CHAPTER 7
Partial Differentiation
From the previous two chapters we know how to differentiate
functions
of one variable
. But many functions in economics depend on several
variables: output depends on both capital and labour, for example. In
this chapter we show how to differentiate
functions of several vari
ables
and apply this to find the
marginal products of labour and
capital, marginal utilities, and elasticities of demand
.
We use
differentials
to find the gradients of isoquants, and hence determine
marginal rates of substitution
.
—
—
1. Partial Derivatives
The derivative of a function of one variable, such as
y
(
x
), tells us the gradient of the function:
how
y
changes when
x
increases. If we have a function of more than one variable, such as:
z
(
x, y
) =
x
3
+ 4
xy
+ 5
y
2
we can ask, for example, how
z
changes when
x
increases but
y
doesn’t change. The answer
to this question is found by thinking of
z
as a function of
x
, and differentiating, treating
y
as if it were a constant parameter:
∂z
∂x
= 3
x
2
+ 4
y
This process is called partial differentiation. We write
∂z
∂x
rather than
dz
dx
, to emphasize that
z
is a function of another variable as well as
x
, which is being held constant.
∂z
∂x
is called the partial derivative of
z
with respect to
x
∂z
∂x
is pronounced “partial dee z by dee x”.
Similarly, if we hold
x
constant, we can find the partial derivative with respect to
y
:
∂z
∂y
= 4
x
+ 10
y
Remember from Chapter 4 that you can think of
z
(
x, y
) as a “surface” in 3 dimensions.
Imagine that
x
and
y
represent coordinates on a map, and
z
(
x, y
) represents the height of
the land at the point (
x, y
). Then,
∂z
∂x
tells you the gradient of the land as you walk in the
direction of increasing
x
, keeping the
y
coordinate constant.
If
∂z
∂x
>
0, you are walking
uphill; if it is negative you are going down. (Try drawing a picture to illustrate this.)
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7. PARTIAL DIFFERENTIATION
Exercises
7.1
:
Find the partial derivatives with respect to
x
and
y
of the functions:
(1)
f
(
x, y
) = 3
x
2

xy
4
(3)
g
(
x, y
) =
ln
x
y
(2)
h
(
x, y
) = (
x
+ 1)
2
(
y
+ 2)
Examples
1.1
:
For the function
f
(
x, y
) =
x
3
e

y
:
(i) Show that
f
is increasing in
x
for all values of
x
and
y
.
∂f
∂x
= 3
x
2
e

y
Since 3
x
2
>
0 for all values of
x
, and
e

y
>
0 for all values of
y
, this derivative is
always positive:
f
is increasing in
x
.
(ii) For what values of
x
and
y
is the function increasing in
y
?
∂f
∂y
=

x
3
e

y
When
x <
0 this derivative is positive, so
f
is increasing in
y
.
When
x >
0,
f
is
decreasing in
y
.
1.1. Secondorder Partial Derivatives
For the function in the previous section:
z
(
x, y
) =
x
3
+ 4
xy
+ 5
y
2
we found:
∂z
∂x
= 3
x
2
+ 4
y
∂z
∂y
= 4
x
+ 10
y
These are the
firstorder
partial derivatives. But we can differentiate again to find
second
order
partial derivatives. The second derivative with respect to
x
tells us how
∂z
∂x
changes as
x
increases, still keeping
y
constant.
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 Spring '10
 Vines
 Economics, Derivative, Supply And Demand, ... ..., Isoquant

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