Example 5
Let
f
(
x, y
):=
!
1
if
y
=
x
2
,x
%
=0
,
0
otherwise.
Given a direction
v
=(
v
1
,v
2
)
, the line
L
through
0
in the direction
v
intersects
the parabola
y
=
x
2
only in
0
and in at most one point. Hence, if
t
is very
small,
f
(0 +
tv
1
,
0+
tv
2
)=0
.
It follows that
!
f
!
v
(0
,
0) = lim
t
!
0
f
(0 +
tv
1
,
tv
2
)
"
f
(0
,
0)
t
=l
im
t
!
0
0
"
0
t
.
However,
f
is not continuous in
0
,s
inc
e
f
"
x, x
2
#
=1
&
1
as
x
&
0
, while
f
(
x,
0) = 0
&
0
as
x
&
0
.
Exercise 6
f
(
x, y
$
x
2
y
x
4
+
y
2
if
(
x, y
)
%
=(0
,
0)
,
0
if
(
x, y
)=(0
,
0)
.
Find all directional derivatives of
f
at
0
and prove that
f
is not continuous at
0
.
5
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View Full DocumentThe previous examples show that in dimension
N
!
2
partial derivatives do
not give the same kind of results as in the case
N
=1
.T
oso
lv
eth
i
sp
rob
lem
,w
e
introduce a stronger notion of derivative, namely, the notion of di
!
erentiability.
We recall that a function
T
:
R
N
"
R
is
linear
if
T
(
x
+
y
)=
T
(
x
)+
T
(
y
)
for all
x
,
y
#
R
N
and
T
(
s
x
sT
(
x
)
for all
s
#
R
and
x
#
R
N
.W
r
i
t
e
x
=
!
N
i
=1
x
i
e
i
. Then by the linearity of
T
,
T
(
x
T
"
N
#
i
=1
x
i
e
i
$
=
N
#
i
=1
x
i
T
(
e
i
)
.
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 Spring '14
 Derivative, lim

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