# Showing $f$ is not differentiable at $(0,0)$ even though the partial derivatives exist.

$$f\begin{pmatrix}x\\y\end{pmatrix}:=\begin{cases}\frac{x^2y}{x^6+2y^2}& (x,y)^T\neq(0,0)^T\\0&(x,y)^T=(0,0)^T\end{cases}$$

I want to find out if $$f$$ is continuously partially differentiable at $$(0,0)^T$$.

The partial derivates at $$(0,0)^T$$ are both equal to $$0$$. So if $$f$$ were differentiable at $$(0,0)^T$$ we would have

$$\lim_{x,y\to 0}\frac{f(x,y)}{\sqrt{x^2+y^2}}=0$$ Using polar coordinates I end up with

$$\lim_{r\to 0}\frac{r^3 \cos^2 \phi \sin \phi}{(r^6 \cos^6 \phi +2 r^2 \sin^2\phi)r}$$

Which does't converge to $$0$$. So this means $$f$$ is not differentiable at $$(0,0)^T$$ and therefore not continuously partially differentiable at $$(0,0)^T$$.

Is that correct?

When you wrote$$\lim_{x,y\to 0}\frac{f(x,y)}{\sqrt{x^2+y^2}},$$you meant$$\lim_{x,y\to 0}\frac{f(x,y)}{\sqrt{x^2+y^2}}=0.$$Otherwise, it is correct.
• Thanks! When checking differentiability at $(0,0)$ can I simply do it like I did it or do I have to first check if the function is continuous at $(0,0)$? – conrad Nov 10 '18 at 19:30