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kJnKp6-04

# kJnKp6-04 - ASSIGNMENT 4 SOLUTIONS MAT 473 SPRING 2012...

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ASSIGNMENT 4 · SOLUTIONS MAT 473 · SPRING 2012 Exercise 4.2. Consider the function f : R 2 R defined by f ( x, y ) = xy x 2 y 2 x 2 + y 2 if ( x, y ) = (0 , 0) 0 if ( x, y ) = (0 , 0). Prove that all second-order partial derivatives of f exist, but the conclusion to Clairaut’s Theorem still fails. Proof. Of course, the problem is at (0 , 0). For ( x, y ) = (0 , 0), by direct calculation (product and quotient rules) we have D 1 f ( x, y ) = · · · = y ( x 4 y 4 + 4 x 2 y 2 ) ( x 2 + y 2 ) 2 , and in particular D 1 f (0 , y ) = y for y = 0. For D 1 f (0 , 0) we need to use the definition: D 1 f (0 , 0) = lim h 0 f ( h, 0) f (0 , 0) h = lim h 0 0 h = 0 . Thus, D 11 f (0 , 0) = lim h 0 D 1 f ( h, 0) D 1 f (0 , 0) h = lim h 0 0 h = 0 and D 21 f (0 , 0) = lim h 0 D 1 f (0 , h ) D 1 f (0 , 0) h = lim h 0 h h = 1 . Similarly, for ( x, y ) = (0 , 0) we have D 2 f ( x, y ) = · · · = x ( x 4 y 4 4 x 2 y 2 ) ( x 2 + y 2 ) 2 , so that D 2 f ( x, 0) = x for x = 0, and D 2 f (0 , 0) = lim h 0 f (0 , h ) f (0 , 0) h = lim h 0 0 h = 0 . Thus D 22 f (0 , 0) = lim h 0 D 2 f (0 , h ) D 2 f (0 , 0) h = lim h 0 0 h = 0 but also D 12 f (0 , 0) = lim h 0 D 2 f ( h, 0) D 2 f

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