TEXTBOOK ON MATHEMATICAL ECONOMICS FOR CU , BU CALCUTTA , SOLVED EXERCISES , STUDY MATERIAL

MATHEMATICAL
ECONOMICS
A GUIDE TO UNDER-
GRADUATE STUDENTS
SOURAV SIR’S CLASSES
2016
Windows

1
MATHEMATIAL ECONOMICS by SOURAV DAS
MATHEMATICAL
ECONOMICS
A Guide to
Under-Graduate students
By
SOURAV SIR’S CLASSES
98367 93076

2

3
PREFACE
Read it, because you need it..
I know its very boring to read a preface but I have given some
clue to crack the eco Honours puzzle which you are finding a
little hard to crack( Obviously some of you are born genious and
know everything already!)
I have tried to show you the most common and easy ways of
getting a grip to Mathematical Economics. Students, whether
from Economics background or from Science both find this
paper a bit challenging. As teachers say, ‘this is the most scoring
paper’, pressure keeps rising on them in order to perform best in
the exam.
Main problems which students face are,
1) Mixing theoretical economics knowledge to mathematical
grounds. Most of the times they end up with memorizing
some types and practice a lot of solved examples to get an
idea of the topic but this creates a problem when the
question paper is a little bit tough or out of the good old
system.

4
The Solution is, don’t just see the solved examples along
with that check the theoretical solutions given in a,b,c,d form
not in numerical. Although apparently it seems boring but in the
long run it gives you the mileage.
2) Some students who didn’t have Economics in class XI-XII ,
think that they are lacking behind those who have already
studied it .
The solution is, Well, I cannot say it is totally wrong , but the
syllabus for the B.Sc Economics course is set in a way that
even a non Economics background student will not found it
tough or out of his or her reach. At the beginning may be they
will have some checkpoints but after 3/4 months they get into
business with flying colours.
3) Another big question for them is whether or not to buy a
book or more than 1 or 2 or 3 books?
I recommend unlike your school days graduation will need
a larger book bank. For a paper or for a topic or even for a
theorem or typical type of problem you may need a book ,
so obviously unless you are a millionaire or planning to
open a library at home get yourself a membership card of
your college library along with a private library where you
can get all the books you need.
For Calcutta based students I prefer

5
a.Ramakrishna Mission Institute Of Culture Library,
Golpark.,033-2465-2531
b.National Library, 033-2479-1400
c.Advaita Ashram Library near Dihi Entally Road
(033-2284-0210).
4) Will notes be enough to cover the syllabus or we need to
study the book only?
I say a combination of both is the classic solution. Ask your
professor about the books you can read according to your
grasping power because some books are written in a lucid
and easy to understand way and some follows a very short
cut and advanced manner which will make you feel scary
about the topic.
After 5/6 months when you have checked some books then
you will see you have developed an idea of , how to study and
from where to study.
I prefer for Mathematical Economics
a. Alpha c. Chiang

6
b. Henderson and Quandt
c. David Romer
d. Jaydeb Sarkhel
For Statistics
a. N.G.Das
b. Rathie and Mathie
c. S.P.Gupta
d. Parimal Mukherjee( Advanced Book)
For Microeconomics
a. Pindyck and Rubinfeld
b. Koutsoyiannis
c. Sampat Mukherjee
d. Taha ( Game Theory)
e. Varian ( Intermediate Microeconomics)
For Macroeconomics
a. Mankiw( Study this book for concept development
but you can’t produce answers from this one)
b. Sampat Mukherjee
c. Hall and Taylor
d. David Romer

7
5) As Statistics a brand new subject don’t be afraid of it or
don’t take it too easily.
Solve as much as you can alongwith that theory part must be
made strong.
Make a good group of people who can study with you , help you
with notes. In first , second year you can do internship at banks
as a student of Economics Honours, can attain Economics
summits organized by different colleges in Kolkata like St.
Xavier’s college, Presidency college. Visit different universities
and have a check on their curriculum, way of entrance exam ,
interview process by visiting their canteens and meeting with
seniors or talking to professors in their off time(well, this needs
a little bit courage from your part, but in long run you will get a
larger package of benefits).
Remember, Economics Honours is the best honours to study .
Not because I have studied it or you have got into it but because
this will allow to Study Mathematics , Statistics , Econometrics
along with Economics in a way where you have to perform best
in all and as a result of that you are becoming one of the most
wanted educated person in the educational or professional field.

8
Happy Hunting,
Warm (at 500
Celsius) regards,
Sourav Das
M.Sc., Research Fellow
98367-93076
Table of contents
 Introduction and instructions
 1. Review of some basic logic, matrix algebra, and calculus
o 1.1 Logic
o 1.2 Matrices and solutions of systems of simultaneous equations

9
o 1.3 Intervals and functions
o 1.4 Calculus: one variable
o 1.5 Calculus: many variables
o 1.6 Graphical representation of functions
 2. Topics in multivariate calculus
o 2.1 Introduction
o 2.2 The chain rule
o 2.3 Derivatives of functions defined implicitly
o 2.4 Differentials and comparative statics
o 2.5 Homogeneous functions
 3. Concavity and convexity
o 3.1 Concave and convex functions of a single variable
o 3.2 Quadratic forms
 3.2.1 Definitions
 3.2.2 Conditions for definiteness
 3.2.3 Conditions for semidefiniteness
o 3.3 Concave and convex functions of many variables
o 3.4 Quasiconcavity and quasiconvexity
 4. Optimization
o 4.1 Introduction
o 4.2 Definitions
o 4.3 Existence of an optimum
 5. Optimization: interior optima
o 5.1 Necessary conditions for an interior optimum
o 5.2 Sufficient conditions for a local optimum
o 5.3 Conditions under which a stationary point is a global optimum
 6. Optimization: equality constraints
o 6.1 Two variables, one constraint
 6.1.1 Necessary conditions for an optimum
 6.1.2 Interpretation of Lagrange multiplier
 6.1.3 Sufficient conditions for a local optimum
 6.1.4 Conditions under which a stationary point is a global optimum
o 6.2 n variables, m constraints
o 6.3 Envelope theorem
 7. Optimization: the Kuhn-Tucker conditions for problems with inequality constraints
o 7.1 The Kuhn-Tucker conditions
o 7.2 When are the Kuhn-Tucker conditions necessary?
o 7.3 When are the Kuhn-Tucker conditions sufficient?
o 7.4 Nonnegativity constraints
o 7.5 Summary of conditions under which first-order conditions are necessary and
sufficient

10
 8. Differential equations
o 8.1 Introduction
o 8.2 First-order differential equations: existence of a solution
o 8.3 Separable first-order differential equations
o 8.4 Linear first-order differential equations
o 8.5 Phase diagrams for autonomous equations
o 8.6 Second-order differential equations
o 8.7 Systems of first-order linear differential equations
 9. Difference equations
o 9.1 First-order equations
o 9.2 Second-order equations
1. Review of some basic logic, matrix algebra, and calculus
 1.1 Logic
 1.2 Matrices and solutions of systems of simultaneous equations
 1.3 Intervals and functions
 1.4 Calculus: one variable

11
 1.5 Calculus: many variables
 1.6 Graphical representation of functions
1.1 Logic
Basics
When making precise arguments, we often need to make conditional statements, like
if the price of output increases then a competitive firm increases its output
or
if the demand for a good is a decreasing function of the price of the good and the
supply of the good is an increasing function of the pricethen an increase in supply at
every price decreases the equilibrium price.
These statements are instances of the statement
if A then B,
where A and B stand for statements. We may alternatively write this general statement
as
A implies B,
or, using a symbol, as
A ⇒ B.
Yet two more ways in which we may write the same statement are
A is a sufficient condition for B,
and
B is a necessary condition for A.
(Note that B comes first in the second of these two statements!!)

12
Important note: The statement A ⇒ B does not make any claim about whether B is
true if A is NOT true! It says only that if A is true, then B is true. While this point
may seem obvious, it is sometimes a source of error, partly because we do not always
adhere to the rules of logic in everyday communication. For example, when we say
"if it's fine tomorrow then let's play tennis" we probably mean both "if it's fine
tomorrow then let's play tennis" and "if it's not fine tomorrow then let's not play
tennis" (and maybe also "if it's not clear whether the weather is good enough to play
tennis tomorrow then I'll call you"). When we say "if you listen to the radio at 8
o'clock then you'll know the weather forecast", on the other hand, we donot mean also
"if you don't listen to the radio at 8 o'clock then you won't know the weather forecast",
because you might listen to the radio at 9 o'clock or check on the web, for example.
The point is that the rules we use to attach meaning to statements in everyday
language are subtle, while the rules we use in logical arguments are absolutely clear:
when we make the logical statement "if A then B", that's exactly what we mean---no
more, no less.
We may also use the symbol "⇐" to mean "only if" or "is implied by". Thus
B ⇐ A
is equivalent to
A ⇒ B.
Finally, the symbol "⇔" means "implies and is implied by", or "if and only if". Thus
A ⇔ B
is equivalent to
A ⇒ B and A ⇐ B.
If A is a statement, we write the claim that A is not true as
not(A).
If A and B are statements, and both are true, we write
A and B,
and if at least one of them is true we write
A or B.

13
Note, in particular, that writing "A or B" includes the possibility that both statements
are true.
Two rules
Rule 1
If the statement
A ⇒ B
is true, then so too is the statement
(not B) ⇒ (not A).
The first statement says that whenever A is true, B is true. Thus if B is false, A
must be false---hence the second statement.
Rule 2
The statement
not(A and B)
is equivalent to the statement
(not A) or (not B).
Note the "or" in the second statement! If it is not the case that both A is true
and B is true (the first statement), then either A is not true or B is not true.
Quantifiers
We sometimes wish to make a statement that is true for all values of a variable. For
example, denoting the total demand for tomatoes at the price p byD(p), it might be
true that
D(p) > 100 for every price p in the set S.
In this statement, "for every price" is a quantifier.
Important note: We may use any symbol for the price in this statement: "p" is a
dummy variable. After having defined D(p) to be the total demand for tomatoes at the
price p, for example, we could write
D(z) > 100 for every price z in the set S.

14
Given that we just used the notation p for a price, switching to z in this statement is a
little odd, BUT there is absolutely nothing wrong with doing so! In this simple
example, there is no reason to switch notation, but in some more complicated cases a
switch is unavoidable (because of a clash with other notation) or convenient. The
point is that every statement of the form
A(x) for every x in the set Y,
where x is any symbol, has exactly the same content.
Another type of statement we sometimes need to make is
A(x) for some x in the set Y,
or, equivalently,
there exists x in the set Y such that A(x).
"For some x" (alternatively "there exists x") is another quantifier, like "for every x";
my comments about notation apply to it.
1.1 Exercises on logic
1. A, B, and C are statements. The following theorem is true:
if A is true and B is not true then C is true.
Which of the following statements follow from this theorem?
a. If A is true then C is true.
b. If A is not true and B is true then C is not true.
c. If either A is not true or B is true (or both) then C is not true.
d. If C is not true then A is not true and B is true.
e. If C is not true then either A is not true or B is true (or both).
2. A and B are statements. The following theorem is true:

15
A is true if and only if B is true.
Which of the following statements follow from this theorem?
a. If A is true then B is true.
b. If B is true then A is true.
c. If A is not true then B is not true.
d. If B is not true then A is not true.
3. Let G be a group of people. Assume that
for every person A in G, there is a person B in G such that A knows a friend
of B.
Is it true that
for every person B in G, there is a person A in G such that B knows a friend
of A?
1.2 Matrices and solutions of systems of simultaneous equations
Matrices
I assume that you are familiar with vectors and matrices and know, in particular, how
to multiply them together. (Do the first few exercises to check your knowledge.)
The determinant of the 2 × 2 matrix
a b
c d
is ad − bc. If ad − bc ≠ 0, the matrix is nonsingular; in this case its inverse is
1
d −b .

