10 1. ADVANCED DERIVATIVES
Problem 1.20 For each of the following functions find f
xy
.
1. f .x; y/ D x
2
y Cx
3
y Cy sin.x/
2. g.x; y/ D x sin.y/ C x tan
1
.x/
3. h.x; y/ D xy
2
C x
2
y Cxy
4. r.x; y/ D
x.y C1/
5
C 1
2
5. s.x; y/ D x sin.xy/
6. q.x; y/ D y cos.xy/
7. a.x; y/ D
x
y
C
y
x
C y
3
8. b.x; y/ D .1 C x C y C xy/
4
Problem 1.21 Find f
x
, f
y
, f
xx
, f
xy
, and f
yy
if f .x; y/ D
x
2
x
2
C y
2
.
Problem 1.22 Find g
x
, g
y
, g
xx
, g
xy
, and g
yy
if g.x; y/ D tan
1
.xy C 1/.
Problem 1.23 Rind f
xx
, f
xxy
, and f
xxyy
if h.x; y/ D sin.xy/.
Problem 1.24 Find z
x
and z
y
if z D .x
2
C 1/
y
.
Problem 1.25 Find z
x
and z
y
if z
xy
D 2.
Problem 1.26 Find z
x
and z
y
if z D
x
3
.x C 1/
2
.x 1/
3
y
2
.y 1/
3
.y C1/
5
.
1.2 THE GRADIENT AND DIRECTIONAL DERIVATIVES
We are now ready to look at some of the opportunities that are available once we understand
partial derivatives. At any point on the surface that forms the graph of z D f .x; y/ there are
an infinite number of directions and so an infinite number of rates at which the function is
changing. Pick a direction, and the function has a rate of change in that direction.
It turns out that there is some order to this richness of directions and rates of growth, in the
form of a simple formula for the direction in which the function is growing fastest.
1.2. THE GRADIENT AND DIRECTIONAL DERIVATIVES 11
Knowledge Box 1.3
If z D f .x; y/ is a function of two variables, then
rf .x; y/ D
f
x
; f
y
is called the gradient of f .x; y/ . e gradient of a function points in the direction
it is growing most quickly; the rate of growth is the magnitude of the gradient.
Example 1.27 Find the gradient of the function f .x; y/ D x
2
C y
2
C 3xy.
Solution:
Using the formula given, rf .x; y/ D
.
2x C 3y; 3x C 2y
/
.
˙
We can ask much more complex questions about the gradient than simply computing its value.
Example 1.28 At what points is the function
g.x; y/ D sin.x/ Ccos.y/
changing the fastest in its direction of maximum increase?
Solution:
is question wants us to maximize the magnitude of the gradient.
First compute the gradient:
rg.x; y/ D
.
cos.x/; sin.y/
/
e magnitude of this is
q
cos
2
.x/ C sin
2
.y/
Since x and y vary independently, the answer is simply those points that make cos
2
.x/ and
sin
2
.y/ both one.
So, the answer is those points .x; y/ such that
x D n and y D
2m C1
2
12 1. ADVANCED DERIVATIVES
where n and m are whole numbers.
˙
e graph in Figure 1.3 might help you understand Example 1.28.
-8
-6
-4
-2
0
2
4
6
8
-8
-6
-4
-2
0
2
4
6
8
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
f(x,y)=sin(x)+cos(y)
Figure 1.3: A graph of z D sin.x/ C cos.y/ for 8 x; y 8.
Example 1.29 Find the gradient of
z D x
y
Solution:
e partial derivative with respect to x is not hard because y is treated as a constant – so
z
x
D yx
y1
. e partial derivative with respect to y is trickier. It uses the formula for the
derivative of a constant to a variable power which gives z
y
D x
y
ln.x/.
is makes the gradient
rz D
yx
y1
; x
y
ln.x/
˙
1.2. THE GRADIENT AND DIRECTIONAL DERIVATIVES 13
What is the physical meaning of the gradient? We have already noted that it points in the
direction in which a surface grows fastest away from the point where the gradient is computed
– the steepest uphill slope away from the point. e magnitude of the gradient also gives of the
rate of fastest growth. Another fact is that the negative of the function
rf .x; y/
is the steepest downhill slope away from the point.
In other words, the negative of the gradient is the direction that a ball, starting at rest,
will roll.
-3
-2
-1
0
1
2
3
4
5
-3
-2
-1
0
1
2
3
4
5
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
f(x,y)=2xy/(x^2+y^2+1)
Figure 1.4: A graph of H.x; y/ D
2x C y
x
2
C y
2
C 1
for 3 x; y 5
Example 1.30 Suppose that the function H.x; y/ D
2x C y
x
2
C y
2
C 1
(Figure 1.4) describes the
height of a surface. Which direction will a ball placed at the point .1; 2/ roll?
14 1. ADVANCED DERIVATIVES
Solution:
We need to compute the negative of the gradient of H.x; y/.
H
x
.x; y/ D
.x
2
C y
2
C 1/.2/ .2x Cy/.2x/
.x
2
C y
2
C 1/
2
H
x
.1; 2/ D
12 8
36
D 1=9
H
y
.x; y/ D
.x
2
C y
2
C 1/.1/ .2x Cy/.2y/
.x
2
C y
2
C 1/
2
H
y
.1; 2/ D
6 8
36
D 1=18
So the ball rolls in the direction of the vector
rH.1; 2/ D
1
9
;
1
18
˙
Notice that, if we were designing a game, then we could use a well-chosen equation to give us
a height map for the surface, and the gradient could be used to tell which ways balls would roll
and water would flow.
Visualization helps us understand functions. is leads to the question: what does a gradient
look like? e gradient of a function f .x; y/ assigns a vector to each point in space. at means
that we could get an idea of what a gradient looks like by plotting the vectors of the gradient on
a grid of points.
Example 1.31 Let f .x; y/ D x
2
C y
2
. For all points .x; y/ with coordinates in the
set f˙2; ˙1:5; ˙1; ˙0:5; 0g, plot the point and the vector starting at that point in the direction
rf .x; y/.
Solution:
Since rf .x; y/ D .2x; 2y/ the vectors are:
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