Econ 308
Handouts
#1 Calculus Refresher
Differentiation
TO BEGIN WITH:
Function, Y = F (X)
Y - dependent variable whose value is determined by the value of X - the independent
variable as dictated by the functional form represented by F, where F could be linear, polynomial, logarithmic
etc. Similarly, multiple layers of dependence, Y = F ( Z (X)) in this case Z is a function of X, and Y is a
function of Z, therefore, Y is also a function of X.
PARTIAL DERIVATIVE - a measure of the change in a dependent variable due to an infinitesimal change in
an independent variable, holding all other variables constant. The economics student should recognize this
definition as MARGINAL change.
DIFFERENCE QUOTIENT - rate of change in a dependent variable for a discrete change in an independent
variable.
Let Y = F (X), if initial value of Xo, then Yo = F (Xo)
Let ΔX = X1 - Xo, then X1 = Xo + ΔX
Therefore, Y1 = F (X1) = F (Xo + ΔX)
Combining together, ΔY = Y1 - Yo = F (Xo + ΔX) - F (Xo)
Dividing both sides by ΔX results in ΔY / ΔX, the change in Y as a per unit change in X, or in graphical terms,
slope
Y f( Xo + X ) - f( Xo )
=
DIFFERENCE QUOTIENT
X
X
Example :
Y = 3x2 - 4
Δy / Δx =[ 3(x + Δx)2 - 4 - (3x2 - 4)] / Δx
Δy / Δx =[ 3x2 +6x Δx + 3 Δ x2 - 4 - 3x2 + 4)] / Δx
Δy / Δx =[ 6x Δ x + 3Δx2] / Δx
Δy / Δx = 6x + 3Δx
The difference quotient is applicable for any change, large or small, in X.
Now as with a partial derivative, as Δx approaches zero for an infinitesimal change, then the rate of change is:
Δy / Δx = 6x
1
GRAPHICAL INTERPRETATION:
Slope, Δy / Δx, can be calculated between two points along a
function, or at a given point.
F(X)
Y
For example along F(X) - between Xo to X1 the slope is
measured by the slope of the line AB. As change in X gets
B

