinewhether each corresponds to a maximum or minimum, we need to lookat the second derivative y′′(x) = 6x− 24. Evaluating this expressionat x = 2 and x = 6, we find that y′′(2) = −12 < 0 and y′′(6) = 12 > 0.Thus x = 2 is a local maximum and x = 6 is a local minimum.

6.2 Exercises

Exercise 6.1. Find the value of x that maximizes y(x) = 2x−x2−2.

Section 7: A Few Words about Integration 52

Exercise 6.2. Determine a local maximum and a local minimum forthe function y(x) = 1

3x3 − 2x2 + 3x.

7 A Few Words about IntegrationWe do not presume that you have studied the integral or integrationtechniques. Occasionally we find it convenient to employ integralnotation and to use one basic fact about integration.

One way we use integral notation is to represent the area under acurve. If we have a positive function y = y(x), as in Figure 9, then∫ b

a

y(x) dx (9)

gives the area below y(x) and above the horizontal axis from x = ato x = b. Just as the derivative is defined as a limit, the integral is alimit of sums of the areas of rectangles as the base becomes arbitrarilysmall.13

13If you read the integral sign as an S for “sum,” then the notation suggests thatthe integral is a limit of sums of terms like y(x) multiplied by small changes in x.


x

y

y

a b

∫b

a

y(x) dx

Figure 9: The integral of a positive function is just the area beneath it.

There are many techniques used to calculate the value of inte-grals. We will use nothing more than simple geometry, restricting ourattention to calculating areas of rectangles and triangles.

Finding an integral involves the reverse process of finding a deriva-tive. In physics we differentiate distance to get velocity but then in-tegrate (or “sum up”) velocity to get distance traveled. In economics

These products are the areas of rectangles which, added together, approximatethe area under the curve.


x

y′

y′

x

y(x)− y(0)

Figure 10: The integral of the derivative y′ is the original function y(x) except

for a constant.

we differentiate total cost to get marginal cost; we need to integratemarginal cost to get back to total cost. We could say that distanceis an antiderivative of velocity and total cost is an antiderivative ofmarginal cost.


We need a way of writing this idea that a function is the integral(or area under) its derivative. We write

y(x) =∫ x

0

y′(s) ds + y(0). (10)

We make three points about equation (10), which is illustratedin Figure 10 and which represents one version of the FundamentalTheorem of Calculus. First, we have to be careful about the notation.In equation (9) the value of the integral is a number and not a functionof x. In equation (10), however, the integral does not depend on sbut does depend on x.

Second, we are restricting our attention to nonnegative values ofthe two variables. Thus we find the area under the positive curve y′

between zero and some prescribed positive value of x.Third, since the derivative of a constant is zero, adding a con-

stant to the total function y produces the same marginal function y′.Thus we cannot hope to find the constant in the total function from

Section 8: Chapter Quiz 56

information about the marginal function alone. The y(0) term inequation (10) provides the appropriate constant.

Example 7.1. Consider the function y′(x) = b as represented in thegraph on the right of Figure 1. The area under this curve from 0 to xis bx (the base x times the height b). Using equation (10) and the factthat y(0) = a, we get y(x) = a + bx. This is the total function in thegraph on the left of Figure 1.

Example 7.2. Consider the function y′(x) = a+bx as represented inthe graph on the right of Figure 2. Consider an arbitrary value of x.The area under this function from 0 to x is the sum of a rectanglewith base x and height a and a triangle with base x and height bx.Thus the area is ax+(1/2)bx2. This is the function y(x) as shown onthe right of Figure 2; when differentiated it gives y′(x).

8 Chapter QuizClick on the “Begin Quiz” button to begin, choose your answers, clickon the “End Quiz” button to see how many questions you answered


correctly, and click on the “Correct” button to see the correct answers.

1. If y′ is positive and decreasing, then y is

decreasing with decreasing slope

decreasing with increasing slope

increasing with decreasing slope

increasing with increasing slope

2. If Revenue R depends on quantity sold Q, how might we writemarginal revenue?

dQ

dx

dx

dQR′(Q) Q′(R)

3. If y(x) = −mx−n, what is y′(x)?

mnx−n+1 mnx−n−1 −mnx−n+1 −mnx−n+1

4. If y(x) =√

3x, what is dy/dx?


