# Differentiability of $f(x)=\frac{x|y|}{\sqrt{x^2+y^2}}$ at $(0,0)$

Prove that$$\begin{array}\\&f(x)&=\begin{cases}\frac{x|y|}{\sqrt{x^2+y^2}}&\text{ if }(x,y) \ne (0,0)\\0&\text{ otherwise }\end{cases}\end{array}$$ is differentiable in $$(0,0)$$.

Proof: We know that $$f$$ is differentiable in $$(0,0)$$, if all partial derivatives exist and are continuous in $$(0,0)$$. We have $$\frac{\partial f}{\partial x}=\lim_{h\to\ 0}\frac{f(h,0)}{h}=0$$ $$\frac{\partial f}{\partial y}=\lim_{h\to\ 0}\frac{f(0,h)}{h}=0$$ So the partial derivatives exist.

I am now stuck at proving the continuity of the partial derivatives in $$(0,0)$$. Can anyone help me with that?

## 1 Answer

Since $\partial_x f(0,0) = \partial_y f(0,0) = 0$ and $f(0,0) = 0$, by definition your function is differentiable at $(0,0)$ if and only if $$\lim_{(h,k) \to (0,0)} \varphi(h,k) = 0, \quad\text{where}\quad \varphi(h,k) := \frac{f(h, k)}{\sqrt{h^2+k^2}} = \frac{h |k|}{h^2+k^2}, \ (h,k)\neq (0,0).$$ But $\varphi(h,h) = \frac{h |h|}{2h^2} = \frac{1}{2} \text{sign} h$, hence your function $f$ is not differentiable at the origin.