Partial Differentation

8/10/2019 Partial Differentation

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EC115 - Methods of Economic AnalysisLecture 3

Partial Differentiation

Renshaw - Chapter 14

University of Essex - Department of Economics

Week 18

Domenico Tabasso (Universityof Essex - Department of Economics)

Lecture 3 Week 18 1 / 35
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Topics for this week

Graphical and economic interpretation: the slope of thelevel curve

Second order partial derivatives

Economic Application: Utility Functions

Cross partials and Youngs Theorem


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Introduction

Recall that a function of several variables is a relation

between some independent variables x1, x2, x3,..., xnand some dependent variable zsuch that:

z=f(x1, x2, x3,...xn)

specifies the value ofzgiven the values ofx1, x2, x3,...xn.

A typical function we encounter in economics is the

Cobb-Douglas function:

u=U(x,y) =Axy1

where A andare positive constants and

[0, 1].


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Today we will analyze these type of functions further byasking the following questions:

How does zchange when x1 changes assumingx2, x3, ..., xn are constant?

How does x1 changes when x2 change assuming

z, x3,..., xn are constant?The answers to these questions:

will help us graph functions of the form z=f(x,y);

andhave interesting economic interpretations.

Domenico Tabasso (Universityof Essex - Department of Economics)Lecture 3 Week 18 4 / 35
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What is a Partial Derivative?

Let us start with a Macro example:

One of the most important questions inmacroeconomics is what determines the growth rate ofa countrys real income.

Suppose that the growth rate, GR, is determined by the following linear

relation:

GRt= 0 + 1It+ 2HCt+ 3INSTt+ 4Techt+ ...+

We would like to know the impact of each of these variables.



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That is, by how much could we improve the growthperformance of an economy if we only

increasing the capital stock (investment), It?

improved the quality of education, HC?

improved our institutions (provide property rights,stable governments. etc.), INST?

improved our technology through innovations, Tech?

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To do this, first we have to analyze the effects of eachcomponent independently of the others, and then, secondwe have to analyze the effects of the components taken asa whole.

Partial derivativesallow us to perform this task.



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Slope of a surface

Consider the utility function:

z=U(x,y) =x1/2y1/3.

and fix the consumption ofy=27.

Recall that this implies we are analyzing the set ofcoordinates that have the property(x, 27, z). That is,we are looking at the graph of the function:

z=U(x, 27) =3x1/2.

The partial derivative ofzwith respect to xwheny=27 is the slope of a straight line tangent to thisfunction at any point (x, z) on its graph.

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14

12

10

8

4

6

2

0

1 2 3 4 5 6 7 8 9 10 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0 2 1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 3 0 3 1

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Original function: z=U(x,y) =x1/2y1/3

Now fix the consumption ofx=16.

That is, we are looking at the graph of the function:

z=U(16,y) =4y1/

3

.

The partial derivative ofzwith respect to y whenx=16,is the slope of a straight line tangent to this

function at any point (y, z) on its graph.

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What does it mean?

Why do we say that it is a measure of the slope? A derivative is a measure of change. For example, how much does zchanges when y changes?

As the change in ygets smaller and smaller we findthat the change in zcan be approximated better andbetter by the change in ymultiplied by the slope of thetangent.

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Suppose that we wanted to know how much does zchanges when we increase y, from y0 to y1,when holding xconstant at 25:

z0 = U(16,y0) =4y1/30

z1 = U(16,y1) =4y1/31

Define the change in yto be y=y1 y0, so that:z1 z0 = U(16,y0+ y) U(16,y0)

z

y =

U(16,y0+ y)U(16,y0)

y

limy0

z

y =

U(16,y)

y .

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A graphical representation


S d b bl fi d h d f h
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So in order to be able to find the derivative ofz withrespect to xwe have to assume that y is just a constant(i.e. any number) and then apply the standard derivation

rules. Vice versa in case we take the derivative with respectto y. So for example:

Find the partial derivatives zx

and zy

of:

z=3x3 +2x2y+ y2 + y

z

x=9x2 +4xy

Why? Because y is a constant! So when x is multiplied by y, its

derivative will be multiplied by y; when y is just added, then itdisappears from the derivative.

z

y=2x2 +2y+1

As before: Now x is a constant!

