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S
econd
O
rder
P
artial
D
erivatives
Previously we have taken the partial derivative of a function
f
(
x
,
y
). But those partial
derivatives were themselves functions and so we can take their partial derivatives.
The SecondOrder Partial Derivatives of
z
=
f
(
x
,
y
)
2
2
()
x
x
zz
x
x
f
f
xx
x
2
yx
y
x
f
f
xy
x
y
2
x
y
x
y
f
f
y
x
2
2
yy
y
y
f
f
y
Example 1:
Compute the secondorder partial derivatives of
f
(
x
,
y
) =
x
2
y
+ 5
x
sin(
y
).
Solution:
Notice that
f
x
(
x
,
y
) = 2
xy
+ 5sin(
y
) and
f
y
(
x
,
y
) =
x
2
+ 5
x
cos(
y
). Thus, we have that
f
xx
(
x
,
y
) = 2
y
f
xy
(
x
,
y
) = 2
x
+ 5cos(
y
)
f
yx
(
x
,
y
) = 2
x
+ 5cos(
y
)
f
yy
(
x
,
y
) = –5
x
sin(
y
).
Notice that in Example 1 above, we have that
f
xy
(
x
,
y
) =
f
yx
(
x
,
y
). Indeed, this is
typically always the case. Thus, it does not matter if we take the partial derivative with
respect to
x
first or with respect to
y
first.
The Equality of Mixed Partial Derivatives
If
f
xy
(
x
,
y
) and
f
yx
(
x
,
y
) are continuous at (
a
,
b
), an interior point of their
domain, then
(,)
xy
yx
f
ab
f
.
1
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View Full DocumentWe can visualize this statement in the following diagram.
z
=
f
x,y
∑
z
∑
x
∑
∑
y
f
x,y
x,y
xy
x,y
=
yx
x,y
xx
x,y
yy
x,y
∑
Figure 1: Equality of Mixed Partial Derivatives
Example 2:
Find all secondorder partial derivatives of
f
(
x
,
y
) = ln(3
x
+ 5
y
).
Solution:
Notice that
3
(, )
35
x
fxy
x
y
and
5
y
x
y
. Thus, we have that
12
2
39
3
(
3 5)
9
(
(
)
xx
fx
y
x y
x
x
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 Spring '07
 Hohnhold
 Calculus, Derivative

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