A
D
smaller - from Xo to X2, the slope now measured be the slope
C

E

of line AC
For an infinitesimal change in X, suppose around point A,
then the slope is measured by the slope of the tangent line at
X0
X2
X1
point A, in essence, Δy / Δx is given as the slope of line DE
A derivative is where the difference quotient is for an infinitesimal change in x, i.e., basically the slope of
the tangent line.
Rules of Differentiation - We are interested in partial differentials. This is where we examine the effects of a
change of one independent variable on a dependent variable, holding all other independent variables constant. .
- This symbol () represents a partial derivative.
1) Constant Function Rule
If Y = F ( X ) = K, where K is a constant, then  Y /  X = 0.
2) Power Function Rule
If Y = F ( X ) = Xn, Then  Y /  X = nXn-1
3) Sum Difference Rule
If Y = F ( X )  G (X) , Then  Y /  X = F ' (X)  G ' (X)
Where F ' (X) =  F(X) /  X and G ' (X) =  G (X) /  X
Example: Y = Ax2 + Bx + C
Let F ( X ) = Ax2, and G (X) = Bx + C
F ' (X) = 2Ax, and G ' (X) = B, therefore
 Y /  X = 2Ax + B
2
4) Product Rule
If Y = F (X) * G(X) , Then  Y /  X = F ' (X) G (X) + F (X) G ' (X)
Example: y = (2x + 3)(3x2), F (X) = (2x + 3) and G(X) = (3x2)
F ' (X)= 2 and G ' (X) = 6x, therefore
 Y / X = 2(3x2) + 6x(2x + 3)
= 6x2 + 12x2 + 18x = 18x2 + 18x = 18x(x+1)
5) Quotient Rule
If Y = F (X)/ G(X) , Then  Y /  X = {F ' (X) G (X) - F (X) G ' (X) } / G 2(X)
Example: y = (2x - 3) / (x + 1), F (X) = (2x - 3) and G(X) = (x + 1)
G ' (X) = 1, and G 2(X) = (x + 1)2
F ' (X) = 2,
 Y /  X = {2(x + 1) - (2x - 3)1} / (x + 1)2
= 5 / (x + 1)2
6) Chain Rule
If Y = F (Z(X)) Then  Y /  X =  Y /  Z *  Z /  X
Example: y = 3(2x +5)2, Then Y (Z) = 3z2 and Z(X) = 2x + 5
 Z /  X = 2 and  Y /  Z = 6z
Therefore  Y /  X =  Y /  Z * Z /  X = 2 * 6z = 12(2x + 5)
More Examples:
1) y = (9x2 - 2)(3x + 1)
Product Rule:
Where
Y = F (X)*G(X)
 Y /  X = F ' (X) G (X) + F (X) G ' (X)
F (X) = (9x2 - 2) & G(X) = (3x + 1)
Therefore F ' (X) = 18x
& G ' (X) = 3
Substituting into Product Rule Equation
 Y /  X = 18x (3x + 1) + 3 (9x2 - 2)
= 54 x2 + 18x + 27 x2 - 6
= 81 x2 + 18x - 6
3
2) y = (3x + 11)(6x2 - X)
Y = F (X)*G(X)
Product Rule:  Y /  X = F ' (X) G (X) + F (X) G ' (X)
Where
Therefore
F (X) = (3x + 11 ) & G(X) = (6x2 - x)
F ' (X) = 3
&
G ' (X) = 12x -1
Substituting into Product Rule Equation
 Y /  X = 3 (6x2 - X) + (3x + 11)(12x -1)
= 18x2 - 3x + 36 x2 -3x + 132x -11
= 54 x2 + 126x -11
3) y = x2 / (4x + 6)
 Y /  X = F (X) / G (X)
Quotient Rule:  Y /  X = {F ' (X) G (X) - F (X) G ' (X) } / G 2(X)
Where
F (X) = x2
&
Therefore F ' (X) = 2x
G (X) = (4x + 6)
& G ' (X) = 4
Substituting into Quotient Rule Equation
 Y /  X = {2x (4x + 6) - 4 x2 } / (4x + 6)2
= {8x2 + 12 x - 4 x2 } / (4x + 6)2
= {4x2 + 12 x} / (4x + 6)2
Alternative Solution to #3
Use a Combination of the Chain Rule and Product Rule
Rewrite Y = x2 / (4x + 6) As y = x2 (4x + 6) -1 using exponents instead of division
The problem can now be stated in terms:
Y = F (X)*G (Z(X)) a Combination of the Chain Rule and Product Rule
Use Product Rule:  Y /  X = F ' (X) G (Z(X)) + F (X) G ' (Z(X))
Where F (X) = x2 & G (Z(X)) = (4x + 6)-1
Therefore F ' (X) = 2x and use the chain rule to solve G ' (Z(X))
Where G(Z) = z -1 and Z(X) = 4x + 6
So
G ' (Z(X)) =  G /  Z *  Z / X
Where  G /  Z = -1z
-2
&  Z / X = 4
Therefore G ' (Z(X)) =  G /  Z *  Z / X = -4 z
4
-2
= -4 (4x + 6)-2
Substituting into the Product Rule Formula
 Y /  X = 2x* (4x + 6)-1 + x2 * (-4) (4x +6)
-2
Multiply by One { (4x + 6)1 (4x + 6)-1 }
2x* (4x + 6)-1 *{ (4x + 6)1 (4x + 6)-1 } + X2 * (-4) (4x +6)
-2
Combine Terms and Factor out
= { 2 x(4x + 6) - 4 x2 } / (4x +6)
2
= { 8x2 + 12x - 4 x2 } / (4x +6) 2
= { 4x2 + 12x } / (4x +6) 2
Practice, Practice, Practice
More practice problems
Use the chain rule to find Y/X for the following:
Y = (3X2 - 13)3
Y = (8X3 - 5X2)9
Y = (aX2 + b)4
Use the product rule to find Y/X for the following:
Y = (9X2 - 2)(3X + 1)
Y = (aX2 - b)(cX3)
Y = (3X + 11)(6X2 - 5X)
Use the quotient rule to find Y/X for the following:
Y = ( X2 + 3) / X
Y = 4X / (X + 5)
Y = (7X4 + 5X3) / (4X5 - 3X2)
Use the concept of a partial differential to find Y/X1 and Y/X2. When solving for Y/X1 treat X2 as
a constant and vice versa.
Y = 2X13 - 11X12X2 + 3X22
Y = 7X1 + 5X1X22 - 9X23
Y = (3X1 - 2X2) / (X12 + 3X2)
5
#2
OPTIMIZATION
Maximizations and Minimizations
Optimization with one choice variable
Y
Y
A
B C
Y
E
D
F F
(A)
X
X
X
(B)
C
X
(C)
In diagram (A) all values for X represent the same value for Y and may be considered either maximums or
minimums.
In figure (B) the function is monotonically increasing, f(x1) > f(x0), if x1 > x0, in other words, it has a positive
slope throughout. Consequently, there is no finite maximum. However, there is a global minimum at point D
Diagram (C) presents both a local maximum and a local minimum at points E and F, respectively. These are
extrema in the immediate neighborhood. These being local extrema does not guarantee that they are global
extrema. In other words, E and F are optimums, maximum and minimum, respectively, over the range in the
diagram. However, we don=t know the values of Y for values of X beyond what is portrayed, hence it is
possible that there exists other optima.
In economics where we will look at choice over a specified range of activity, this last case will be of primary
interest.
First Derivative Test
Y
Recall that derivatives are defined as rates of change and can be
interpreted as slopes.
E
Where an extremum exists it is necessary, but not sufficient, that
the slope of the function, f ' (xo) = 0. In the diagram the slopes of
the tangents at points E and F are equal to zero. The tangents are
horizontal.
F
X
The slopes of the function at points E and F are equal to zero. This
basically means we have either climbed to the top of a hill or have
descended to the bottom of a valley.
6
Therefore evaluating the first derivative at zero gives a solution for an extremum, either a maximum or a
minimum, but without more information we would not be able to tell which one.
(One exception is possible, f '(xo) = 0 could be an inflection point.)
To determine if these extrema represent maximums or minimums we need to plot the slopes of the function in
the vicinity of these extrema.
Y
For a local maximum as in point E, note
that the slope of the curve to the left is
positively slope and as we move from left
to right the slope falls becomes zero at the
max (E) and then becomes negative. The
slope is plotted below. Another way of
expressing this is that the slope of the
slope is negative or gets smaller as X gets
larger.
E