16
ad − bc −c a
(You can check that the product of the matrix and its inverse is the identity matrix.)
We denote the inverse of the matrix A by A−1.
The determinant of the 3 × 3 matrix
a b c
d e f
g h i
is
Δ = a(ei − h f ) − b(di − g f ) + c(dh − eg).
If Δ ≠ 0 the matrix is nonsingular; in this case its inverse is
1
Δ
D11 −D12 D13
−D21 D22 −D23
D31 −D32 D33
where Dij is the determinant of the 2 × 2 matrix obtained by deleting
the ith column and jth row of the original 3 × 3 matrix. That
is, D11 = ei − h f ,D12 = bi − ch, D13 = b f − ec, D21 = di − f g, D22 = ai − cg, D23 = a f − dc
, D31 = dh − eg, D32 = ah − bg, and D33 = ae − db. (Again, you can check that the
product of the matrix and its inverse is the identity matrix.)
The determinant of the matrix A is denoted |A|.
Let A be an n × m matrix (i.e. a matrix with n rows and m columns). The transpose A'
of A is the m × n matrix in which, for i = 1, ..., n, the ith row of A' is the ith column
of A. In particular, if x is a column vector (n × 1 matrix) then x' is a row vector (1
× n matrix).

17
Solutions of systems of simultaneous equations
Consider a system of two equations in two variables x and y:
ax + by= u
cx + dy= v.
Here are three ways to solve for x and y.
1. Isolate one of the variables in one of the equations and substitute the result into
the other equation. For example, from the second equation we have
y = (v − cx)/d.
Substituting this expression for y into the first equation yields
ax + b(v − cx)/d = u,
which we can write as
(a − bc/d)x + bv/d = u,
so that
x =
u − bv/d
a − bc/d
or
x =
ud − bv
ad − bc
.
To find y we now use the fact that y = (v − cx)/d, to get
y =
va − cu
.

18
ad − bc
2. Use Cramer's rule (due to Gabriel Cramer, 1704-1752). Write the two equations
in matrix form as
3.
a b
c d
x
y
=
u
v
(*)
4. Cramer's rule says that the solutions are given by
x =
u b
v d
ad − bc
5. and
y =
a u
c v
ad − bc
.
6. Note that the denominator in these expressions, ad − bc, is the determinant of
the matrix at the left of (*). The matrix in the numerator of each expression is
obtained by replacing the column in the matrix on the left of (*) that

19
corresponds to the variable for which we are solving with the column vector on
the right of (*).
7. Calculating the determinants in the numerators, we have
x =
ud − bv
ad − bc
8. and
y =
va − cu
ad − bc
.
9. Write the two equations in matrix form as
a b
c d
x
y
=
u
v
10.(as when using Cramer's rule) and solve by inverting the matrix on the left hand
side. The inverse of this matrix is
1
ad − bc
d −b
−c a
11.so we have
x
y
=
1
ad − bc
d −b
−c a
u
v

20
12.so that
x =
ud − bv
ad − bc
13.and
y =
va − cu
ad − bc
.
Which of these three methods is best depends on several factors, including your
ability to remember Cramer's rule and/or how to invert a matrix. If you have to solve
for only one of the variables, Cramer's rule is particularly convenient (if you can
remember it).
For a system of more than two variables, the same three methods are available. The
first method, however, is pretty messy, and unless you are adept at inverting matrices,
Cramer's rule is probably your best bet.
1.2 Exercises on matrix algebra and solving simultaneous
equations
1. Let
A =
4 −1
6 9
2. and

21
B =
0 3
3 −2
3. Find (i) A + B, (ii) 2A − B, (iii) AB, (iv) BA, and (v) A' (the transpose of A).
4. Let
A =
4 −1
6 9
2 3
5. and
B =
0 3
3 −2
6. (i) Is AB defined? If so, find it. (ii) Is BA defined? If so, find it.
7. Given u' = (5,2,3), find u'·u (the scalar product, or inner product).
8. You buy n items in the quantities q1,...,qn at the prices p1,...,pn. Express your
expenditure using (i) Σ notation, (ii) vector notation.
9. Let x' = (x1, x2) and let
A =
a11 a12
a21 a22
10.Find x'Ax.
11.Find the determinants of the matrices A and B in Problem 1.
12.Find the inverse of the matrix A in Problem 1, and verify that it is indeed the
inverse.
13.Find the determinants of the following two matrices.
A =
8 1 3
4 0 1
6 0 3
14.and

22
B =
a b c
b c a
c a b
15.Are either of the matrices A and B in the previous problem nonsingular?
16.Find the inverse of the matrix A in Problem 8.
17.Use Cramer's rule to find the values of x and y that solve the following two
equations simultaneously.
3x − 2y=11
2x + y=12
18.Solve the two equations in the previous problem by using matrix inversion.
19.Use Cramer's rule to find the values of x, y, and z that solve the following three
equations simultaneously.
4x +3y −2z=7
x +y =5
3x + z =4
20.Solve the three equations in the previous problem by using matrix inversion.
1.3 Intervals and functions
Intervals
An interval is a set of (real) numbers between, and possibly including, two numbers.
The interval from a to b is denoted as follows:
 [a, b] if a and b are included (i.e. [a, b] = {x: a ≤ x ≤ b})
 (a, b) if neither a nor b is included (i.e. (a, b) = {x: a < x < b})
 [a, b) if a is included but b is not
 (a, b] if b is included but a is not.
We use the special symbol "∞" ("infinity") in the notation for intervals that extend
indefinitely in one or both directions, as illustrated in the following examples.

23
 (a, ∞) is the interval {x: a < x}.
 (−∞, a] is the interval {x: x ≤ a}.
 (−∞, ∞) is the set of all numbers.
Note that ∞ is not a number, but simply a symbol we use in the notation for intervals
that have at most one endpoint.
The notation (a, b) is used also for an ordered pair of numbers. The fact that it has two
meanings is unfortunate, but the intended meaning is usually clear from the context.
(If it is not, complain to the author.)
The interior of an interval is the set of all numbers in the interval except the endpoints.
Thus the interior of [a, b] is (a, b); the interval (a, b) is the interior also of the intervals
(a, b], [a, b), and (a, b).
We say that (a, b), which does not contain its endpoints, is an open interval, whereas
[a, b], which does contain its endpoints, is a closed interval. The intervals (a, b] and
[a, b) are neither open nor closed.
Functions
A function is a rule that associates with every point in some set, a single point in
another set. The first set is called the domain of the function. A function with
domain A is said to be defined on A.
To specify a function we need to specify the domain and the rule. Here are some
examples.
 Domain: [−2, 1]. Rule: f (x) = x2
.
 Domain: (−∞, ∞). Rule: f (x) = x2.
 Domain: (−1, 1). Rule: f (x) = x if x ≥ 0; f (x) = 1/x if x < 0.
 Domain: union of (0, 1) and (4, 6). Rule: f (x) = √x.
 Domain: set of all pairs of numbers. Rule: f (x, y) = xy.
 Domain: set of n-vectors (x1, ..., xn) for which 0 ≤ xi ≤ 1 for i = 1, ..., n.
Rule: f (x1, ..., xn) = ∑i=1
naixi, where a1, ..., an are nonnegative constants.
(Note that the symbols x and y are arbitrary. We could, for example, define the first
function equally well by f (z) = z2 or f (g) = g2. We generally usex and y for variables,
but we may use any other symbols.)
These examples have two features in common:

24
 the domain is a subset of the set of n-vectors of numbers, for some positive
integer n (where n may of course be 1, as in the first four examples)
 the rule associates a real number with each point in the domain.
A function is not restricted to have these features. For example, the domain of a
function may be a set of complex numbers and the function may associate a set with
each member of its domain. All the functions in this tutorial, however, have the two
features, and the word "function" throughout means a rule that associates a number
with every point in some subset of the set of n-vectors of numbers (where n may be
1). I refer to a function whose domain is a set of 1-vectors (i.e. numbers) as a function
of a single variable, and one whose domain is a set of n-vectors for n ≥ 1 as a function
of many variables. (Note that a function of a single variable is a special case of a
function of many variables.)
The number that a function associates with a given member x of its domain is called
the value of the function at x. As x varies over all points in the domain of a function,
the value of the function may (and generally does) vary. The set of all such values of
the function is called the range of the function. Here are the ranges of the examples
given above.
 Domain: [−2, 1]. Rule: f (x) = x2
. Range: [0, 4].
 Domain: (−∞, ∞). Rule: f (x) = x2
. Range: [0, ∞).
 Domain: (−1, 1). Rule: f (x) = x if x ≥ 0; f (x) = 1/x if x < 0. Range: union of
(−∞, −1) and [0, 1).
 Domain: union of (0, 1) and (4, 6). Rule: f (x) = √x. Range: union of (0, 1) and
(2, √6).
 Domain: set of all pairs of numbers. Rule: f (x, y) = xy. Range: (−∞, ∞).
 Domain: set of n-vectors (x1, ..., xn) for which 0 ≤ xi ≤ 1 for i = 1, ..., n.
Rule: f (x1, ..., xn) = ∑i=1
naixi, where a1, ..., an are nonnegative constants. Range:
(0, ∑i=1
nai).
In formal presentations of mathematical material, the notation f : A → B is used for a
function given by the rule f and the domain A whose range is a subset of B. We
might say, for example, "consider the function f : [0, 1] → ℝ defined by f (x) = √x",
or "for every function f : ℝ → ℝ." (The symbol ℝ denotes the set of all (real)
numbers.) The set B in this notation is called the co-domain or target of the function.
Note that this set is not part of the definition of the function, and may be larger than
the range of the function. When we say, for example, "for every function f : ℝ → ℝ",
we mean every function whose domain is ℝ and whose range is a subset of ℝ.

