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Unformatted text preview: Math 136, Answers to Review Problems for Test 1 1. We calculate D f using the chain rule and the fact that F ( x,y,f ( x,y )) = 0. We define G ( x,y ) = x y f ( x,y ) and note that D g = 1 1 ∂f ∂x ∂f ∂y . We know G is differentiable since its coordinate functions are C 1 (the first two coordinate functions are C ∞ , and the third, f , is C 1 since F and ( x ,y ,z ) satisfy the Implicit Function Theorem). Using the chain rule on the equation F ( x,y,f ( x,y )) = 0, we get = D 0 = D ( F ( x,y,f ( x,y ) ) = D ( F ◦ G ) = D F · D G where · is matrix multiplication. Now, using the fact that D F ( x,y,z ) = ( ∂F ∂x , ∂F ∂y , ∂F ∂z ) we see (1) 0 = ∂F ∂x + ∂F ∂z ∂f ∂x , ∂F ∂y + ∂f ∂y ∂F ∂z . We can solve (1) for ∂f ∂x and ∂f ∂y and this give ∂f ∂x = ∂F ∂x ∂F ∂z and ∂f ∂y = ∂F ∂y ∂F ∂z evaluated at ( x ,y ,f ( x ,y )). Here we used that ∂F ∂z ( x ,y ,z ) 6 = 0 (and z = f ( x ,y )). 2. The function F in this problem satisfies the hypotheses of the Implicit Function Theo rem, so we can assert there is an open neighborhood U of x and an open neighborhood V of y and a C 1 function f : U → V such that F ( x,f ( x )) = F ( x ,y ) for all x ∈ U . (a) So, as there are an infinite number of points in U , and for each x ∈ U , F ( x,f ( x )) = F ( x ,y ), there are an infinite number of points that satisfy F ( x,y ) = F ( x ,y )....
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This note was uploaded on 03/27/2008 for the course MATH 136 taught by Professor Quinto during the Spring '08 term at Tufts.
 Spring '08
 Quinto
 Chain Rule, The Chain Rule

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