Total Differentation

8/10/2019 Total Differentation

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EC115 - Methods of Economic AnalysisSpring Term, Lecture 6

Total Differentiation

Renshaw - Chapter 15

University of Essex - Department of Economics

Week 21

Domenico Tabasso (University of Essex - Department of Economics)

Lecture 6 - Spring Term Week 21 1 / 29
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Introduction

Supposez=f(x1, x2, ..., xn)(a function with morethan one variable).

We use partial differentiation when we want to analyze

howzchanges when we change one of the independentvariables holding the others constant.

This technique is useful in economics as it helps us tounderstand economic concepts such as: the marginal utility from consuming a good; and the marginal product from using an input,

Domenico Tabasso (University of Essex - Department of Economics)

Lecture 6 - Spring Term Week 21 2 / 29
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Total Differentiation

Total differentiation builds on this concept but gives ameasure of the change in z(=f(x1, x2, ..., xn)) when allthe variables x1, x2, ..., xn change at the same time.

This concept is useful in economics as it allows us toanalyze things like: returns to scale in production;

direct and indirect effects an independent variable might have on z; and

the marginal rate of substitution (in consumption) and marginal rate oftechnical substitution (in production).

Domenico Tabasso (University of Essex - Department of Economics)Lecture 6 - Spring Term Week 21 4 / 29
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Generalising the concept of the slope of a surface

Consider any function z=f(x,y)and suppose we are

at some initial position P= (x0,y0, z0)on its surface.

From our analysis on partial derivatives we know that ifwe increase y from y0 to y1, such that dy=y1 y0,

but hold xconstant at x=x0 then the change in z canbe approximated by the slope of the tangent at P inthe(y, z)plane multiplied by the change in y.

This comes from the following mathematical relation:

limdy0

dz

dy =

f(x0,y0)

y .

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By re-arranging that expression we can obtain anapproximate of the change in zthat occurs when y

changes holding constant x:

dzy=f(x0,y0)

y dy.

Similarly if we increased x from x0 to x1, such thatdx=x1 x0, while keepingyconstant then the changeinzcan be approximated by the slope of the tangent at

P in the(x, z)plane. This comes from:

limdx0

dz

dx =

f(x0,y0)

x .

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Again, by re-arranging this expression we can obtain anapproximate of the change in zthat occurs when xchanges holding constant y:

dzx=f(x0,y0)

x dx.

The terms dzx and dzyare calledpartial differentials.

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Now consider the effect ofsimultaneouschanges in xand y. Namely, let x andy change by dx and dy,

respectively.The resulting total change in zwould be given by:

z1 z0=f(x0+dx,y0+dy) f(x0,y0).

For any function z=f(x,y), ifx and ychange by asmall amount then the resulting change in zcan beapproximated by the sum of the partial

differentials:

dz=dzx+dzy=f(x0,y0)

x dx+

f(x0,y0)

y dy.

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We call dz thetotal differential.

If the surface ofz=f(x,y)is a plane, i.e.

z=a+bx+cy, then this approximation is exact since

zx =band zy =c.

However, if the surface ofz=f(x,y) is not a planethen there will be some error in the approximation.

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Example 1

Find the total differential for each of the followingfunctions:

z=x2 +xy+y2 1Solutiondz= (2x+y)dx+ (2y+x)dy

z=x1/4y1/2

Solutiondz=

14 x

3/4y1/2

dx+

12x

1/4y1/2

dy

z= (x2 +y3)1/2

Solutiondz= (1/2)(x2 +y3)1/2(2xdx+3y2dy)

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Ifz=x2 +y2, suppose xincreases from 5 to 5.1 and yincreases from 10 to 10.1. What error occurs when we

use the total differential, dz, of the function tocalculate the resulting change in z?

Computing the variation directly we getz=3.02.

Using the total differential we get:

dz = 2x dx+2y dy

= dz = 2(5)(0.1) +2(10)(0.1) =3

Thusz dz=0.01 which is an error of one third ofone percent.

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Differentiating a function of a function

Up to this point we have considered cases in which the

changes in x and yare independent. That is, a changein xdoes not affect the value ofy (and vice versa).However, in economics there are many applications inwhich a change in xdoes affect the value ofy.

For example, consider the case of a profit maximizingmonopolist that faces the following inverse marketdemand function:

p=D(q),

where pand qdenote the price and quantity of thegood produced by the monopolist.

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Let TR(p, q) =pqdenote the total revenue of themonopolist.

The total change in revenue when both thepriceandquantityincrease is given by

dTR=TR(p, q)

q

dq+TR(p, q)

p

dp.

The inverse demand tells us that if the quantityproduced increases, consumers will not be willing to pay

the same price as before. Price must decrease by:dp

dq=

dD(q)

dq .

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This implies that changes in qarenot independentfrom changes in p.

We must take this into account when analyzing thechange in total revenues.

To do this we can use the total differential:

dTR=TR(p, q)

q dq+

TR(p, q)p

dp

From this we obtain:

dTR

dq =

TR(p, q)

q

dq

dq+

TR(p, q)

p

dp

dq

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But dqdq

=1 so:

dTR

dq =TR(p, q)

q +TR(p, q)

p

dp

dq.