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Another Example

z= (x3 +y2)1/2

This is a more complicated case: We have to apply theChain Rule. So the derivative will be the product of twofunctions: the derivative of theoutside function (i.e.

the power of 1/2) times the derivative of the insideone. So:

z

x = 1

2(x3 +y2)

1/2(3x2) = 3x

2

2x3+y2

zy

= 2y2

x3+y2

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Second order partial derivatives

Now it is easy to see that the first order partial

derivatives tells us if a function is increasing ordecreasing given that we hold constant the rest ofindependent variables.

For example, iff(x0,y)/y>0 for all y

[0, 10],thisimplies that given x=x0,the function f is increasingin yfor values ofythat range from 0 to 10 inclusive.Using the previous exercises we obtained that if

f(x,y) = (x3

+y2

)1/2

then the partial with respect toy is:

f(x0, y)

y = y(x30 + y

2)1/2 >0 for all values ofy>0 given thatx0 0.



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Summary

If we have a have a function oftwovariables, we cancalculatetwopartial derivatives

U(x,y) =xy (1)

=

Ux

=x1y (2)

U

y =xy1 (3)

But each of these derivatives can itself be a function ofxand y. So for each of them we can calculate the derivatives

with respect to x and y!Domenico Tabasso (Universityof Essex - Department of Economics)Lecture 3 Week 18 17 / 35
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Fi d i i f i


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First derivative function

Going back to a previous example:

z= (x3

+y2

)1/2

(12)Fixing x=x0 and performing the partial derivative weobtain a new function in y:

zy

=h(x0,y) =y(x30 +y2)1/2.

By differentiating again with respect to ywe can get

information about how this first partial changes when ychanges holding constant x=x0.

That is, how does h change when ychanges (i.e., whatis the slope ofh(x0,y))?


Th f i h
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The function h

Note that hdefines the rate of change ofz when y changes

and x=x0. In this case, we obtain:

h(x0,y)

y =

2z

y2 = (x30 +y

2)1/2 y2(x30 +y2)3/2.

Consider the case ofx=0; then:

h(0,y)

y = (y2)1/2 y2(y2)3/2

= y1 y2y3= y1 y1 =0.

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The interpretation is as follows whenx

=x

0=0 then:first partial: an increase in y increases z.z is anincreasingfunction in y;

second partial: this increase isconstantfor all y.

Note that we could have obtained the same results byputtingx=x0=0 into f(x,y) and then differentiating(f(0,y) = (03 +y2)1/2 =y).


A hi l t ti
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A graphical representation

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Consumer Theory =Marginal Utility andDiminishing Marginal Utility

U(x,y)

x is called theMU

xwhen we hold constant theconsumption ofy.U(x,y)

y is called theMUywhen we hold constant theconsumption ofx.



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Cross Partial Derivatives


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Cross Partial Derivatives

Recall that second order partial derivatives measurehow the slope ofz=f(x,y0) changes when xchanges.

Cross partials are a natural extension as they measurehow the slope ofz=f(x,y0) changes when y changeskeeping xconstant.

Formally cross partials are denoted by:

2f(x,y

)xy or

2f(x,y

)yx .


Youngs Theorem
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Young s Theorem

Definition (Youngs Theorem)

For a function y=f(x1, x2, . . . , xn), with continuous firstand second-order partial derivatives, theorderof

differentiation in computing cross-partials isirrelevant.

2f(x,y)

xy

=2f(x,y)

yx

.


Graphical Representation
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Graphical Representation


The other cross partial derivative
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The other cross partial derivative


Example: Cobb-Douglas Function
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Example: Cobb Douglas Function

For the Cobb-Douglas utility function we have

U(x,y) =x0.5y0.5

U(x,y)

x =

1

2yx0.5 and U(x,y)

y =

1

2xy

0.5

2U(x,y)

x2 = 1

4y0.5x1.5 and

2U(x,y)

y2 = 1

4x0.5y1.5

2U(x,y)xy

= 14x0.5y0.5 =

2U(x,y)yx


More exercises
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More exercises

Find the second and cross partial derivatives of

1 z=3

x3

+2x2y

+y2

+y

z

x =9x2 +4xy

z

y =2x2 +2y+1

2 z= (x3

+y2

)0.5

z

x =

3

2x2(x3 +y2)0.5

z

y =y(x3 +y2)0.5

3 z=100e2x+3y

z

x =200e2x+3y

z

y =300e2x+3y

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Partial Differentation

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