PRIMITIVE FUNCTION
F

X
SLOPE OF ABOVE FUNCTION
+


X
-
For a minimum as in point F, the opposite
occurs. As we move from left to right the
slope increases. It is initially negative,
becomes zero at F and then becomes
positive. Again the slope of the slope is
positive.
Therefore we can determine if an
extremum is a maximum or a minimum by what are called second order conditions. By taking the derivative of
the first order derivatives we can evaluate these second order derivatives at their extremum. If the second order
derivative is negative then the extremum is a maximum. If it is positive then it is a minimum.
EXAMPLE:
Let the primitive function be given as, Y = X3 - 3X + 5.
Differentiating, we get Y/X = 3X2 -3.
Setting this derivative equal to zero is equivalent to finding the value of X where the slope equals zero. In other
words, solving for Y/X = 0, tells us where there is an extremum. These are called first order conditons,
(foc) 3X2 -3 = 0 , extrema occur at X = 1 & X = -1.
In order to determine if these extema are maxima or minima we need to determine the sign of the slope around
X = 1 & X = -1.
We accomplish this by taking what is called the second order conditions, taking the slope of the slopes or taking
the second derivative. The first derivative is Y/X = 3X2 -3. The second derivative  (Y/X) /X =
2Y/X2 = 6X. Evaluating the second derivatives at their extrema, X = 1 & X = -1, tells us whether the
extremum is a maximum or a minimum.
f '' (1) = 6 hence the extremum at x=1 is a minimum
f ''(-1) = -6 hence the extremum at x= -1 is a maximum
7
Optimization with Two or More Choice Variables
The situation with more than one choice variable is conceptually the same as before but gets a more
complicated.
Instead of a two dimensional problem, multiple choice or
independent variables increases the dimensionality of the
problem.
U = f ( X,Y )
U
U2
For example a consumer's utility function composed of two
goods gives rise to a utility surface. The goods X & Y are
measured in the XY plane at the bottom of the surface while
utility is measured on an axis rising perpendicular from the
XY plane.
U1
Uo
Horizontal slices of the utility surface, when collapsed down
into the XY plane, represent indifference curves. One side
note, notice that changing the height of the surface,
stretching it up or squashing it down the u axis does not change the relative rankings of commodity bundles. So
doing would not change the indifference curve mapping. This is the ordinal property of preferences.
With regards to the consumers problem of maximizing utility subject to a budget constraint we need to partition
the surface into those commodity bundles that are feasible and those which are not. This represented by a
vertical slice of the utility surface along the familiar budget line in the XY plane.
The solution to the consumer's problem follows the same logic as any optimization.
The choice of the commodity bundle which maximizes utility is the one at which the slope of the budget plane
is parallel to the XY plane.
This occurs in the diagram (on the next page) at the
intersection of the indifference curve u2 with the budget plane.
As we move along the budget plane where it intersects the
utility surface we climb the surface. At least one direction
points upward. This is analogous to the slope of the surface
pointing upward. At the top of the budget plane all directions
along its intersection with the utility surface point downwards.
At this point the slope of the budget plane is horizontal to the
XY plane. This is the rationale for the first order conditions.
We optimize by differentiating the utility function constrained
by the budget constraint with respect to the choice variables x
& y and set their derivatives equal to zero.
U = f ( X,Y )
U
U2
U1
Uo
Deriving second order conditions is much more difficult than
in the one choice variable case. It requires the use of matrix algebra in its solution. This will be beyond the
scope of this course. The interested student will be supplied with materials if they so desire.
8
9