MATH1211/1011(2)/Asol3/TNK
Department of Mathematics, The University of Hong Kong
MATH1211 (201011, Second semester) Suggested solution to Assignment 3
Note
: Some solutions given below are outline only. You may need to give more details in your solution. The
parts that are typed in
red color
are remarks for you to read; they are not a part of the solution.
1.
(a) For (
x,y
)
6
= (0
,
0),
f
(
x,y
) =
x
3
y

xy
3
x
2
+
y
2
. Hence
f
x
(
x,y
) =
(3
x
2
y

y
3
)(
x
2
+
y
2
)

(
x
3
y

xy
3
)2
x
(
x
2
+
y
2
)
2
=
x
4
y
+ 4
x
2
y
3

y
5
(
x
2
+
y
2
)
2
,
f
y
(
x,y
) =
(
x
3

3
xy
2
)(
x
2
+
y
2
)

(
x
3
y

xy
3
)2
y
(
x
2
+
y
2
)
2
=
x
5

4
x
3
y
2

xy
4
(
x
2
+
y
2
)
2
.
(b) For
y
6
= 0 and
x
= 0, by result of (a) we have
f
x
(0
,y
) =
0 + 0

y
5
(0 +
y
2
)
2
=

y
. To ﬁnd
f
x
(0
,
0) we
evaluate:
f
x
(0
,
0) = lim
h
→
0
f
(
h,
0)

f
(0
,
0)
h
= lim
h
→
0
0

0
h
= 0
.
For
x
6
= 0 and
y
= 0, by result of (a), we have
f
y
(
x,
0) =
x
5

0

0
(
x
2
+ 0)
2
=
x
. To ﬁnd
f
y
(0
,
0), we
evaluate:
f
y
(0
,
0) = lim
k
→
0
f
(0
,k
)

f
(0
,
0)
k
= lim
k
→
0
0

0
k
= 0
.
(c) From result of (b), we have
f
xy
(0
,
0) = lim
k
→
0
f
x
(0
,k
)

f
x
(0
,
0)
k
= lim
k
→
0

k

0
k
=

1
,
f
yx
(0
,
0) = lim
h
→
0
f
y
(
h,
0)

f
y
(0
,
0)
h
= lim
h
→
0
h

0
h
= 1
.
Hence the mixed second order derivatives
f
xy
(0
,
0) and
f
yx
(0
,
0) are not equal. Here Theorem 4.3
does not apply. It is because, from (a), for (
x,y
)
6
= (0
,
0),
f
xy
(
x,y
) =
(
x
4
+ 12
x
2
y
2

5
y
4
)(
x
2
+
y
2
)
2

(
x
4
y
+ 4
x
2
y
3

y
5
)2(
x
2
+
y
2
)2
y
(
x
2
+
y
2
)
4
=
x
6
+ 9
x
4
y
2

9
x
2
y
4

y
6
(
x
2
+
y
2
)
3
.
Along the line
x
= 0, we have lim
y
→
0
f
xy
(0
,y
) = lim
y
→
0

y
6
y
6
=

1. Along the line
y
= 0, we have
lim
x
→
0
f
xy
(
x,
0) = lim
x
→
0
x
6
x
6
= 1. Hence
lim
(
x,y
)
→
(0
,
0)
f
xy
(
x,y
) does not exist, which implies that
f
xy
is not
continuous at (0
,
0). Thus the hypothesis in Theorem 4.3, namely, that all ﬁrst order and second
order partial derivatives are continuous, does not hold.
2.
(a)
f
x
(0
,
0) = lim
h
→
0
f
(
h,
0)

f
(0
,
0)
h
= lim
h
→
0
0

0
h
= 0.
f
y
(0
,
0) = lim
h
→
0
f
(0
,h
)

f
(0
,
0)
h
= lim
h
→
0
0

0
h
=
0.
1
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f
◦
x
(
t
) =
f
(
t,at
) =
(
at
3
(1+
a
2
)
t
2
if
t
6
= 0
0
if
t
= 0
=
(
at
(1+
a
2
)
if
t
6
= 0
0
if
t
= 0
. Since
±
at
1+
a
2
²
t
=0
= 0, we can
combine the two cases of
t
6
= 0 and
t
= 0 together, to get
f
◦
x
(
t
) =
at
1+
a
2
for all
t
∈
R
. Obviously
f
◦
x
is diﬀerentiable, and
D
(
f
◦
x
)(0) =
a
1+
a
2
.
(c) By deﬁnition
Df
(0
,
0) =
³
f
x
(0
,
0)
f
y
(0
,
0)
´
=
³
0 0
´
, and hence
Df
(0
,
0)
D
x
(0) =
³
0 0
´
µ
1
a
¶
= 0,
which is not equal to
D
(
f
◦
x
)(0) as obtained in (b). This is because
f
is actually not diﬀerentiable
at (0
,
0) and hence the chain rule cannot apply. To verify that
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