Math 326A. Midterm Sample.
1. Consider the function
f
(
x, y
) =
x
5
+
y
5
x
3
+
y
3
for (
x, y
)
negationslash
= (0
,
0), with the value
f
(0
,
0) = 0. Show
that
f
(
x, y
) is differentiable at (0
,
0) by verifying that the partial derivatives are defined at
(0
,
0) and that one of them (your choice) is continuous at (0
,
0). (Use the quotient formula
to compute a partial derivative at any point (
x, y
)
negationslash
= (0
,
0).)
Solution:
∂f
∂x

(0
,
0)
=
d
(
f
(
x,
0)
dx

x
=0
=
d
(
x
2
)
dx

x
=0
= 2
x

x
=0
= 0
.
In the same spirit,
∂f
∂y

(0
,
0)
= 0.
To check for continuity of
∂f
∂y
at (0
,
0), we need to see if lim
(
x,y
)
→
(0
,
0)
∂f
∂y
=
∂f
∂y

(0
,
0)
, i.e.
if
lim
(
x,y
)
→
(0
,
0)
∂f
∂y
= 0.
To compute the limit, we have the following formula for
∂f
∂y
which is valid at points (
x, y
)
negationslash
=
(0
,
0):
∂f
∂y
=
5
y
4
(
x
3
+
y
3
)
−
(
x
5
+
y
5
)(3
y
2
)
(
x
3
+
y
3
)
2
.
Because the numerator is homogeneous of degree 7 and the denominator is homogeneous
of degree 6, we have lim
(
x,y
)
→
(0
,
0)
∂f
∂y
= 0, which is the value of
∂f
∂y

(0
,
0)
.
Hence, we have
established the continuity of
∂f
∂y
at (0
,
0). Hence, by our theorem,
f
(
x, y
) is differentiable at
(0
,
0).
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2. Consider the differentiable function
F
(
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 Spring '10
 Silverman
 Derivative, lim, 5 pts, 10 pts

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