25
Graphical illustrations aid the understanding of many functions. A function of a single
variable, for example, may be represented on x−y coordinates by plotting, for each
value of x, the value of f (x) on the y-axis. An example is shown in the following
figure. In this diagram, the small circle indicates a point excluded from the graph: the
value of the function at x0 is y0, whereas the value of the function at points slightly
greater than x0 is y1.
The red line in this figure is called the graph of the function. Techniques for drawing
graphs are discussed in a later section.
Logarithms and exponentials
You need to be comfortable working with the logarithm function and with functions
of the form xy (where y is known as an exponent). In particular, you should know the
following rules.
 xyxz = xy+z
 (xy)z = xyz (so that in particular (xy)1/y = x)
 ln ex
= x and eln x
= x
 a ln x = ln xa (so that ea ln x
= xa)
Continuous functions
A function of a single variable is continuous if its graph has no "jumps", so that it can
be drawn without lifting pen from paper. In more precise terms, a function f is
continuous at the point a if we can ensure that the value f (x) of the function is as
close as we wish to f (a) by choosing x close enough to a. Here is a completely
precise definition for a function of many variables. (The distance between two points
(x1, ..., xn) and (y1, ..., yn) is the Euclidean distance √[∑i=1
n(xi − yi)2].)
Definition
Let f be a function of many variables and let a be a point in its domain. Then f is continuous at a if, for any numb
> 0 such that for any value of x in the domain of f for which the distance between x and a is less than δ, the differe

26
is less than ε. The function f is continuous if it is continuous at every point in its domain.
The function whose graph is shown in the previous figure is not continuous at x0. The
value of the function at x0 is y0, but the values at points slightly larger than x0 is much
larger than y0: no matter how small we choose δ, some points x within the distance δ
of x0 yield values of the function far from y0(= f (x0)).
The following result is useful in determining whether a function is continuous.
Proposition
 If the functions f and g of many variables are continuous at x0 then the function h defined by h(x) = f (x) +
at x0.
 If the functions f and g of many variables are continuous at x0 then the function h defined by h(x) = f (x)g
 If the functions f and g of many variables are continuous at x0 and g(x0) ≠ 0 then the function h defined by
all xwith g(x) ≠ 0 is continuous at x0.
 If the function f of many variables is continuous at x0 and the function g of a single variable is continuous
function hdefined by h(x) = g( f (x)) for all x is continuous at x0.
An implication of the second part of this result is that if f is continuous at x0 then the
function h defined by h(x) = ( f (x))k, where k is a positive integer, is continuous at x0.
A polynomial is a function of a single variable x of the form a0 + a1x + a2x2 + ... + akxk,
where k is a nonnegative integer and a0, ..., ak are any numbers. Because the
function f (x) = x is continuous, all polynomials are continuous.
The next result gives an important property of continuous functions. It says that if the
function f of a single variable on the domain [a, b] is continuous, then f(x) takes on
every value from f(a) to f(b).
Proposition (Intermediate value theorem)
If f is a continuous function of a single variable with domain [a,b] and f (a) ≠ f (b), then for any
number y between f (a) and f (b), or equal to f (a) or f (b), there is a value of x (in [a,b]) such
that f (x) = y.
This result is illustrated in the following figure.

27
In this figure, the set of values from f (a) to f (b) is shown in red; for every value y in
this set, there is a value of x such that f (x) = y. For example, f (x1) = y1. Two points to
note:
 For some values y between f (a) and f (b) there may be more than one value
of x such that y = f (x). For example, in the figure f (x2) = f (x3) = f (x4) = y2.
 The result does not say that for values y less than f (a) or greater than f (b)
there is no x such that f (x) = y. Indeed, in the figure we have f (x5) = y3.
An important implication of the result is that if f (a) is positive and f (b) is negative,
then f (x) = 0 for some x.
Example
Consider the function f defined by f (x) = x4
− 4x2
+ 2. Does the equation f (x) = 0 have a solution between 0 and
polynomial, and thus is continuous. We have f (0) = 2 and f (1) = −1, so the Intermediate Value Theorem implies
question is yes: for some value of x between 0 and 1 we have f (x) = 0.
1.3 Exercises on intervals and functions
1. At what points is the function
f (x) =
x2 − 2 if x ≤ 0
−x2 if 0 < x < 1
x2 − 2 if x ≥ 1
2. continuous?
3. Use the Intermediate Value Theorem to show that the equation x7 − 5x5 + x3 − 1
= 0 has a solution between −1 and 1.

28
1.4 Calculus: one variable
Differentiation
Let f be a function of a single variable defined on an open interval. This function
is differentiable at the point a if it has a well-defined tangent at a. Itsderivative at a,
denoted f '(a), is the slope of this tangent.
Precisely, consider "secant lines" like the one from (a, f (a)) to (a + h, f (a + h)) in
the following figure.
Such a line has slope ( f (a + h) − f (a))/h. The derivative of f at a is defined to be the
limit, if it exists, of this slope as h decreases to zero.
Definition
The function f of a single variable defined on an open interval is differentiable at a if limh→0( f (a + h) − f (a))/h ex
is the derivative of the function f at a, denoted f '(a).
The statement "limh→0( f (a + h) − f (a))/h exists" means, precisely, that there is a
number k such that for every number ε > 0 (no matter how small), we can find a
number δ > 0 such that if |h| < δ then the difference between f (a + h)
− f (a))/h and k is less than ε.

29
If f is differentiable at every point in some interval I, we say that f is "differentiable
on I".
The graph of a function differentiable at a is "smooth" at a. In particular, a function
that is differentiable at a is definitely continuous at a. An example of a function that
is not differentiable at some point is shown in the following figure.
The function f in the figure is not differentiable at a, because the slope of the secant
line from (a, f (a)) to (a + h, f (a + h)) is very different for h > 0 and for h < 0, even
when h is arbitrarily small, so that limh→0( f (a + h) − f (a))/h does not exist.
At any point at which a function has a "kink", it is similarly not differentiable.
The derivative of f at a is sometimes denoted (d f /dx)(a).
Rules for differentiation
The definition of a derivative implies the following formulas for the derivative of
specific functions.
f (x) f '(x)
k 0
kxn knxk−1
ln x 1/x
ex ex
ax axln a

30
cos x −sin x
sin x cos x
tan x 1 + (tan x)2
Three general rules (very important!!):
Sum rule
F (x) = f (x) + g(x): F '(x) = f '(x) + g'(x)
Product rule
F (x) = f (x)g(x): F '(x) = f '(x)g(x) + f (x)g'(x)
Quotient rule
F (x) = f (x)/g(x): F '(x) = [ f '(x)g(x) − f (x)g'(x)]/(g(x))2
Note that if you know that the derivative of (g(x))n
is n(g(x))n−1
g'(x) (an implication of
the "chain rule", discussed later), the quotient rule follows directly from the product
rule: if you write f (x)/g(x) as f (x)(g(x))−1 then the product rule implies that the
derivative is
f '(x)(g(x))−1
− f (x)(g(x))−2
g'(x),
which is equal to [ f '(x)g(x) − f (x)g'(x)]/(g(x))2
.
Example
Let F (x) = x2 + ln x. By the sum rule, F '(x) = 2x + 1/x.
Example
Let F (x) = x2ln x. By the product rule, F '(x) = 2xln x + x2/x = 2xln x + x.
Example
Let F (x) = x2/ln x. By the quotient rule, F '(x) = [2xln x − x2/x]/(ln x)2 = [2xln x − x]/(ln x)2.
Second derivatives
If the function f is differentiable at every point in some open interval I then its
derivative f ' may itself be differentiable at points in this interval. If f ' is

31
differentiable at x then we say that f is twice-differentiable, we call its derivative
at x the second derivative of f at x, and we denote this derivative f "(x).
Integration
Let f be a function of a single variable on the domain [a, b]. The definition of the
("definite") integral of f from a to b, denoted
∫b
a f (z)dz,
is a measure of the area between the horizontal axis and the graph of f ,
between a and b.
Notes
 If f (x) < 0 for some x between a and b, then the corresponding areas (shaded
red in the above figure) count negatively in the integral.
 The integral is defined precisely as the limit, if it exists, of approximations to
the area consisting of sums of the areas of narrow rectangles as the width of
these rectangles approaches zero. If the limit exists then f is integrable. It may
be shown that if f is continuous then it is integrable. If f is not integrable,
then its integral is not defined. (Such functions are fairly exotic. An example is
the function f with domain [0, 1] defined by f (x) = 1 if x is a rational number
and f (x) = 0 if x is an irrational number.)
 Note that the variable z is a dummy variable, and can be replaced by any other
variable. Sometimes it is dropped entirely, and the integral is written simply as
∫b
a f .
The fundamental theorem of calculus shows that integration and differentiation are, in
a sense, inverse operations.

32
Proposition (Fundamental theorem of calculus)
Let f be an integrable function of a single variable defined on [a, b]. Define the function F of a single variable on
F (x) = ∫x
a f (z)dz.
If f is continuous at the point c in [a, b], then F is differentiable at c and
F '(c) = f (c).
Similarly, define the function G on [a, b] by
G(x) = ∫b
x f (z)dz.
If f is continuous at the point c in [a, b], then G is differentiable at c and
G'(c) = − f (c).
If f is continuous on [a, b] and f = F ' for some function F , then
∫b
a f (z)dz = F (b) − F (a).
This result shows us how to calculate the integral of a function f : we need to find a
function that, when differentiated, yields f .
The symbol
∫ f (x)dx,
called the indefinite integral of f , denotes the set of functions F for which F ' = f .
Why "set"? Because if F '(z) = f (z) for all z then for any function H with H(z) = F (z)
+ c, where c is a constant, we also have H'(z) = f (z) for all z. In honor of the
constant c, we sometimes write statements like "∫2x dx = x2 + c", meaning that the
derivative of the function x2 + c, for any value of c, is 2x.
For many functions, finding the indefinite integral is not easy. In fact, the integral of
many functions cannot be written as an explicit formula.
Some integrals that may be expressed simply are
∫xndx = xn + 1/(n+1) + c,
∫exdx = ex + c
and

33
∫(1/x)dx = ln |x| + c.
A useful fact to employ when finding some integrals is that the derivative of ln f (x)
is f '(x)/ f (x) (an implication of the chain rule, discussed later). Thus if you can
express the function you are integrating in the form f '(x)/ f (x), its integral is ln f (x).
For example,
∫
x
x2 + 1
dx = (1/2)ln (x2 + 1) + c.
Integration by parts
Sometimes we can find the indefinite integral of a product of two functions by using
the result that
∫ f (x)g'(x)dx = f (x)g(x) − ∫ f '(x)g(x)dx.
Example
∫xex
dx = xex
− ∫ex
dx = xex
− ex
+ c. Check by differentiating!
1.4 Exercises on one-variable calculus
1. Find the first derivatives of
a. 2x4 + 3x1/2 + 7
b. 4/x2
c. (x + 3)/(x2 + x)
d. xln x
2. At what points is the function |x| differentiable?
3. Find the following indefinite integrals, remembering that the derivative
of eax is aeax (an implication of the chain rule, studied in a later section).
a. ∫x4dx
b. ∫2e−2xdx
c. ∫(x + 3)(x + 1)1/2dx
d. ∫(4x + 2)/(x2
+ x)dx
4. Find the following definite integrals.
a. ∫1
33x1/2dx
b. ∫2
3(e2x + ex)dx

34
c. ∫0
∞
e−rt
dt where r > 0 is a constant
1.5 Calculus: many variables
Open and closed sets
To make precise statements about functions of many variables, we need to generalize
the notions of open and closed intervals to many dimensions.
We want to say that a set of points is "open" if it does not include its boundary. But
how exactly can we define the boundary of an arbitrary set of vectors?
We say that a point x is a boundary point of a set of n-vectors if there are points in the
set that are arbitrarily close to x, and also points outside the set that are arbitrarily
close to x. A point x is an interior point of a set if we can find a (small) number ε such
that all points within the distance ε of xare in the set.
The green point in the following figure, for example, is a boundary point of the (two-
dimensional) gray set because every disk centered at the point, however small,
contains both points in the set and points outside the set. The red point is an interior
point because the blue disk (and all smaller disks, as well as some larger ones) contain
exclusively points in the set.
Here is a more precise definition of the two notions.

