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Unformatted text preview: artial derivative of f with respect
!f
to xi and is denoted ! xi (x0 ) or fxi (x0 ) or Di f (x0 ).
Remark 4 Let F := {t # R : x0 + tv # E } $ R. If we consider the function
!f
of one variable g (t) := f (x0 + tv), t # F , then ! v (x0 ), when it exists, is
!f
!f
simply the derivative of g at t = 0, ! v (x0 ) = g " (0). If ! v (x0 ) exists, then g
is di!erentiable in t = 0 and so it is continuous at t = 0. Thus, the function f
restricted to the line L is continuous at x0 .
In view of the previous remark, one would be tempted to say that if the
directional derivatives at x0 exist and are ﬁnite in every direction, then f is
continuous at x0 . This is false in general, as the following example shows.
Example 5 Let
f (x, y ) := ! 1
0 if y = x2 , x %= 0,
otherwise. Given a direction v = (v1 , v2 ), the line L through 0 in the direction v intersects
the parabola y = x2 only in 0 and in at most one point. Hence, if t is very
small,
f (0 + tv1 , 0 + tv2 ) = 0.
It follows that
!f
f (0 + tv1 , 0 + tv2 ) " f (0, 0)
0"0
(0, 0) = lim
= lim
= 0.
t!0
t!0
!v
t
t
"
#
However, f is not continuous in 0, since f x, x2 = 1 & 1 as x & 0, while
f (x, 0) = 0 & 0 as x & 0.
Exercise 6 Let
f (x, y ) := $ x2 y
x4 +y 2 0 if (x, y ) %= (0, 0) ,
if (x, y ) = (0, 0) . Find all directional derivatives of f at 0 and prove that f is not continuous at
0....
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 Spring '14
 Geometry

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