In addition since TR=pq then TR(p,q)q =pandTR(p,q)

p =q.Therefore:

dTR

dq

=p(q) +qdp(q)

dq

=MR(q),

where MR(q)equals marginal revenue function whichtells us total revenue when quantity changes.

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For a monopolist this can be decomposed into twoeffects:

Direct effect= When she sells more she gets revenuefrom the added output she sells. The direct effect is given by:

TR(p, q)

q

dq

dq

=p(q)

Indirect effect= But selling more lowers the price so sheloses revenue on the output she was already selling. Theindirect effect is given by:

TR(p, q)

p

dp

dq =q

dp(q)

dq

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Moreover, the trade-off faced by the monopolist isaffected by the elasticity of the market demand.

The price elasticity of demand shows the percentagechange in quantity demanded over the percentagechange in price:

p=

dqq

/

dpp

= dq

dppq


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Chain Rule

We sometimes encounter functions that are interrelated

in the following way:

z=f(y, r) where y=g(x) andr=h(x).

In these cases, we can obtain the change in zwhen xchanges by applying a similar logic as before:

dz = f(y, r)

y

dy+f(y, r)

r

dr

dz

dx =

f(y, r)

y

dg(x)

dx +

f(y, r)

r

dh(x)

dx ,


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Example 2

Use the differential dzto find the total derivative dzdx

, of the

following functions:z=x3 +y2, where y=x2

Solution dzdx

= z

x

dx

dx

+ z

y

dy

dx

=

3x2 + 2y 2x=3x2 +4x3


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Example 2

Use the differential dzto find the total derivative dzdx

, of the

following functions:z=x3 +y2, where y=x2

Solution dzdx

= z

x

dx

dx

+ z

y

dy

dx

=

3x2 + 2y 2x=3x2 +4x3

z=uv where u=3x+2 and v=3x2

Solution dzdx

= zududx

+ zvdvdx

=

v 3 + u 6x= 9x2 + (3x+2)6x


I li it F ti Th
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Implicit Function Theorem

Note that the functions that we have been studying are

given by:z=f(x1, x2, ..., xn),

where z is the dependent variable and x1, ..., xn are the

independent variables. In this case, we say that z is anexplicit function of the independent variables.

However, we might encounter cases in which z is givenas an implicit function of the dependent variablesdefined by an equation of the form:

G(x1, x2,..., xn, z) =0.


F l h i
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For example, the equation:

2x+y 5=0

gives us yas an implicit function ofx.


F l h i
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2x+y 5=0


This equation can be easily re-written in an explicitformat as:

y=5 2x


F l th ti
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2x+y 5=0



y=5 2xAnother example is:

y5 5xy+4x2 =0


F l th ti
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2x+y 5=0



y=5 2xAnother example is:

y5 5xy+4x2 =0

The first equation can be easily solved to give yas anexplicit function ofx. However, it is not possible to dothe same with the last equation.

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Even if we cannot make a function explicit, we stillwould like to know its properties. In particular, wewould like to know if:

1 a function y=f(x) can be obtained from its implicit formG(x, y) =0, at least around some point (x0, y0)

2 if the function f(x) is differentiable at x=x0.

The implicit function theorem gives conditions that

guarantee a positive answer to these questions


Consider the implicit function G (x y ) 0 and define a
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Consider the implicit function G(x,y) =0 and define anew variable zsuch that:

z=G(x,y) or G(x,y) z=0.

We can now apply total differentiation to this functionand obtain:

dz=G

xdx+G

ydy

We then set dz=0 to reflect the fact that z is aconstant and equal to zero when the given equation is

satisfied. This gives:

0=G

xdx+

G

ydy

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By rearranging the terms we get:

dy

dx =

G

x

/

G

y

The implicit function theorem says that ifG/y=0,at least around (x0,y0), there exists a continuous anddifferentiable function y=f(x) around the point(x0,y0) such that y0=f(x0).


Application: Indifference Curves
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Application: Indifference Curves

For any utility function u=U(x,y), we have thatu0=U(x,y)along the indifference curve since theindifference curve is the set of all combinations ofx,ythat give the same utility.

Note that by defining u0=U(x,y)we are defining animplicit function between x and y. Hence along theindifference curve:

du0=U(x,y)

x dx+

U(x,y)

y dy=0.

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Furthermore, by applying total differentiation onceagain to the above expression, we can obtain conclusion

with respect to:

d2y

dx2 =

1

U3y U2yUxx 2UxUyUxy+U

2xUyy

Uy, Uxdescribe the first order partial derivatives.

Uxx, Uyy and Uxydescribe the second order partial andcross partial derivatives.

What conditions guarantee that the indifference curves

of U(x,y) are decreasing and convex? [difficult]


Example: the Cobb-Douglas
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p C ug

Consider the Cobb-Douglas utility function

U(x,y) =x0.5y0.5.We have that for any utility level u0=U(x,y)>0 theindifference curve is given by:

y= u0x0.5

1/0.5

=u20

x .

We can obtain the change in ygiven a change in xalong the indifference curve:

dy

dxu=u0

=MRSx,y= u

0x2

0.

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Total Differentation

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