35
Definition
 A point x is a boundary point of a set S of vectors if for every number ε > 0 (however small), at least one po
ofx is in S, and at least one point within the distance ε of x is outside S.
 A point x is an interior point of a set S of vectors if there is a number ε > 0 such that all points within the d
of S.
Example
The set of interior points of the set [a, b] is (a, b) (as we saw previously), and the boundary points are a and b.
Example
The set of interior points of the set (a, b) is (a, b) (as we saw previously), and the boundary points are a and b. Not
boundary points of the set are not members of the set.
We may now define open and closed sets.
Definition
 The set S is open if every point in S is an interior point of S.
 The set S is closed if every boundary point of S is a member of S.
Note that a set may be neither open nor closed (consider the interval [0, 1)).
Example
The boundary points of the set [a, b] are a and b, so this set is closed.
Example
The interior of the set (a, b) is (a, b), the set itself, so the set is open.
Example
The set of interior points of the set {(x, y): x + y ≤ c, x ≥ 0, and y ≥ 0} is {(x, y): x + y < c, x > 0, and y > 0}, and the
{(x, y): x + y = c, x = 0, or y = 0}. Thus the set is closed.
Informally, a set defined by weak inequalities (≤ and ≥) is closed, whereas one
defined by strict inequalities (< and >) is open.
Example

36
Consider the set of all real numbers. This set has no boundary points, and is thus both open and closed.
Differentiability
The derivative of a function of a single variable at a point is a good linear
approximation of the function around the point. If no good linear approximation exists
at some point x (as is the case if the graph of the function has a "kink" at x), then the
function is not differentiable at x.
The definition of differentiability for a function of many variables, of which I do not
give a precise statement, captures the same idea: a function of many variables is
differentiable at a point if there exists a good linear approximation of the function
around the point. Like the graph of a differentiable function of a single variable, the
graph of a differentiable function of many variables is "smooth", with no "kinks".
In economic theory we often assume that the functions in our models (production
functions, utility functions, ...) are differentiable.
Partial derivatives
Let f be a differentiable function of n variables. (The value of f at the point (x1,
..., xn), for example, is f (x1, ..., xn).) Suppose that we fix the values of all variables
except xi. Specifically, suppose that xj = cj for all j ≠ i. Denote by g the resulting
function of the single variable xi. Precisely,
g(xi) = f (c1, ..., ci−1, xi, ci+1, ..., cn) for all xi.
The derivative of g at the point ci is called the partial derivative of f with respect to
its ith argument at the point (c1, ..., cn). We may characterize the process of obtaining
this partial derivative by saying that we "differentiate f with respect to its ith
argument holding all the other arguments fixed".
Following common practice in economics, I usually denote the value of the partial
derivative of f with respect to its ith argument at the point (c1, ...,cn) by f 'i(c1, ..., cn).
This notation does not work well for a function that itself has a subscript. Suppose, for
example, that we are working with the functions gj for j = 1, ..., m, each of which is a
function of n variables. How do we denote the partial derivative of gj with respect to
its kth argument? I write g'ji, but this notation is not elegant. In the mathematical
literature, the notation Di f (c1, ...., cn) is used for the value of the partial derivative
of f with respect to its ith argument at the point (c1, ..., cn). This notation is elegant
and may be used unambiguously for any function. However, it is not common in
economics.

37
The notation (∂ f /∂xi)(c1, ..., cn) is also sometimes used. Although this notation is
clumsy in using six symbols (∂ f /∂xi) where three (Di f ) suffice, it is often used by
economists, and I sometimes follow this practice.
Occasionally the argument of a function may be more conveniently referred to by its
name than its index. If I have called the arguments of f by the names w and p, for
example (writing f (w, p)), I may write f p(w, p) for the value of the partial derivative
of f with respect to its second argument at the point (w, p).
Example
Let f (x1, x2) = (x1)3ln x2. Then f '1(x1, x2) = 3(x1)2ln x2 and f '2(x1, x2) = (x1)3/x2.
Let f be a function of n variables. The fact that each of the n partial derivatives
of f exists for all values of the argument of f does not imply that f is differentiable.
In fact, it does not even imply that f is continuous! (See the example if you are
curious.) However, if all the partial derivatives of f exist and are continuous
functions, then f is differentiable, and in fact its derivative is continuous. This result
justifies our defining a function to becontinuously differentiable if all its partial
derivatives exist and all these partial derivatives are continuous functions. (Absent the
result, this terminology would be misleading, because it does not contain the word
"partial", though it concerns properties of the partial derivatives.)
The derivative of f 'i with respect to its jth argument evaluated at (x1, ..., xn) is
denoted f "ij(x1, ..., xn), and is called a "cross partial". The following result shows that
for functions that satisfy relatively mild conditions, the order in which the
differentiations are performed does not matter.
Proposition (Young's theorem)
Let f be a differentiable function of n variables. If each of the cross partials f "ij and f "ji exists and is continuous a
setS of values of (x1, ..., xn) then
f "ij(x1, ..., xn) = f "ji(x1, ..., xn) for all (x1, ..., xn) in S.
The condition in this result is satisfied for any polynomial, and for all the other
differentiable functions that appear in this tutorial.
1.5 Exercises on multivariate calculus

38
1. Determine whether each of the following sets is open, closed, both open and
closed, or neither open nor closed.
a. {(x, y): x2
+ y2
< 1}
b. {x: x is an integer}
c. {(x, y): 0 < x < 1 and y = 0}.
2. For each of the following functions, find the partial derivatives f '1, f '2,
and f "12.
a. f (x1, x2) = 2x1
3 + x1x2
b. f (x1, x2) = (x1 + 2)/(x2 + 1)
3. For the production function f (K, L) = 9K1/3
L2/3
, find the marginal products
of K and L (i.e. the partial derivatives of the function with respect to Kand with
respect to L).
1.6 Graphical representation of functions
Diagrams are very helpful in solving many mathematical problems involving
functions. They are especially helpful in solving optimization problems, which occur
throughout economic theory. Learning how to graphically represent functions will
help enormously in understanding the material in this tutorial.
Functions of a single variable
A function of a single variable is most usefully represented by its graph.
Linear functions
A linear function of a single variable has the form
f (x) = ax + b.
(Such a function is sometimes called "affine" rather than linear, the term "linear"
being reserved by some mathematicians for functions of the form f (x) = ax.) The
graph of this function is a straight line with slope a; its value when x = 0 is b. Two
examples are shown in the following figure.

39
Quadratic functions
A quadratic function of a single variable has the form
f (x) = ax2 + bx + c
with a ≠ 0. The graph of such a function takes one of the two general forms shown in
the following figure, depending on the sign of a.
The derivative of the function is zero when x = −b/2a, at which point the value of the
function is c − b2/4a. The function may be written as
a(x + b/2a)2 + c − b2/4a.
We can see from this expression that for any number z, the value of the function is the
same at z − b/2a as it is at −z − b/2a. That is, the graph of the function is symmetric
about a vertical line at x = −b/2a. The steepness of the graph depends on the
parameters a and b: the derivative of the function at the point x is 2ax + b.

40
Example
Sketch the graph of 2x2
− 8x + 6. The coefficient of x2
is positive, so the graph is U-shaped. The derivative is zero a
value of the function is −2. The value of the function at x = 0 is 6. The function is sketched in the following figure.
Reciprocal function
The function
f (x) = c/x
has one of the forms shown in the following figure, depending on the sign of c.
For c > 0 the graph is symmetric about the line with slope 1 through the origin; for c <
0 it is symmetric about the line with slope −1 through the origin.

41
The curves in these diagrams are called "rectangular hyperbolae".
Exponential function
The function
f (x) = ex
is positive for all values of x, has positive derivatives of every order, and takes the
value 1 at x = 0. Its graph is shown in the following diagram.
Other functions
One way to get an idea of the shape of the graph of a arbitrary function of a single
variable is to ask some plotting software on a computer to draw the graph. One such
piece of software is available on the web. (Note that this software plots multiple
functions: you enter a comma-separated list delimited by brackets---for example
"[x^2, sin(x), exp(x)]".) If you don't have a computer handy, it is usually helpful to
examine several characteristics of the function:
 the points (if any) at which its first derivative is zero
 its value at the points at which its first derivative is zero
 its value when x = 0
 the points at which its value is zero
 the behavior of its derivative.
Example
Sketch the function 2/x + x2. The derivative of this function is −2/x2 + 2x, which is zero when x = 1, at which point
positive. The value of the function at x = 1 is 3. As x decreases to 0 the value of the function increases without bou
value decreases without bound. The value of the function is zero when x = −21/3
, and the derivative is negative for a
Putting this information together we get the following figure.

42
Functions of two variables
The graph of a function of two variables is a surface in three dimensions. This surface
may be represented in a perspective drawing on a piece of paper, but for many
functions the drawing (a) is difficult to execute and (b) hides some features of the
function---only parts are visible. Computer software allows one to construct such
drawings easily, from many different viewpoints, solving both problems. But even so,
another way of looking at a function of two variables is useful. This method, which
involves the construction of a topographic map of the function, may easily be carried
out by hand for many functions.
Level curves
Let f be a function of two variables, and c a constant. The set of pairs (x, y) such that
f (x, y) = c
is called the level curve of f for the value c.
Example
Let f (x, y) = x2 + y2 for all (x, y). The level curve of f for the value 1 is the set of all pairs (x, y) such that x2 + y2 =
set is shown in the following figure.

43
Example
Let f (x, y) = a(x − b)2 + c(y − d)2. Each level curve of this function is an ellipse centered at (b, d). If a = c the ellips
previous example). If a < c it is elongated horizontally; if a > c then the ellipse is elongated vertically. Examples ar
figure.
Example
Let f (x, y) = xy for all (x, y). The level curve of f for the value 1 is the set of all pairs (x, y) such that xy = 1, or, eq
of this set for −2 ≤ x ≤ 2 and −2 ≤ y ≤ 2 is the union of the sets of points on the orange lines in the following figure.

44
By drawing the level curves of a function for various values, we construct a
topographic map of the function that gives us a good picture of the nature of the
function. If the values we choose are equally spaced (e.g. 1, 2, 3, or 10, 20, 30) then
curves that are close indicate regions in which the rate of change of the function is
large, whereas curves that are far apart indicate regions in which the rate of change of
the function is small.
Example
Let f (x, y) = x2 + y2 for all (x, y). A collection of level curves of f is shown in the following figure. The number be
of the function to which it corresponds. We see that the graph of the function is a bowl with sides whose slopes inc
the center.
Example
Let f (x, y) = xy for all (x, y). A collection of level curves of f is shown in the following figure. The number beside
the function to which it corresponds. (The level curve for the value 0 consists of the axes.)
Note that a level curve is defined as a set, and indeed some level curves of some
functions are not "curves" at all.
Consider, for example, the function f defined by f (x, y) = 1 for all (x, y). (That is, the
value of f is 1 for all values of x and y.) The level curve of this function for the value

45
2 is empty (there are no values of (x, y) such that f (x, y) = 2) and the level curve for
the value 1 is the set all all points (x,y).
In less extreme examples, some but not all level curves are sets. Consider a conicle
terraced mountain, with terraces at heights of 10 and 30. Some level curves of this
function are shown in the following figure. The level curves corresponding to heights
of 10 and 30 are thick, whereas those corresponding to heights of 0 and 20 are not.
Contour lines on topographic maps of regions of the world are never thick because the
earth is nowhere exactly flat (even in Manitoba).
Economists call the level curves of a utility function indifference curves and those of a
production function isoquants.
1.6 Exercises on graphical representation of functions
1. Sketch the following functions of a single variable for the domain −2 ≤ x ≤ 2.
a. f (x) = x2
b. f (x) = x3
c. f (x) = −x4
d. f (x) = 2 − x − x2
e. f (x) = 1/x
f. f (x) = (x2 − 1)2
2. Sketch some level curves of the following functions of two variables.
a. f (x, y) = x + 2y
b. f (x, y) = (x + 2y)2
c. f (x, y) = x2 + 2y2
d. f (x, y) = xy2
3. Sketch the following sets in the plane.

