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 deFned by f ( x,y )= xy x 2 y 2 x 2 + y 2 if ( x,y ) ° =(0 , 0) 0i f ( 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). ±or ( 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 . ±or D 1 f (0 , 0) we need to use the deFnition: D 1 f (0 , 0) = lim h 0 f ( h, 0) f (0 , 0) h =l im 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 =l im 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 =l im 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 =l im h
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kJnKp6-04 - ASSIGNMENT 4 SOLUTIONS MAT 473 SPRING 2012...

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