32√

3x

23√

3x

3√3x

12√

3x

5. If y(x) = (2 + x3)5, then y′(x) is

5(x3)4

5(2 + x3)45(2 + x3)4(2 + 3x2)5(2 + x3)4(3x2)

6. Another way to write x−.75 is as

x3/4 x4/3 1x3/4

1x4/3

7. What is the derivative of lnx?

lnx − lnx1x

− 1x

8. If y(x) = u(x) · v(x), then y′(x) isdu

dv· dv

dx

dv

du· du

dxu

dv

dx− v

du

dxu

dv

dx+ v

du

dx

9. If y = a− bx, thendx

dyequals

−b −1b

a1b


10. The function y(x) = x − x2 achieves its maximum at an x valueof

.5 1 2 5

Answers to Quizzes 60

Answers to Quizzes

Answers to Section Quiz 4.3

1. b 2. c 3. a 4. c 5. c 6. a 7. d 8. b 9. c 10. d

Answers to Section Quiz 5.6

1. b 2. a 3. d 4. d (The easiest way to differentiate y =√

x/x isto recognize it as y = x1/2/x1 = x−1/2 so that y′(x) = − 1

2x−3/2.) 5.b 6. c 7. a 8. b 9. d 10. a

Answers to Chapter Quiz 8

1. c 2. c 3. b 4. a (chain rule) 5. d (chain rule) 6. c 7. c8. d (product rule) 9. b (derivative of an inverse function) 10. a(stationary point)


Solutions to ExercisesExercise 3.1.

The marginal tax rate is written as dT/dx or T ′(x). The units ofthe marginal tax rate are the units of T (dollars per year) divided bythe units of x (dollars per year). Thus everything cancels out, andthere are no units! The marginal tax rate is a pure number, such as30%. Return to Exercise 3.1


x

y′

y′

Figure 11: The graph of y′(x) for Exercise 3.2.

Exercise 3.2.The graph of the derivative y′ is shown in Figure 11. The func-

tion y is always positive (it lies above the horizontal axis) and increas-ing (it slopes up). The function y′ is positive but it decreases at firstand then increases. The second derivative y′′ (not shown) is negativeat first and then becomes positive. It is always increasing.

Return to Exercise 3.2


Exercise 3.3.Marginal cost first decreases and then increases. (Marginal cost

is always positive as well. If this were not true, the function y wouldnot be a reasonable cost function.) Return to Exercise 3.3


Exercise 3.4.No. See Figures 2 and 3 for cases where this is clearly false. The

function y will be a straight line only for the case (as in Figure 1)where the derivative is a horizontal line. Return to Exercise 3.4


Exercise 3.5.The second derivative measures the slope of the slope, i.e., it mea-

sures the curvature of the function at the particular point. If thesecond derivative is positive, then the function looks like a convexfunction near the point (as in the diagram on the left of Figure 2). Ifthe second derivative is negative, then the function looks like a con-cave function near the point (as in the diagram on the left of Figure 3).



Exercise 6.1.First note that y′(x) = 2−2x, so stationary points satisfy 2−2x =

0 or x∗ = 1. The second derivative, y′(x) = −2 < 0 everywhere, sothe function is maximized at x∗ = 1. Return to Exercise 6.1


Exercise 6.2.The first derivative is y′(x) = x2−4x+3, so the stationary points

are solutions to the equation x2 − 4x + 3 = 0. This is a quadraticequation with solution given by

x =4±

√16− 122

=4± 2

2= 2± 1.

Thus there are two stationary points, x = 1 and x = 3. To determinewhether each corresponds to a maximum or minimum, we need to lookat the second derivative y′′(x) = 2x − 4. Evaluating this expressionat x = 1 and x = 3, we find that y′′(1) = −2 < 0 and y′′(3) = 2 > 0.Thus x = 1 is a local maximum and x = 3 is a local minimum.


References 68

References[1] Binger, Brian R. and E. Hoffman, Microeconomics with Calculus,

2nd ed., Addison-Wesley, Reading, MA, 1998. 7

[2] Hughes-Hallett, Deborah, et. al., Calculus, 2nd ed., John Wiley &Sons, New York, 1998. 6

[3] Landsburg, Steven E., Price Theory and Applications, 5th ed.,South-Western, Cincinnati, 2002. 7

[4] Mankiw, N. Gregory, Principles of Economics, 2nd ed., The Dry-den Press, Fort Worth, 2001. 5

[5] Nicholson, Walter, Microeconomic Theory: Basic Principles andExtensions, 7th ed., The Dryden Press, Fort Worth, 1998. 7

[6] Story, Donald P., e-calculus, April 3, 2000, http://www.math.uakron.edu/~dpstory/e-calculus.html. Available: February20, 2002. 6




References 69

[7] Thompson, Silvanus P., Calculus Made Easy, Macmillan, London,1910. 6

[8] Varian, Hal R., What Use is Economic Theory? August,1989, http://www.sims.berkeley.edu/~hal/Papers/theory.pdf. Available: February 20, 2002. 9

[9] Varian, Hal R., Intermediate Microeconomics: A Modern Ap-proach, 5th ed., W. W. Norton & Co., New York, 1999. 7

http://www.sims.berkeley.edu/~hal/Papers/theory.pdf



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