46
a. {(x,y): x + 2y ≤ 4, x ≥ 0, and y ≥ 0}
b. {(x,y): x2
+ 4y2
≤ 4}
c. {(x,y): xy ≤ 4, 0 ≤ x ≤ 4, 0 ≤ y ≤ 4}
2. Topics in multivariate calculus
 2.1 Introduction
 2.2 The chain rule

47
 2.3 Derivatives of functions defined implicitly
 2.4 Differentials and comparative statics
 2.5 Homogeneous functions
2.1 Introduction to multivariate calculus
Economic theory consists of models designed to improve our understanding of
economic phenomena. Many of these models have the following structure: each
member of a set of economic agents optimizes given some constraints and, given the
optimal actions, variables adjust to reach some sort of equilibrium.
Consider, for example, a model in which the agents are profit-maximizing firms.
Suppose that there is a single input that costs w per unit, and that a firm transforms
input into output using a (differentiable) production function f and sells the output
for the price p. A firm's profit when it uses the amount x of the input is then
p f (x) − wx.
As you know, if the optimal amount of the input is positive then it satisfies the "first-
order condition"
p f '(x) − w = 0.
Further, under some conditions on f this equation has a single solution and this
solution maximizes the firm's profit. Suppose these conditions are satisfied. For any
pair (w, p), denote the solution of the equation by z(w, p). Then the condition
p f '(z(w, p)) − w = 0 for all (w, p)
defines z(w, p) implicitly as a function of w and p.
What can we say about the function z? Is it increasing or decreasing in w and p? How
does the firm's maximized profit change as w and p change?
If we knew the exact form of f we could answer these questions by solving
for z(w, p) explicitly and using simple calculus. But in economic theory we don't
generally want to assume that functions take specific forms. We want our theory to
apply to a broad range of situations, and thus want to obtain results that do not depend
on a specific functional form. We might assume that the function f has some
"sensible" properties---for example, that it is increasing---but we would like to impose
as few conditions as possible. In these circumstances, in order to answer the questions

48
about the dependence of the firm's behavior on w and p we need to find the
derivatives of the implicitly-defined function z. Before we do so, we need to study
the chain rule and derivatives of functions defined implicitly, the next two topics. (If
we are interested only in the rate of change of the firm's maximal profit with respect
to the parameters w and p, not in the behavior of its optimal input choice, then
the envelope theorem, studied in a later section, is useful.)
Having studied the behavior of a single firm we may wish to build a model of an
economy containing many firms and consumers that determines the prices of goods.
In a "competitive" model, for example, the prices are determined by the equality of
demand and supply for each good---that is, by a system of equations. In many other
models, an equilibrium is the solution of a system of equations. To study the
properties of such an equilibrium, another mathematical technique is useful.
I illustrate this technique with an example from macroeconomic theory. A simple
macroeconomic model consists of the four equations
Y=C + I + G
C= f (Y − T)
I=h(r)
r=m(M)
where Y is national income, C is consumption, I is investment, T is total taxes, G is
government spending, r is the rate of interest, and M is the money supply. We assume
that M, T, and G are "parameters" determined outside the system (by a government,
perhaps) and that the equilibrium values of Y,C, I and r satisfy the four equations,
given M, T, and G.
We would like to impose as few conditions as possible on the functions f , h, and m.
We certainly don't want to assume specific functional forms, and thus cannot solve for
an equilibrium explicitly. In these circumstances, how can we study how the
equilibrium is affected by changes in the parameters? We may use the tool
of differentials, another topic in this section.
2.2 The chain rule

49
Single variable
You should know the very important chain rule for functions of a single variable:
if f and g are differentiable functions of a single variable and the function F is
defined by F (x) = f (g(x)) for all x, then
F '(x) = f '(g(x))g'(x).
This rule may be used to find the derivative of any "function of a function", as the
following examples illustrate.
Example
What is the derivative of the function F defined by F (x) = ex1/2? If we define the functions f and g by f (z) = ez and
have F (x) = f (g(x)) for all x. Thus using the chain rule, we have F '(x) = f '(g(x))g'(x) = eg(x)(1/2)x−1/2 = (1/2)ex1/2x−1/2.
Example
What is the derivative of the function F defined by F (x) = log x2
? If we define the functions f and g by f (z) = log
have F (x) = f (g(x)), so that by the chain rule we have F '(x) = f '(g(x))g'(x) = (1/x2)2x = 2/x.
More importantly for economic theory, the chain rule allows us to find the derivatives
of expressions involving arbitrary functions of functions. Most situations in
economics involve more than one variable, so we need to extend the rule to many
variables.
Two variables
First consider the case of two variables. Suppose that g and h are differentiable
functions of a single variable, f is a differentiable function of two variables, and the
function F of a single variable is defined by
F (x) = f (g(x), h(x)) for all x.
What is F '(x) in terms of the derivatives of f , g, and h? The chain rule says that
F '(x) = f '1(g(x), h(x))g'(x) + f '2(g(x), h(x))h'(x),
where f i' is the partial derivative of f with respect to its ith argument. (This
expression is sometimes referred to as the total derivative of F with respect to x.)
An extension
We can extend this rule. If f , g, and h are differentiable functions of two variables
and the function F of two variables is defined by

50
F (x, y) = f (g(x, y), h(x, y)) for all x and y
then
F 'x(x, y) = f '1(g(x, y), h(x, y))gx'(x, y) + f '2(g(x, y), h(x, y))hx'(x, y),
and similarly for F 'y(x, y).
More generally, we have the following result.
Proposition
If gj
is a differentiable function of m variables for j = 1, ..., n, f is a differentiable function of n variables, and the f
defined by
F (x1, ..., xm) = f (g1(x1, ..., xm), ..., gn(x1, ..., xm)) for all (x1, ..., xm)
then
F 'j(x1, ..., xm) = ∑i=1
n f i'(g1(x1, ..., xm), ..., gn(x1, ..., xm))gi
j'(x1, ..., xm),
where gi
j' is the partial derivative of gi with respect to its jth argument.
Example
Consider a profit-maximizing firm that produces a single output with a single input. Denote its (differentiable) prod
price of the input by w, and the price of the output by p. Suppose that its profit-maximizing input when the prices a
maximized profit is
π(w, p) = p f (z(w, p)) − wz(w, p).
How does this profit change if p increases?
Using the chain rule we have
π'p(w, p) = f (z(w, p)) + p f '(z(w, p))z'p(w, p) − wz'p(w, p)
or
π'p(w, p) = f (z(w, p)) + z'p(w, p)[p f '(z(w, p)) − w].
But we know that if z(w, p) > 0 then p f '(z(w, p)) − w = 0, which is the "first-order condition" for maximal profit. T
have
π'p(w, p) = f (z(w, p)).
In words, the rate of increase in the firm's maximized profit as the price of output increases is exactly equal to its o

51
Example
A consumer trades in an economy in which there are n goods. She is endowed with the vector e of the goods, and f
demand for any good i depends on p and the value of her endowment given p, namely p·e (the inner product of p a
alternatively write as ∑j=1
npjej). Suppose we specify her demand for good i by the function f of n + 1 variables with
What is the rate of change of her demand for good i with respect to pi?
The function f has n + 1 arguments---the n elements p1, ..., pn of the vector p and the number p·e. The der
to pi is 0 for j ≠ i and 1 for j = i, and the derivative of p·e with respect to pi is ei. Thus by the chain rule, th
consumer's demand with respect to pi is
f 'i(p, p·e) + f 'n+1(p, p·e)ei.
Leibniz's formula
We sometimes need to differentiate a definite integral with respect to a parameter that
appears in the integrand and in the limits of the integral. Suppose that f is a
differentiable function of two variables, a and b are differentiable functions of a single
variable, and the function F is defined by
F (t) = ∫b(t)
a(t) f (t, x) dx for all t.
What is F '(t)? By the logic of the chain rule, it is the sum of three components:
 the partial derivative of the integral with respect to its top limit, times b'(t)
 the partial derivative of the integral with respect to its bottom limit, times a'(t)
 the partial derivative of the integral with respect to t, holding the limits fixed.
By the Fundamental Theorem of Calculus, the partial derivative of the integral with
respect to its top limit is f (t, b(t)), and the partial derivative of the integral with
respect to its bottom limit is − f (t, a(t)). As for the last term, you might correctly
guess that it is
∫b(t)
a(t) f t'(t, x) dx,
the integral of the partial derivative of the function. We have the following result,
discovered by Gottfried Wilhelm von Leibniz (1646-1716).
Proposition (Leibniz's formula)
Let f be a differentiable function of two variables, let a and b be differentiable functions of a single variable, and d
F (t) = ∫b(t)
a(t) f (t, x) dx for all t.

52
Then
F '(t) = f (t, b(t))b'(t) − f (t, a(t))a'(t) + ∫b(t)
a(t) f 1'(t, x) dx.
As with other expressions obtained by the chain rule, we can interpret each of its
parts. If t changes then the limits of the integral change, and the value of the
function f changes at each point x. The change in the integral can thus be
decomposed into three parts:
 the part due to the change in b(t), namely f (t, b(t))b'(t)
 the part due to the change in a(t), namely − f (t, a(t))a'(t) (if a(t) increases then
the integral decreases)
 the part due to the change in the values of f (t, x), namely ∫b(t)
a(t) f t'(t, x) dx.
Example
The profit of a firm is π(x) at each time x from 0 to T. At time t the discounted value of future profit is
V(t) = ∫T
tπ(x)e−r(x−t)dx,
where r is the discount rate. Find V'(t).
Use Leibniz's rule. Define a(t) = t, b(t) = T, and f (t, x) = π(x)e−r(x−t). Then a'(t) = 1, b'(t) = 0, and f 1'(t, x) =
V'(t) = −π(t)e−r(t−t) + ∫T
tπ(x)re−r(x−t)dx = −π(t) + rV(t).
The first term reflects the fact that the future is shortened when t increases and the second term reflects the fact tha
profit at any given time is obtained sooner, and is thus worth more.
2.2 Exercises on the chain rule
1. Find the derivatives of the following functions of a single variable.
a. f (x) = (3x2 − 1)3.
b. f (x) = xe2x. (Remember that ln(ax) = xln a, so that ax = exln a
.)
c. f (x) = 2x
+ x2
.
d. f (x) = ln x2.
e. f (x) = sin bx, where b is a constant.
2. Define the function F of two variables by F (x, y) = f (g(x, y), h(x, y)) for all
(x, y), where f (s, t) = st2, g(x, y) = x + y2, and h(x, y) = x2y. Use the chain rule to
find F 'x(x, y) and F 'y(x, y).

