21-356 Lecture 2

# F f simply the derivative of g at t 0 v x0 g 0

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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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