130
CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES
3.3
3.3.1
Partial Derivatives
Quick Calculus I Review
dy
You will recall that if y = f (x), then the derivative of f , denoted f 0 (x) or dx
d
or dx f (x) is the instantaneous rate of change of f with respect to x. We list
some facts about the derivative students should know:
1. f 0 (a) = lim
x!a
f (x) f (a)
x a
= lim
h!0
f (a+h) f (a)
h
2. If f is di¤erentiable at a point a ( f 0 (a) exists) then f is continuous at a.
3. f 0 (a) is the slope of the tangent to y = f (x) at x = a.
4. If f 0 (x) > 0 on an interval, then f is increasing on that interval.
5. If f 00 (x) > 0 on an interval, then f is concave up on that interval ( f is
increasing at an increasing rate).
We extend the notion of the derivative to functions of several variables.
3.3.2
De…nitions and Interpretations of Partial Derivatives
If the derivative is the rate of change, a …rst question which arises when dealing
with functions of two or more variable is "rate of change with respect to which
variable"? In other words, when we study how z = f (x; y) is changing, what
kind of change are we talking about? Is it a change with respect to x, to y,
to both? In Calculus I, the graph of a function y = f (x) was often associated
with a path along which you were walking. If the path was climbing, then the
derivative (rate of change of y with respect to x) was positive. If the path was
‡at, then the derivative (rate of change) was 0. Otherwise, it was negative. We
can carry this analogy to functions of several variables. The graph of a function
of two variables, z = f (x; y), is a surface in 3-D. It looks more like a real terrain
than the graph of a function of one variable did. Carrying the analogy, we
could say that as we walk along a path, if we are climbing then the derivative is
positive, if we are going downhill then the derivative is negative and if that path
is ‡at, then the derivative is 0. What is more complex in the case of functions of
several variables is that we can be walking in many (in…nitely many) directions.
We …rst look at what happens if we are walking in a direction parallel to the
x-axis or the y-axis. We will then see what happens when we are walking in
any direction.
Consider the surface given by z = f (x; y) and a point P (a; b; f (a; b)) on the
surface. If we intersect this surface with a plane parallel to the xz-plane through
P , then the equation of this plane is y = b (see …gure 3.5). The intersection of
this plane with the surface is a curve which only depends on x, call it g (x). Its
3.3. PARTIAL DERIVATIVES
131
equation is z = g (x) = f (x; b). Along this curve, the rate of change of z with
respect to x is
g 0 (x)
=
=
g (x + h) g (x)
h
f (x + h; b) f (x; b)
lim
h!0
h
lim
h!0
Geometrically, this corresponds to the slope of this the curve. Similarly, if we
intersect the surface with a plane parallel to the yz-plane through P , then the
equation of this plane is x = a. The intersection of this plane with the surface is
a curve which only depends on y, call it q (y). Its equation is z = q (y) = f (a; y).
The slope of this curve is q 0 (y). and
q 0 (y)
q (y + h) q (y)
h
f (a; y + h) f (a; y)
= lim
h!0
h
=
lim
h!0
Using the idea above, we de…ne the partial derivatives of f .
De…nition 225 (partial derivatives) Let f be a function of two variables.
1. The partial derivative of f with respect to x, denoted fx (x; y) is de…ned
to be:
f (x + h; y) f (x; y)
fx (x; y) = lim
h!0
h
2. The partial derivative of f with respect to y, denoted fy (x; y) is de…ned to
be:
f (x; y + h) f (x; y)
fy (x; y) = lim
h!0
h
There are other notations for the partial derivatives.
De…nition 226 Let z = f (x; y). Then:
1. fx (x; y) = fx =
@f
@x
=
@
@x f
(x; y) =
@z
@x
= f1 = D 1 f = D x f
2. fy (x; y) = fy =
@f
@y
=
@
@y f
(x; y) =
@z
@y
= f2 = D2 f = Dy f
Remark 227 The subscripts 1 and 2 represent the …rst and second variable
which are x and y. It becomes more relevant when we deal with functions having
n variables, for large n.
Looking at the de…nition of fx (x; y), we see that the variable y is not changing. Only x is changing. So, we are computing how f (x; y) is changing with
respect to x. Geometrically, if we consider the curve C1 at the intersection of
the surface z = f (x; y) with the plane y = c where c is a constant, then such a
curve is simply a function of x since along the curve y = c. The slope of this
132
CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES
Figure 3.5: The slope of the curve at the intersection of the surface z = f (x; y)
with the plane y = 1 is the fx (x; y)
curve is fx (x; y) (see …gure 3.5). To be more precise, the slope of this curve
would be fx (x; c). Similarly, fy (x; y) is the slope of the curve C2 at the intersection of the surface z = f (x; y) and the plane x = c. To be more precise, the
slope of this curve would be fy (c; y). Going back to the analogy with walking
on a terrain, fx (x; y) corresponds to the slope of the path along a direction
parallel to the x-axis and fy (x; y) corresponds to the slope of the path along a
direction parallel to the y-axis : In other words, the partial derivatives fx (a; b)
and fy (a; b) give the slope at (a; b) of C1 and C2 where C1 is the curve through
(a; b) at the intersection of z = f (x; y) and the plane y = b and C2 is the curve
through (a; b) at the intersection of z = f (x; y) and the plane x = a.