53
3. Define the function F of two variables by F (x, y) = f (g(x, y), h(k(x))),
where f , g, h, and k are differentiable functions. Find the partial derivative
of F with respect to x in terms of the partial derivatives of f , g, h, and k.
4. Define the function F of two variables by F (p, q) = p f (p, q, m(p, q)),
where f and m are differentiable functions. Find an expression for the partial
derivative of F with respect to p in terms of the partial derivatives of f and m.
5. Define the function F of two variables by F (x, y) = h( f (x), g(x, y)),
where f , g, and h are differentiable functions. Find the derivative of F with
respect to x in terms of the partial derivatives of f , g, and h.
6. Define the functions U and V of two variables by U(x, y) = F ( f (x) + g(y)) for
all (x, y), and V(x, y) = ln[U'x(x, y)/U'y(x, y)] for all (x, y), where f , g, and F are
twice-differentiable functions. Find V"xy(x, y).
7. Let
y = F (x1(p), ..., xn(p), p) − p·x(p),
where F and xi for i = 1, ..., n are differentiable functions, p is an n-vector, x(p)
denotes the vector (x1(p), ..., xn(p)), and p·x(p) denotes the inner product
of p and x(p). Find the derivative of y with respect to pi, given pj for j ≠ i. (Use
the notation ∂xj/∂pi for the partial derivative of xjwith respect to pi.)
8. The amount x of some good demanded depends on the price p of the good and
the amount a the producer spends on advertising: x = f (p, a), with f 'p(p, a) < 0
and f 'a(p, a) > 0 for all (p, a). The price depends on the weather, measured by
the parameter w, and the tax rate t: p =g(w, t), where g'w(w, t) > 0 and g't(w, t) <
0 for all (w, t). The amount of advertising depends only on t: a = h(t), with h'(t)
> 0. If the tax rate increases does the demand for the good necessarily increase
or necessarily decrease, or neither?
9. Let
H(r) = ∫0
g(r) e−rt f (t) dt,
where g and f are differentiable functions. Find H'(r).
10.Let
H(t) = ∫t
t−Tg(x)e−δ(t−x) dx,

54
where g is a differentiable function and δ and T are constants. Find the
derivative H'(t).
11.Suppose that the amount of some good demanded depends on the price p of the
good and the price q of another good; both these prices depend on two
parameters, α and β (e.g. the weather, the rate of government subsidy). You
observe that ∂x/∂α > 0, ∂p/∂α < 0, ∂q/∂α > 0, and |∂p/∂α| > |∂q/∂α|. You have a
theory that x = f (p(α,β),q(α,β)), where f 1'(p,q) > 0 and f '2(p,q) > 0 for all
(p,q). Are your observations consistent with your theory? Are your
observations consistent with a theory that imposes the stronger restriction
that f 1'(p,q) > f '2(p,q) > 0 for all (p,q)?
12.A firm faces uncertain demand D and has existing inventory I. The firm wants
to choose its stock level Q to minimize the value of the function
g(Q) = c(Q − I) + h∫0
Q(Q−D) f (D)dD + p∫Q
a(D−Q) f (D)dD,
where c, I, h, p, and a are positive constants with p > c, and f is a nonnegative
function that satisfies ∫0
a f (D)dD = 1 (so that it can be interpreted as a
probability distribution function). (The first term represents the cost of the new
stock; the second term represents the cost of overstocking; and the third term
represents the cost of understocking (you miss sales, and the customers who are
turned away may not come back in the future).)
a. Find g'(Q) and g''(Q) and show that g''(Q) > 0 for all Q.
b. Define F (Q*) = ∫0
Q* f (D)dD, where Q* is the stock level that
minimizes g(Q). Use the "first-order" condition g'(Q*) = 0 to find F (Q*)
(the probability that demand D does not exceed Q*) in terms of the
parameters p, c, and h. (Hint: Use the fact that ∫0
Q* f (D)dD + ∫Q*
a f (D)dD =
∫0
a f (D)dD = 1.)
2.3 Derivatives of functions defined implicitly
One parameter
The equilibrium value of a variable x in some economic models is the solution of an
equation of the form
f (x, p) = 0,

55
where f is a function and p is a "parameter" (a given number). In such a case, we
would sometimes like to know how the equilibrium value of xdepends on the
parameter. For example, does it increase or decrease when the value of the parameter
increases?
Typically, we make assumptions about the form of the function f ---for example, we
might assume that it is increasing in x and decreasing in p---but do not assume that it
takes a specific form. Thus typically we cannot solve explicitly for x as a function
of p (i.e. we cannot write g(p) = something).
We say that the equation
f (x, p) = 0 for all p
defines x implicitly as a function of p. We may emphasize this fact by
writing f (x(p), p) = 0 for all p.
Before trying to determine how a solution for x depends on p, we should ask whether,
for each value of p, the equation has a solution. Certainly not all such equations have
solutions. The equation x2 + 1 = 0, for example, has no (real) solution. Even a
single linear equation may have no solution in the relevant range. If, for example, the
value of x is restricted to be nonnegative number (perhaps it is the quantity of a good),
then for p > 0 the equation x+ p = 0 has no solution.
If a single equation in a single variable has a solution, we may be able to use
the Intermediate Value Theorem to show that it does. Assume that the function f is
continuous, the possible values of x lie between x1 and x2, and for some value of p we
have f (x1, p) < 0 and f (x2, p) > 0, or alternatively f (x1, p) > 0 and f (x2, p) < 0. Then
the Intermediate Value Theorem tells us that there exists a value
of x between x1 and x2 for which f (x, p) = 0. (Note that even if these conditions are not
satisfied, the equation may have a solution.)
If we cannot appeal to the Intermediate Value Theorem (because, for example, f is
not continuous, or does not satisfy the appropriate conditions), we may be able to
argue that a solution exists by appealing to the particular features of our equation.
Putting aside the question of whether the equation has a solution, consider the
question of how a solution, if one exists, depends on the parameter p. Even if we
cannot explicitly solve for x, we can find how a solution, if it exists, depends on p---
we can find the derivative of x with respect to p---by differentiating the equation that
defines it. The principle to use is simple: if you want to find a derivative, differentiate!

56
Differentiating both sides of the equation f (x(p), p) = 0, using the chain rule, we get
f '1(x(p), p)x'(p) + f '2(x(p), p) = 0,
so that
x'(p) = − f '2(x(p), p)/ f '1(x(p), p).
Notice that even though you cannot isolate x in the original equation, after
differentiating the equation you can isolate the derivative of x, which is what you
want.
This calculation tells you, for example, that if f is an increasing function of both its
arguments ( f '1(x, p) > 0 and f '2(x, p) > 0 for all (x, p)), then x is a decreasing function
of p.
Application: slopes of level curves
The equation f (x, y) = c of the level curve of the differentiable function f for the
value c defines y implicitly as a function of x: we can write
f (x, g(x)) = c for all x.
What is g'(x), the slope of the level curve at x? If we differentiate both sides of the
identity f (x, g(x)) = c with respect to x we obtain
f 1'(x, g(x)) + f 2'(x, g(x))g'(x) = 0,
so that we can isolate g'(x):
g'(x) = −
f 1'(x, g(x))
f 2'(x, g(x))
or, in different notation,

57
dy
dx
= −
f 1'(x, y)
f 2'(x, y)
.
In summary, we have the following result.
Proposition
Let f be a differentiable function of two variables. The slope of the level curve of f for the value f (x0, y0) at the p
−
f 1'(x0, y0)
f 2'(x0, y0)
.
We deduce that the equation of the tangent to the level curve at (x0, y0) is
y − y0 = −
f 1'(x0, y0)
f 2'(x0, y0)
·(x − x0).
(Remember that the equation of a line through (x0, y0) with slope m is given
by y − y0 = m(x − x0).) Thus the equation of the tangent may alternatively be written as
f 1'(x0, y0)(x − x0) + f 2'(x0, y0)(y − y0) = 0,
or
( f 1'(x0, y0), f 2'(x0, y0))
x − x0
y − y0
= 0.
The vector ( f 1'(x0, y0), f 2'(x0, y0)) is called the gradient vector and is denoted ∇ f (x0, y0).
Let (x, y) ≠ (x0, y0) be a point on the tangent at (x0, y0). Then the vector

58
x − x0
y − y0
is parallel to the tangent. The previous displayed equation, in which the product of this
vector with the gradient vector is 0, shows that the two vectors are orthogonal (the
angle between them is 90°). Thus the gradient vector is orthogonal to the tangent, as
illustrated in the following figure.
One can compute the second derivative of the level curve as well as the first
derivative, by differentiating once again.
Many parameters
Suppose that the equilibrium value of the variable x is the solution of an equation of
the form
f (x, p) = 0,
where p is a vector of parameters---p = (p1, ..., pn), say. By differentiating the equation
with respect to pi, holding all the other parameters fixed, we may determine
how x varies with pi.
Recording the dependence of x on p explicitly in the notation, we have
f (x(p), p) = 0 for all p.
Differentiating this identity with respect to pi we have
f 'x(x(p), p)x'i(p) + f 'pi(x(p), p) = 0
so that

59
x'i(p) = −
f 'pi(x(p), p)
f 'x(x(p), p)
.
Example
Consider the competitive firm studied previously that uses a single input to produce a single output with the differe
function f , facing the price w for the input and the price p for output. Denote by z(w, p) its profit-maximizing inpu
know thatz(w, p) satisfies the first-order condition
p f '(z(w, p)) − w = 0 for all (w, p).
How does z depend on w and p?
Differentiating with respect to w the equation that z(w, p) satisfies we get
p f ''(z(w, p))z'w(w, p) − 1 = 0.
Thus if f ''(z(w, p)) ≠ 0 then
z'w(w, p) =
1
p f ''(z(w, p))
.
We know that f ''(z(w,p)) ≤ 0 given that z(w, p) is a maximizer, so that if f ''(z(w, p)) ≠ 0 we conclude that z'w(w, p)
the input price increases, the firm's optimal output decreases.
A similar calculation yields
z'p(w, p) = −
f '(z(w, p))
p f ''(z(w, p))
,
which for the same reason is positive.
2.3 Exercises on derivatives of functions defined implicitly

60
1. Suppose that 2x2
+ 6xy + y2
= c for some constant c. Find dy/dx.
2. Suppose that the functions f and g are differentiable and g( f (x)) = x for all
values of x. Use implicit differentiation to find an expression for the
derivative f '(x) in terms of the derivative of g.
3. The demand and the supply for a good both depend upon the price p of the
good and the tax rate t: D = f (p,t) and S = g(p,t). For any given value of t, an
equilibrium price is a solution of the equation f (p,t) = g(p,t). Assume that this
equation defines p as a differentiable function of t. Find ∂p/∂t in terms of the
partial derivatives of f and g.
4. The demand for a good both depends upon the price p of the good and the tax
rate t: D = f (p,h(t)). The supply of the good depends on the price: S = g(p). For
any given value of t, an equilibrium price is a solution of the equation f (p,h(t))
= g(p). Assume that this equation defines pas a differentiable function of t. Find
∂p/∂t in terms of the derivatives of f , g, and h.
5. Let f (x, y) = 2x2 + xy + y2.
a. Find the equation of the tangent at (x, y) = (2, 0) to the level curve
of f that passes through this point.
b. Find the points at which the slope of the level curve for the value 8 is 0.
6. Let D = f (r, P) be the demand for an agricultural product when the price
is P and the producers' total advertising expenditure is r; f is decreasing in P.
Let S = g(w, P) be the supply, where w is an index of how favorable the
weather has been; g is increasing in P. Assume thatg'w(w, P) > 0. An
equilibrium price satisfies f (r, P) = g(w, P); assume that this equation
defines P implicitly as a differentiable function of r andw. Find ∂P/∂w and
determine its sign.
7. The equilibrium value of the variable x is the solution of the equation
f (x, α, β) + g(h(x), k(α)) = 0,
where α and β are parameters and f , g, h, and k are differentiable functions.
How is the equilibrium value of x affected by a change in the parameter α
(holding β constant)?
8. The equilibrium value of the variable x is the solution of the equation
f (x, g(x, α), β) + h(x, β) = 0,
where α and β are parameters and f , g, and h are differentiable functions. How
is the equilibrium value of x affected by a change in the parameter α (holding β
constant)?
9. The equilibrium value of the variable x depends on the parameters (a1, ..., an):

61
f (a1, ..., an, x) = 0.
Find the rate of change of x with respect to ai for any i = 1, ..., n.
10.The value of y is determined as a function of t by the equation
∫t
1 f (x,y)dx = 0,
where f is a differentiable function. Find dy/dt.
11.The function g is defined implicitly by the condition F ( f (x,y),g(y)) = h(y).
Find the derivative g'(y) in terms of the functions F , f , g, and hand their
derivatives.
12.Suppose that x is implicitly defined as a function of the parameter t by the
equation f (x)/ f '(x) − x = t, where f is a twice-differentiable function.
Find dx/dt.

