A4f_soln - Math 237 Assignment 4 Due: Friday, Oct 17th 1....

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Math 237 Assignment 4 Due: Friday, Oct 17th 1. Find a function f ( x, y ) such that f xy (0 , 0) and f yx (0 , 0) both exist, but f xy (0 , 0) 6 = f yx (0 , 0). Prove you are correct. Solution: One example is f ( x, y ) = ± xy ( x 2 - y 2 ) x 2 + y 2 if ( x, y ) 6 = (0 , 0) 0 if ( x, y ) = (0 , 0) . . Observe that f x (0 , y ) = lim h 0 f ( h, y ) - f (0 , y ) h = lim h 0 hy ( h 2 - y 2 ) h 2 + y 2 h = lim h 0 y ( h 2 - y 2 ) h 2 + y 2 = - y f y ( x, 0) = lim h 0 f ( x, h ) - f ( x, 0) h = lim h 0 xh ( x 2 - h 2 ) x 2 + h 2 h = lim h 0 x ( x 2 - h 2 ) x 2 + h 2 = x Thus f xy (0 , y ) = - 1 for all y and f yx ( x, 0) = 1 for all x . Hence f xy (0 , 0) = - 1 and f yx (0 , 0) = 1. 2. Let f ( u, v ) = uv 2 where u = u ( x, y ) = x sin y and v = v ( x, y ) = e xy . Find f xy . Solution: By the chain rule, we have f x = f u u x + f v v x = v 2 (sin y ) + 2 vuye xy . Then, applying the chain rule again we get f xy = ∂f x ∂u u y + ∂f x ∂v v y + ² ∂f x ∂y ³ u,v,x = 2 vye xy ( x cos y ) + [2
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This note was uploaded on 04/18/2010 for the course MATH 237 taught by Professor Wolczuk during the Spring '08 term at Waterloo.

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A4f_soln - Math 237 Assignment 4 Due: Friday, Oct 17th 1....

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