This de…nition extends to functions of more than two variables. In general,
we have:
De…nition 228 Let f (x1 ; x2 ; :::; xn ) be a function of n variables. Then,
fxi (x1 ; x2 ; :::; xn ) = lim
h!0
f (x1 ; x2 ; :; xi + h; ::; xn )
h
f (x1 ; x2 ; :; xi ; ::; xn )
3.3. PARTIAL DERIVATIVES
3.3.3
133
Computation
Since partial derivatives are rates of change with respect to one variable only, we
can use the rules of di¤erentiation from Calculus I. More speci…cally, we have:
Proposition 229 Rules to …nd partial derivatives of z = f (x; y)
1. To …nd fx , regard y as a constant and di¤ erentiate f (x; y) with respect to
x.
2. To …nd fy , regard x as a constant and di¤ erentiate f (x; y) with respect to
y.
Remark 230 In the process outlines above, you can use all the rules of di¤ erentiation from Calculus I.
Example 231 Find
@f
@x
and
@f
@y
for f (x; y) = x2 + y 2 + 5xy.
1.
@f
@x
=
@
@x
x2 + y 2 + 5xy = 2x + 5y
2.
@f
@y
=
@
@y
x2 + y 2 + 5xy = 2y + 5x
Example 232 Find
@f
@x
and
@f
@y
x
1+y
for f (x; y) = sin
1.
@f
@x
=
@
sin
@x
=
cos
=
cos
=
@
sin
@y
=
cos
=
cos
x
1+y
x
1+y
x
1+y
@
x
@x 1 + y
1
1+y
(chain rule)
2.
@f
@y
Example 233 Find
@z
@x
x
1+y
x
@
1 + y @y
x
1+y
x
(chain rule)
1+y
!
x
(1 + y)
2
if z is de…ned implicitly by
x2 + y 2 + z 2 = 4
134
CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES
Since z is de…ned implicitly, we must use implicit di¤ erentiation.
@
x2 + y 2 + z 2
@x
@z
2x + 2z
@x
@z
@x
Example 234 Find
@z
@x
=
@
4
@x
=
0
x
z
=
if z is de…ned implicitly by
x3 + y 3 + z 3 + 6xyz = 1
Since z is de…ned implicitly, we must use implicit di¤ erentiation.
@
x3 + y 3 + z 3 + 6xyz
@x
@z
@z
3x2 + 3z 2
+ 6yz + 6xy
@x
@x
@z
3z 2 + 6xy
@x
@z
@x
@z
@x
3.3.4
=
@
1
@x
=
0
3 x2 + 2yz
=
3 x2 + 2yz
3 (z 2 + 2xy)
x2 + 2yz
z 2 + 2xy
=
=
Higher Order Derivatives
If f is a function in two variables, so are fx and fy . So, we can di¤erentiate
then. Their partial derivatives will also be functions in several variables, so we
can di¤erentiate then again. Thus, we have the following:
@
@x
@
=
@y
@
=
@y
@
=
@x
(fx )x
= fxx = f11 =
(fy )y
= fyy = f22
(fx )y
= fxy = f12
(fy )x
= fyx = f21
@f
@x
@f
@y
@f
@x
@f
@y
@2f
@2z
=
@x2
@x2
2
@ f
@2z
=
=
@y 2
@y 2
2
@ f
@2z
=
=
@y@x
@y@x
@2f
@2z
=
=
@x@y
@x@y
=
It is similar for higher order derivatives.
Example 235 Find the second order partial derivatives for z = f (x; y) = x3 +
y 3 + x2 y 2
3.3. PARTIAL DERIVATIVES
135
1. fx = 3x2 + 2y 2 x
2. fy = 3y 2 + 2x2 y
3. fxx = 6x + 2y 2
4. fyy = 6y + 2x2
5. fxy = 4yx
6. fyx = 4xy
Remark 236 You will note that the mixed partials are the same. This is not an
accident. Clairaut, a French mathematician (1713-1765) proved the following
theorem:
Theorem 237 (Clairaut’s Theorem) Suppose that f is de…ned on a disk D
which contains the point (a; b). If the functions fxy and fyx are both continuous
on D, then
fxy (a; b) = fyx (a; b)
3.3.5
Di¤erentiability and Continuity
For functions of one variable, di¤erentiability at x = a simply means that the
derivative exists at x = a, that is lim f (a+h)h f (a) exists. For functions of several
h!0
variables, there are several partial derivatives to consider. We give the result as
a theorem without proof.
