Prove that $f\circ g$ is differentiable in t=0, but the chain rule doesn't work.

Given $f(x,y)=\begin{cases}\frac{x^2y}{x^2+y^2} &\text{ if }(x,y) \neq (0,0)\\\\ 0&\text{ if }(x,y)=(0,0)\end{cases}$

and let $g(t)=(t,-2t).$ Prove that $f\circ g$ is differentiable in t=0, but the chain rule doesn't work.

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Prove that $f \circ g$ is differentiable is easy. But I don't know how to prove that the chain rule doesn't work. – Henfe Jan 5 '13 at 19:52
Well, the chain rule would apply if $f$ and $g$ were differentiable. So you'd better show that at least one of them is not differentiable at the relevant point. – Chris Eagle Jan 5 '13 at 19:55

I interpret the question like this: Since $f\circ g(t) = -\frac25t$, we have that $(f\circ g)'(t) = -\frac25$.

On the other hand, if the chain rule would be applicable, we would get

$$(f\circ g)'(0) = f'_x(0,0)\cdot 1 + f'_y(0,0) \cdot (-2) = 0,$$

since $f'_x(0,0) = f'_y(0,0) = 0$ as noted by vesszabo in his answer.

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$g$ is differentiable (it's easy, I hope :) ). So consider $f$. If it was diff'able at $0$, then we could write $$f(x,y)-f(0,0)=A (x-0)+B(y-0)+R(x,y),$$ where $$\lim_{(x,y)\to(0,0)} \frac{R(x,y)}{\sqrt{x^2+y^2}}=0.$$ A theorem says, in this case $A=D_1 f(0,0)$, $B=D_2 f(0,0)$, where $D_j$ denotes the partial derivative with respect to the $j$-th variable. Now using the definition of partial derivative we have $$A=0,\quad B=0.$$ So we get $f(x,y)=R(x,y)$. Investigate the limit assumption. Introducing polar coordinates, $x=r\cos(\varphi)$, $y=r\sin(\varphi)$ we obtain $$\lim_{(x,y)\to(0,0)} \frac{R(x,y)}{\sqrt{x^2+y^2}}=\lim_{r\to 0+} \frac{r^3\cos^2(\varphi)\sin(\varphi)}{r^3}=\cos^2(\varphi)\sin(\varphi),$$ which, in general, is not $0$.

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