62
2.4 Differentials and comparative statics
Introduction
We may use the tool of implicit differentiation to study the dependence of a
variable x on a list p of parameters when the variable is defined by an equation
like f (x, p) = 0. Many models in economic theory involve several variables that
satisfy several equations simultaneously. In such cases, another (closely related)
method is useful.
Suppose that we have two variables, x and y, and two parameters, p and q, and for any
values of p and q the values of the variables satisfy the two equations
f (x, y, p, q)= 0
g(x, y, p, q)= 0.
These two equations implicitly define x and y as functions of p and q. As in the case
of a single equation, two questions arise:
 Do the equations have solutions for x and y for any given values of p and q?
 How do the solutions change as p or q, or possibly both, change?
Existence of a solution
We have seen that even a single equation in a single variable may not have a solution,
but that if it does, we may be able to use the Intermediate Value Theorem to show that
it does. Generalizations of the Intermediate Value Theorem, which I do not discuss,
can be helpful in showing that a collection of equations in many variables has a
solution.
A useful rough guideline for a set of equations to have a unique solution is that the
number of equations be equal to the number of variables. This condition definitely is
neither necessary nor sufficient, however. For example, the single equation x2 = −1 in
a single variable has no solution, while the single equation x2 + y2 = 0 in two variables
has a unique solution ((x, y) = (0, 0)). But there is some presumption that if the
condition is satisfied and the equations are all "independent" of each other, the system
is likely to have a unique solution, and if it is not satisfied there is little chance that it
has a unique solution.

63
Differentials
Now consider the question of how a solution, if it exists, depends on the parameters. A
useful tool to address this question involves the notion of adifferential.
Let f be a differentiable function of a single variable. If x increases by a small
amount from a to a + Δx, by how much does f (x) increase? A function of a single
variable is approximated at a by its tangent at a. Thus if Δx is very small then the
approximate increase in f (x) is
f '(a)Δx
(where f '(a) is of course the derivative of f at a).
For any change dx in x we define the differential of f (x) as follows.
Definition
Let f be a function of a single variable. For any real number dx, the differential of f (x) is
f '(x)dx.
By the argument above, if dx is small then the differential f '(x)dx is approximately
equal to the change in the value of f when its argument increases or decreases
by dx from x.
If f is a function of two variables, it is approximated by its tangent plane: for (x, y)
close to (a, b) the approximate increase in f (x, y) when xchanges by Δx and y changes
by Δy is
f 1'(a, b)Δx + f 2'(a, b)Δy.
For a function of many variables, the differential is defined as follows.

64
Definition
Let f be a function of n variables. For any real numbers dx1, ..., dxn, the differential of f (x1, ..., xn) is
f '1(x1, ..., xn)dx1 + ... + f 'n(x1, ..., xn)dxn.
As in the case of a function of a single variable, if dxi is small for each i = 1, ..., n, then
the differential f '1(x1, ..., xn)dx1 + ... + f 'n(x1, ..., xn)dxn is approximately equal to the
change in the value of f when each argument xi changes by dxi.
To find a differential we may simply find the partial derivatives with respect to each
variable in turn. Alternatively we can use a set of rules that are analogous to those for
derivatives. Denoting the differential of the function f by d( f ), we have:
d(a f + bg)=ad f + bdg
d( f ·g)=gd f + f dg
d( f /g)=(gd f − f dg)/g2
if z = g( f (x, y)) then dz = g'( f (x, y))d f
Comparative statics
Start with the simplest case: a single equation in one variable and one parameter:
f (x, p) = 0 for all x,
where x is the variable and p the parameter. We have previously seen how to
use implicit differentiation to find the rate of change of x with respect top. We may
reach the same conclusion using differentials. The differential of the left-hand side of
the equation is
f '1(x, p)dx + f '2(x, p)dp.
When p changes, the value of f (x, p) must remain the same for the equation f (x, p) =
0 to remain satisfied, so for small changes dx in x and dp in pwe must have,
approximately,
f '1(x, p)dx + f '2(x, p)dp = 0.
Rearranging this equation we have

65
dx
dp
= −
f 2'(x, p)
f 1'(x, p)
.
The entity on the left-hand side is the quotient of the small quantities dx and dp, not a
derivative. However, we can in fact show that the right-hand side is the derivative
of x with respect to p, as we found previously.
This technique may be extended to systems of equations. Suppose, for example, that
the variables, x and y, satisfy the following two equations, wherep and q are
parameters, as in the opening section above:
f (x, y, p, q)= 0
g(x, y, p, q)= 0.
Assume that the functions f and g are such that the two equations define two solution
functions
x*(p, q) and y*(p, q).
That is, f (x*(p, q), y*(p, q), p, q) = 0 and g(x*(p, q), y*(p, q), p, q) = 0 for all p and
all q.
How do x* and y* depend on the parameters p and q? Assuming that the functions x*
and y* are differentiable, we can answer this question by calculating the differentials
of the functions on each side of the two equations defining them. If the changes
in p and q are small, then the differentials must be equal, so that the equations
defining x* and y* remain satisfied. That is,
f 1' · dx + f 2' · dy + f 3' · dp + f 4' · dq= 0
g1' · dx + g2' · dy + g3' · dp + g4' · dq= 0.
(To make these equations easier to read, I have omitted the arguments of the partial
derivatives.)
To find the changes dx and dy in x and y necessary for these equations to be satisfied
we need to solve the equations for dx and dy as functions of dpand dq, the changes in
the parameters. (See the page on matrices and solutions of systems of simultaneous
equations if you have forgotten how.) We obtain

66
dx =
−g2' · ( f 3' · dp + f 4' · dq) + f 2' · (g3' · dp + g4' · dq)
f 1' · g2' − f 2' · g1'
and
dy =
g1' · ( f 3' · dp + f 4' · dq) − f 1' · (g3' · dp + g4' · dq)
f 1' · g2' − f 2' · g1'
.
Now, to determine the impact on x and y of a change in p, holding q constant, we
set dq = 0 to get
dx =
(−g2' · f 3' + f 2' · g3') · dp
f 1' · g2' − f 2' · g1'
and
dy =
(g1' · f 3' − f 1' · g3') · dp
f 1' · g2' − f 2' · g1'
.
We can alternatively write the first equation, for example, as
∂x
∂p
=
−g2' · f 3' + f 2' · g3'
f 1' · g2' − f 2' · g1'
.
If we make some assumption about the signs of the partial derivatives of f and g, this
expression may allow us to determine the sign of ∂x/∂p---that is, the direction in
which the equilibrium value of x changes when the parameter p changes.
This technique allows us also to study the change in a variable when more than one
parameter changes, as illustrated in the following economic example.
Example
Consider the macroeconomic model

67
Y=C + I + G
C= f (Y − T)
I=h(r)
r=m(M)
where the variables are Y (national income), C (consumption), I (investment) and r (the rate of interest), and the pa
supply), T (the tax burden), and G (government spending). We want to find how the variables change with the para
Take differentials:
dY= dC + dI + dG
dC= f '(Y − T)(dY − dT)
dI= h'(r)dr
dr= m'(M)dM
We need to solve for dY, dC, dI, and dr in terms of dM, dT, and dG. The system is too big to use Cramer's rule easi
to proceed step-by-step.
From the last two equations we have
dI = h'(r)m'(M)dM.
Now substitute for dI in the first equation to get
dY − dC=h'(r)m'(M)dM + dG
f '(Y − T)dY − dC= f '(Y − T)dT
You can solve this system for dY and dC. For example,
dY =
h'(r)m'(M)
1− f '(Y − T)
dM −
f '(Y − T)
1− f '(Y − T)
dT +
1
1− f '(Y − T)
dG.
Thus, for example, if T changes while M and G remain constant (so that dM = dG = 0), then the rate of change of Y
∂Y
∂T
=
− f '(Y − T)
1 − f '(Y − T)
.

68
That is, if 0 < f '(z) < 1 for all z then Y decreases as T increases. Further, we can deduce that if T and G increase by
the change in Y is
dY =
1 − f '(Y − T)
1 − f '(Y − T)
dT = dT.
That is, an equal (small) increase in T and G leads to an increase in Y of the same size.
2.4 Exercises on differentials
1. Find the differentials of the following.
a. z = xy2 + x3.
b. z = a1x1
2 + ... + anxn
2 (where a1, ..., an are constants).
c. z = A(α1x1
−ρ
+ ... + αnxn
−ρ
)−1/ρ
(where A, ρ, and α1, ..., αn are constants). [This is
a constant elasticity of substitution function.]
2. Consider the system of equations
xu3 + v=y2
3uv − x=4
a. Take the differentials of both equations and solve for du and dv in terms
of dx and dy.
b. Find ∂u/∂x and ∂v/∂x using your result in part (a).
3. The equilibrium values of the variables x, y, and λ are determined by the
following set of three equations:
U1'(x, y)=λp
U2'(x, y)=λq
px + qy=I
4. where p, q, and I are parameters and U is a twice differentiable function. Find
∂x/∂p. [Use Cramer's rule to solve the system of equations you obtain.]
5. The equilibrium values of the variables Y, C, and I are given by the solution of
the three equations
Y= C + I + G
C= f (Y, T, r)
I= h(Y, r)

69
6. where T, G, and r are parameters and f and h are differentiable functions. How
does Y change when T and G increase by equal amounts?
7. An industry consists of two firms. The optimal output of firm 1 depends on the
output q2 of firm 2 and a parameter α of firm 1's cost function: q1= f (q2,α),
where f 1'(q2,α) < 0 and f 2'(q2,α) > 0 for all q2 and all α. The optimal output of
firm 2 depends on the output q1 of firm 1: q2 =g(q1), where g'(q1) < 0. The
equilibrium values of q1 and q2 are thus determined as the solution of the
simultaneous equations
q1 = f (q2, α)
q2 = g(q1).
8. Is this information sufficient to determine whether an increase in α increases or
decreases the equilibrium value of q1?
9. The variables x and y are determined by the following pair of equations:
f (x)= g(y)
Ay + h(x)= β
10.where f , g, and h are given differentiable functions, β is a constant, and A is a
parameter. By taking differentials, find ∂x/∂A and ∂y/∂A.
11.The optimal advertising expenditure of politician 1 depends on the
spending s2 of politician 2 and a parameter α: s1 = f (s2, α), where 0 < f 1'(s2, α) <
1 and f 2'(s2, α) < 0 for all s2 and all α. The optimal expenditure of politician 2
depends on the spending s1 of politician 1 and a parameter β: s2 = g(s1, β), where
0 < g'1(s1, β) < 1 and g'2(s1, β) < 0. The equilibrium values of s1 and s2 are given
by the solution of the simultaneous equations
s1 = f (s2, α)
s2 = g(s1, β).
12.Does an increase in α (holding β constant) necessarily increase or necessarily
decrease the equilibrium value of s1?
13.The equilibrium outputs q1 and q2 of two firms satisfy
q1 = b(q2, c1)
q2 = b(q1, c2),
14.where b is a differentiable function that is decreasing in each of its arguments
and satisfies b1'(q, c) > −1 for all q and c, and c1 and c2 are parameters.
a. Find the differentials of the pair of equations.