Theorem 238 Consider the function f (x; y). If the partial derivatives fx and
fy are continuous on an open region D, then f is di¤ erentiable at every point
in D.
Another important di¤erence between functions of one variable and functions of several variables is related to the relationship between di¤erentiability
and continuity. For functions of one variable, if f 0 (a) exists then f has to be
continuous at a. For functions f of several variables, it is possible for fx (a; b)
and fy (a; b) to exist and for f not to be continuous at (a; b). For f to be continuous we also need to know that the partials fx and fy are continuous at (a; b)
in other words that f is di¤erentiable at (a; b). We have the following theorem:
Theorem 239 If f (x; y) is di¤ erentiable at (a; b) that is if fx and fy exist and
are continuous at (a; b) then f is also continuous at (a; b).
3.3.6
Partial Di¤erential Equations
A partial di¤erential equation is an equation which involves an unknown function and some of its partial derivatives. Such equations arise in many applications in physics, chemistry, economics, ... We mention two such equations
here.
136
CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES
The Heat Equation:
@2u @2u
+ 2
@x2
@y
@u
=k
@t
u (x; y; t) gives the heat of an isotropic, homogeneous plate as a function
of the position on the plate: (x; y) and time: t. k is a constant which
depends on the medium. In 3 D, the heat equation becomes
@2u @2u @2u
+ 2 + 2
@x2
@y
@z
@u
=k
@t
In this case u (x; y; z; t) gives the heat of an isotropic, homogeneous 3 D
object as a function of the position on the object: (x; y; z) and time: t.
Laplace Equation:
@2u @2u
+ 2 =0
@x2
@y
Solutions of this equation are called harmonic functions. They play a
role in problems related to heat conduction, ‡uid ‡ow.
The wave equation is:
2
@2u
2@ u
=
a
@t2
@x2
It describes the motion of a waveform. Examples of waveforms include an
ocean wave, a sound wave, a wave traveling along a vibrating string. The
equation written above corresponds to the vibration of a string. u (x; t)
gives the displacement of a string x units from one end of the string at
time t. In the case of an ocean wave, the function would be of the form
u (x; y; t) and the wave equation would be
@2u
= a2
@t2
@2u @2u
+ 2
@x2
@y
The constant a is related to the vibrating medium.
Example 240 Show that u (x; y) = e
@u
@x
@2u
@x2
x
sin y satis…es Laplace equation.
=
e
x
sin y
=
e
x
sin y
=
e
x
cos y
and
@u
@y
@2u
@y 2
=
e
x
sin y
3.3. PARTIAL DERIVATIVES
137
Therefore
@2u @2u
+ 2
@x2
@y
x
= e
=
sin y
=
uxx
=
sin y
0
Example 241 Show that u (x; t) = sin (x
ux
x
e
at) satis…es the wave equation.
cos (x
at)
sin (x
at)
a cos (x
at)
and
ut
=
utt
=
2
a sin (x
at)
Therefore
utt
a2 uxx
a2 sin (x
=
=
3.3.7
at)
a2 ( sin (x
at))
0
Partial Derivatives with Maple
Let f denote a function of one of more variables. Maple can …nd the partial
derivatives of any order of f . The table below explains the syntax to use.
Partial Syntax
Other Syntax
fx
di¤(f; x)
fxx
di¤(f; x; x)
di¤(f; x$2)
fxxx
di¤(f; x; x; x)
di¤(f; x$3)
fxy
di¤(f; x; y)
fxxy
di¤(f; x; x; y)
fyxyx
di¤(f; x; y; x; y)
See the accompanying Maple worksheet for examples.
3.3.8
We use
Partial Derivatives With Scienti…c Notebook or Workplace
@
@2 @2
@2
@
or
for …rst order partial derivatives. We use
,
,
,
@x
@y
@x2 @y 2 @y@x
@2
for second order derivatives and so on.
@x@y
Example 242 Find
@f
@x
@f
@y
and
@ sin
for f (x; y) = sin
x
1+y
@x
=
x
1+y
1
x
cos
y+1
y+1
138
CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES
and
@ sin
x
1+y
@y
=
x
(y + 1)
2
cos
x
y+1
which is what we found earlier when we did it by hand.
Example 243 Find all the second order derivatives of f (x; y) = x3 + y 3 + x2 y 2
@ 2 x3 + y 3 + x2 y 2
= 2y 2 + 6x
@x2
@ 2 x3 + y 3 + x2 y 2
= 2x2 + 6y
@y 2
@ 2 x3 + y 3 + x2 y 2
= 4xy
@y@x
@ 2 x3 + y 3 + x2 y 2
= 4xy
@x@y
3.3.9
Assignment
Do odd # 1 - 49, 57, 61, 65, 69, 71 at the end of 11.3 in your book.