21-356 Lecture 6

14 friday january 21 2011 4 higher order derivatives

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Unformatted text preview: an write U = A ' A1 , where A and A1 are open and disjoint. Since A is nonempty and U is connected, necessarily A1 must be empty, since otherwise we would negate the fact that U is connected. Hence, U = A, that is, f (x) = c for all x $ U . 14 Friday, January 21, 2011 4 Higher Order Derivatives Let E ! RN , let f : E " R and let x0 # E . Let i # {1, . . . , N } and assume !f that there exists the partial derivatives ! xi (x) for all x # E . If j # {1, . . . , N } and x0 is an accumulation point of E $ L, where L is the line through x0 in !f the direction ej , then we can consider the partial derivative of the function ! xi with respect to xj , that is, ! " ! !f !2f =: . ! xj ! xi ! xj ! xi Note that in general the order in which we take derivatives is important. Example 25 Let f (x, y ) := If y %= 0, then !f ! (x, y ) = !x !x ! # y 2 arctan x y 0 x y arctan y 2 3 = and !f ! (x, y ) = !y !y ! y , x2 + y 2 x y arctan y 2 " =y 2 1+ 1 ! $ %2 !x x y x = 2y arctan + y 2 y 2 = 2y arctan " if y %= 0, if y = 0. x xy &2 , y x + y2 1+ !" x y 1 ! $ %2 !y x y !" x y while at points (x0 , 0) we have: !f f (x0 + t, 0) & f (x0 , 0) 0&0 (x0 , 0) = lim = lim = 0, t!0 t!0 !x t t t2 arctan xt0 & 0 !f f (x0 , 0 + t) & f (x0 , 0) (x0 , 0) = lim = lim t!0 t!0 !y t t x0 = lim t arctan & = 0, t!0 t where we have used the fact that arctan xt0 is bounded and t " 0. Thus, & & y3 xy 2 !f !f if y %= 0, 2y arctan x & x2 +y2 if y %= 0, x2 +y 2 y (x, y ) = (x, y ) = !x !y 0 if y = 0, 0 if y = 0. 15...
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