70
b. Find the effect on the values of q1 and q2 of equal increases
in c1 and c2 starting from a situation in which c1 = c2 and an equilibrium in
which q1 = q2.
15.The equilibrium values of the variables Y and r are given by the solution of the
two equations
I(r)= S(Y)
aY + L(r)= M
16.where a > 0 and M are parameters, I is an increasing differentiable function,
and S and L are decreasing differentiable functions. How do Y andr change
when M increases (holding a constant)?
17.Consider a market containing two goods. Denote the prices of these goods
by p and q. Suppose that the demand for each good depends on p,q, and the
amount of advertising expenditure a on good 1, and that the supply of each
good depends only on the price of that good. Denoting the demand functions
by x and y and the supply functions by s and t, for any given value of a a
market equilibrium is a pair (p, q) or prices such that
x(p, q, a)= s(p)
y(p, q, a)= t(q).
18.How does the equilibrium price p of good 1 change as a changes?
19.Assume that x'p < 0, x'q > 0, x'a > 0, s' > 0, y'q < 0, y'p > 0, y'a < 0, and t' > 0. (What
are the economic interpretations of these assumptions?) Assume also that
(x'p − s')(y'q − t') − x'qy'p > 0 for all (p, q, a). Does the equilibrium price of good
1 necessarily increase if a increases?
2.5 Homogeneous functions
Definition
Multivariate functions that are "homogeneous" of some degree are often used in
economic theory. A function is homogeneous of degree k if, when each of its
arguments is multiplied by any number t > 0, the value of the function is multiplied

71
by tk
. For example, a function is homogeneous of degree 1 if, when all its arguments
are multiplied by any number t > 0, the value of the function is multiplied by the same
number t.
Here is a precise definition. Because the definition involves the relation between the
value of the function at (x1, ..., xn) and it value at points of the form (tx1, ..., txn)
where t is any positive number, it is restricted to functions for which (tx1, ..., txn) is in
the domain whenever t > 0 and (x1, ..., xn) is in the domain. (Some domains that have
this property are: the set of all real numbers, the set of nonnegative real numbers, the
set of positive real numbers, the set of all n-tuples (x1, ..., xn) of real numbers, the set
of n-tuples of nonnegative real numbers, and the set of n-tuples of positive real
numbers.)
Definition
A function f of n variables for which (tx1, ..., txn) is in the domain whenever t > 0 and (x1, ..., xn) is in the domain is
f (tx1, ..., txn) = tk f (x1, ..., xn) for all (x1, ..., xn) in the domain of f and all t > 0.
Example
For the function f (x1, x2) = Ax1
ax2
b with domain {(x1, x2): x1 ≥ 0 and x2 ≥ 0} we have
f (tx1, tx2) = A(tx1)a(tx2)b = Ata+bx1
ax2
b = ta+b f (x1, x2),
so that f is homogeneous of degree a + b.
Example
Let f (x1, x2) = x1 + x2
2, with domain {(x1, x2): x1 ≥ 0 and x2 ≥ 0}. Then
f (tx1, tx2) = tx1 + t2
x2
2
.
It doesn't seem to be possible to write this expression in the form tk(x1 + x2
2) for any value of k. But how do we prov
of k? Suppose that there were such a value. That is, suppose that for some k we have
tx1 + t2x2
2 = tk(x1 + x2
2) for all (x1, x2) ≥ (0, 0) and all t > 0.
Then in particular, taking t = 2, we have
2x1 + 4x2 = 2k(x1 + x2
2) for all (x1, x2).
Taking (x1, x2) = (1, 0) and (x1, x2) = (0, 1) we thus have
2 = 2k
and 4 = 2k
,

72
which is not possible. Thus f is not homogeneous of any degree.
In economic theory we often assume that a firm's production function is homogeneous
of degree 1 (if all inputs are multiplied by t then output is multiplied by t). A
production function with this property is said to have "constant returns to scale".
Suppose that a consumer's demand for goods, as a function of prices and her income,
arises from her choosing, among all the bundles she can afford, the one that is best
according to her preferences. Then we can show that this demand function is
homogeneous of degree zero: if all prices and the consumer's income are multiplied
by any number t > 0 then her demands for goods stay the same.
Partial derivatives of homogeneous functions
The following result is sometimes useful.
Proposition
Let f be a differentiable function of n variables that is homogeneous of degree k. Then each of its partial derivativ
homogeneous of degree k − 1.
Proof
The homogeneity of f means that
f (tx1, ..., txn) = tk f (x1, ..., xn) for all (x1, ..., xn) and all t > 0.
Now differentiate both sides of this equation with respect to xi, to get
t f 'i(tx1, ..., txn) = tk f 'i(x1, ..., xn),
and then divide both sides by t to get
f 'i(tx1, ..., txn) = tk−1 f 'i(x1, ..., xn),
so that f 'i is homogeneous of degree k − 1.
Application: level curves of homogeneous functions
This result can be used to demonstrate a nice result about the slopes of the level
curves of a homogeneous function. As we have seen, the slope of the level curve of
the function F through the point (x0, y0) at this point is
−
F 1'(x0, y0)
.

73
F 2'(x0, y0)
Now suppose that F is homogeneous of degree k, and consider the level curve
through (cx0, cy0) for some number c > 0. At (cx0, cy0), the slope of this curve is
−
F 1'(cx0, cy0)
F 2'(cx0, cy0)
.
By the previous result, F '1 and F '2 are homogeneous of degree k−1, so this slope is
equal to
−
ck−1F 1'(x0, y0)
ck−1F 2'(x0, y0)
= −
F 1'(x0, y0)
F 2'(x0, y0)
.
That is, the slope of the level curve through (cx0, cy0) at the point (cx0, cy0) is exactly
the same as the slope of the level curve through (x0, y0) at the point (x0, y0), as
illustrated in the following figure.
In this figure, the red lines are two level curves, and the two green lines, the tangents
to the curves at (x0, y0) and at (cx0, xy0), are parallel.
We may summarize this result as follows.
Let F be a differentiable function of two variables that is homogeneous of some
degree. Then along any given ray from the origin, the slopes of the level curves
of F are the same.

74
Euler's theorem
A function homogeneous of some degree has a property sometimes used in economic
theory that was first discovered by Leonhard Euler (1707-1783).
Proposition (Euler's theorem)
The differentiable function f of n variables is homogeneous of degree k if and only if
∑i=1
nxi f i'(x1, ..., xn) = k f (x1, ..., xn) for all (x1, ..., xn). (*)
Condition (*) may be written more compactly, using the notation ∇ f for the gradient
vector of f and letting x = (x1, ..., xn), as
x·∇ f (x) = k f (x) for all x.
Proof
I first show that if f is homogeneous of degree k then (*) holds. If f is homogeneous of degree k then
f (tx1, ..., txn) = tk f (x1, ..., xn) for all (x1, ..., xn) and all t > 0.
Differentiate each side of this equation with respect to t, to give
x1 f '1(tx1, ..., txn) + x2 f '2(tx1, ..., txn) + ... + xn f 'n(tx1, ..., txn) = ktk−1
f (x1, ..., xn).
Now set t = 1, to obtain (*).
I now show that if (*) holds then f is homogeneous of degree k. Suppose that (*) holds. Fix (x1, ..., xn) and
single variable by
g(t) = t−k f (tx1, ..., txn) − f (x1, ..., xn).
We have
g'(t) = −kt−k−1
f (tx1, ..., txn) + t−k
∑i=1
n
xi f 'i(tx1, ..., txn).
By (*), we have
∑i=1
ntxi f i'(tx1, ..., txn) = k f (tx1, ..., txn),
so that g'(t) = 0 for all t. Thus g(t) is a constant. But g(1) = 0, so g(t) = 0 for all t, and hence f (tx1, ..., txn) = tk f (x1, .
homogeneous of degree k.
Example
Let f (x1, ..., xn) be a firm's production function; suppose it is homogeneous of degree 1 (i.e. has "constant returns to

75
shows that if the price (in terms of units of output) of each input i is its "marginal product" f 'i(x1, ..., xn), then the to
∑i=1
n
xi f i'(x1, ..., xn)
is equal to the total output, namely f (x1, ..., xn).
2.5 Exercises on homogeneous functions
1. Determine whether or not each of the following functions is homogeneous, and
if so of what degree.
a. 3x + 4y.
b. 3x + 4y − 2.
c. (√x + √y + √z)/(x + y + z).
2. Determine whether or not each of the following functions is homogeneous, and
if so of what degree.
a. 2x2 + xy.
b. x2
+ x3
.
c. [g(x1, ..., xn)]p where g is homogeneous of degree n.
3. Is the function 30x1/2
y3/2
− 2x3
/y homogeneous of any degree? If so, which
degree? (If not, give a complete argument why not.)
4. The function f (x1, ..., xn), with domain equal to the set of n-tuples of positive
numbers, is homogeneous of degree k and f (x1, ..., xn) > 0 for all (x1, ..., xn) in
the domain. Is there any value of k for which the function g defined by g(x1,
..., xn) = ln f (x1, ..., xn) is homogeneous of some degree?
5. Suppose that f (x1, ..., xn) is homogeneous of degree r. Show that each of the
following functions h(x1, ..., xn) is homogeneous, and find the degree of
homogeneity.
a. h(x1, ..., xn) = f (x1
m, ..., xn
m) for some number m.
b. h(x1, ..., xn) = [ f (x1, ..., xn)]p for some number p.
6. Is the sum of two homogeneous functions necessarily homogeneous?
7. Is the product of two homogeneous functions, with possibly different degrees
of homogeneity, necessarily homogeneous? Necessarily not homogeneous?
8.
a. Is the function x1/2y + x3/2 homogeneous of any degree? (If so, which
degree?)
b. A firm's (differentiable) production function f (x1, ..., xn) is homogeneous
of degree 1 (where xi is the amount of input i). Assume that the price (in

TEXTBOOK ON MATHEMATICAL ECONOMICS FOR CU , BU CALCUTTA , SOLVED EXERCISES , STUDY MATERIAL

TEXTBOOK ON MATHEMATICAL ECONOMICS FOR CU , BU CALCUTTA , SOLVED EXERCISES , STUDY MATERIAL

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to TEXTBOOK ON MATHEMATICAL ECONOMICS FOR CU , BU CALCUTTA , SOLVED EXERCISES , STUDY MATERIAL

Similar to TEXTBOOK ON MATHEMATICAL ECONOMICS FOR CU , BU CALCUTTA , SOLVED EXERCISES , STUDY MATERIAL (20)

More from SOURAV DAS

More from SOURAV DAS (20)

Recently uploaded

Recently uploaded (20)

TEXTBOOK ON MATHEMATICAL ECONOMICS FOR CU , BU CALCUTTA , SOLVED EXERCISES , STUDY MATERIAL