-1-
Partial Differentiation
Suppose you want to forecast the weather this weekend in Los Angeles.
You construct a formula for the temperature as a function of several
environmental variables, each of which is not entirely predictable. Now
you would like to see how your weather forecast would change as one
particular environmental factor changes, holding all the other factors
constant. To do this investigation, you would use the concept of a partial
derivative.
Let the temperature T depend on variables x and y, T=f(x y). The rate of
change of f with respect to x (holding y constant) is called the partial
derivative of f with respect to x and is denoted by fx(x y). Similarly, the
rate of change of f with respect to y is called the partial derivative of f
with respect to y and is denoted by fy(x y).
We define
Do you see the similarity between these and the limit definition of a
function of one variable?
Example
Let f(x y) = xy2 .
In practice, we use our knowledge of single-variable calculus to compute
partial derivatives. To calculate
, you view y as a constant and
differentiate f(x y) with respect to x:
y2 as expected since
, and fy(x y)=2xy since
[y2]=2y
Exercise; Find both partial derivatives for
A. f(x,y) = xy sin x
B.
x+y
f(x,y) =
x-y

Let z=f(x y).
The partial derivative fx(x y) can also be written as:
Similarly, fy(x y) can also be written as

The partial derivative fy(x y) evaluated at the point (x0 y0) can be
expressed in several ways:
fx(x0 y0)
.
There are analogous expressions for fy(x0 y0).
Examples
1. Find
f
f
for f(x, y) = x y cos xy + sin xy
and
x
y
2. Find
f
f
2y
for f(x, y) =
and
x
y
y  cos x
3. Find
f
f
y
for f(x, y) = tan 1
and
x
x
y
4. If Z = x ey then prove that x
Z Z

x y
5. If Z = x +yz then prove that x
w
w
w
y
z
 2w
x
y
z
Geometrical Meaning
Suppose the graph of z=f(x y) is the surface
shown. Consider the partial derivative of f
with respect to x at a point (x0 y0). Holding
y constant and varying x, we trace out a
curve that is the intersection of the surface
with the vertical plane y=y0.
The partial derivative fx(x0 y0) measures the
change in z per unit increase in x along this
curve. That is, fx(x0 y0) is just the slope of
the curve at (x0 y0). The geometrical
interpretation of fy(x0 y0) is analogous.
Notes

Functions of More than Two Variables
For g(x y z), the partial derivative gx(x y z) is calculated by
holding y and z constant and differentiating with respect to x. The
partial derivatives gy(x y z) and gz(x y z) are calculated in an
analogous manner.
Higher partial derivative
To get the higher derivatives it is found that we must use the first
derivative to get the second one if we differentiate
Z
we have two
x
options we can differentiate it with respect to x or differentiate it with
respect to y. if we differentiate it with respect to x we get
 Z
( ) or
x x
2Z
what we call it 2 or Zxx and it is defined by
x
=
f ( x  x, y)  f x ( x, y)  2 Z
 Z
( ) = lim x
x 0
x x
x
x 2
and with respect to y we get
=
 Z
( ) or Zxy and it is defined by
y x
f ( x, y  y )  f x ( x, y )  2 Z
 Z
( ) = lim x
y 0
y
y x
yx
Similarly we can get the first derivate with respect to y to get the second
derivative with respect to x or y so we get
f y ( x, y  y)  f y ( x, y)  2 Z
 Z
= ( ) = lim
y 0
y y
y
y 2
and
f y ( x  x, y )  f y ( x, y )
2Z
 Z
= ( ) = lim
x 0
x
xy x y
Theorem (the mixed derivative theorem)
2Z
Z Z  2 Z
,
,
and
are defined
x y xy
yx
If f(x, y) and its partial derivatives
throughout an open region containing a point (a, b) and are all
continuous at (a, b) then
=
e.g.: z = x2y2 + 3xy, →
2Z 2Z
xy yx
2Z
Z
Z
=2xy2 +3y,
=2x2y + 3x,
=2y2,
2
x
y
x
2
2
2Z
2and  Z =  Z = 4xy +3:
=2x
y 2
xy yx
And suppose that f(x,y,z) = xy - 2yz
is a function of three variables, then we can define the partial derivatives in much the
same way as we defined the partial derivatives for three variables.
We have
fx = y
fy = x - 2z
and
fz = -2y
Example: The Heat Equation
Suppose that a building has a door open during a snowy day. It can be shown that the
equation Ht = c2Hxx
models this situation where H is the heat of the room at the
point x feet away from the door at time t. Show that H = e-t cos(x/c) satisfies this
differential equation.
Solution We have
Ht = -e-t cos(x/c)
Hx = -1/c e-t sin(x/c)
Hxx = -1/c2 e-t cos(x/c)
So that
c2Hxx = -e-t cos(x/c)
And the result follows.
Examples
Find the second partial derivatives for
1. Z = x2 (1+y2 )
2. Z = x ey – y ex
3. show hat the function f(x, y) =
x
is harmonic except at origin
x  y2
2