Differentiability of the function $f(x,y)=\begin{cases} \frac{xy}{x^2+y^2}, & \text{if $(x,y)\ne(0,0)$}\\ 0, & \text{otherwise} \end{cases}$

Question

Consider the function $f(x,y)=\begin{cases} \frac{xy}{x^2+y^2}, & \text{if $(x,y)\ne(0,0)$}\\ 0, & \text{otherwise} \end{cases}$

Question: Is it differentiable everywhere? Does it have partial derivatives everywhere?

My attempt:

$f$ is differentiable on $\Bbb R^2\backslash\{(0,0)\}$ because it is the ratio between two differentiable functions and the denominator doesn't vanish. Is it still the case in $(0,0)$?

If we write $x=r\cos\theta$ and $y=r\sin\theta$ then the function becomes

$\begin{cases} \frac{r^2\cos\theta\sin\theta}{r^2} & \text{if } r\ne0 & \\ 0 &\text{if } r=0 \end{cases}$ = $\begin{cases} \cos\theta\sin\theta & \end{cases}$

We realize that $f$ doesn't depend on the radius, so the fact that $f(1,1)=1/2\ne-1/2=f(-1,1)$ so we can approach zero by decreasing the radius and preserving the values $1/2$ and one hand and $-1/2$ on the other. So the function is discontinuous at zero, so not differentiable.

Does it have partial derivatives everywhere?

$$\frac{\partial f}{\partial x}=\lim_{h\to\ 0}\frac{f(h,0)}{h}=0$$ $$\frac{\partial f}{\partial y}=\lim_{h\to\ 0}\frac{f(0,h)}{h}=0$$ So the partial derivatives exist in zero and elsewhere.

Answer 1

Yes you are right the function is not continuous and then it is not differentiable at $(x,y)=(0,0)$, since differentiability $\implies$ continuity.

Remarks:

• as a simpler alternative, to show that $f(x,y)$ is not continuos at the origin, it suffices to consider the paths with $x=0$ and $x=y$.
• the existence of the partial derivative is not sufficient to guarantee differentiability indeed for that we need also that at least one partial derivative is also continuos at that point.