# Contradict the chain rule [closed]

1. Let $$f ( x , y ) = \left\{ \begin{array} { c l } { \frac { x y ^ { 2 } } { x ^ { 2 } + y ^ { 2 } } \text { if } ( x , y ) \neq ( 0,0 ) } \\ { 0 \quad \text { if } ( x , y ) = ( 0,0 ) } \end{array} \right.$$ (a) Show that $$\frac { \partial f } { \partial x } ( 0,0 ) = 0$$ and $$\frac { \partial f } { \partial y } ( 0,0 ) = 0$$

(b) For $$g ( t ) = ( t , 2 t )$$ , show that $$( f \circ g ) ^ { \prime } ( 0 ) = \frac { 4 } { 5 } .$$

Does this contradict the Chain Rule? Why? Can anyone help me answer the question of whether it contradicts the chain rule? I know the f and g are differentiable at a certain point. Thanks! Please help me?

## closed as off-topic by Brian Borchers, Alexander Gruber♦Feb 10 at 5:54

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• “Differentiable at a certain point.” Which point might that be? There’s only one point in each function’s domain that’s relevant here. – amd Feb 10 at 5:13

Chain rule can be applied if $$f$$ is differentiable as function from $$\mathbb R^{2} \to \mathbb R$$. It is not enough if the partial derivatives exist at $$(0,0)$$.