$$f(x,y)=\begin{cases}\dfrac{y^3}{x^2+y^2} &(x,y) \neq \ \mathbb{(0,0)}\\ 0 & (x,y)=(0,0) \\ \end{cases}$$
Evaluate $f_x(0,0)$ and $f_y(0,0)$ and $D_\overrightarrow{u}f(0,0)$
I tried directly taking the derivative to no avail (obviously) so then I tried to use the definition of partial derivative which also left me without a correct solution. Also I have proved that it is continuous (Sertoz Theorem) but how would I prove that it is also differentiable at $(0,0)$?
Best Answer
Using polar coordinates $$\begin{gather}x = \rho\cos{\varphi}, \\ y= \rho\sin{\varphi}.\end{gather}$$ we have $${\frac{f(x,\,y) - f(0,\,0)}{\sqrt{x^2+y^2}}} ={\frac{f(\rho\cos{\varphi},\,\rho\sin{\varphi}) - 0}{\rho}} = {\frac{\rho^3\sin^3{\varphi}}{\rho^2\rho}} = \sin^3 \varphi$$
Since this can take arbitrary values when $\varphi$ changes, it means that the limit do not exists and so $Df(0,\,0)$ do not exists, i.e $f$ not differentiable in $(0,0)$.