# Check on differentiability of a function

I need a check on the following exercise:

Consider the following function $$f(x,y)= \begin{cases} \frac{x^3 y}{x^2 + y^4} \quad (x,y) \ne (0,0) \\ 0\quad \quad (x,y)=(0,0) \end{cases}$$

determine where the function is differentiable.

I checked the function is continuous over the whole $$\mathbb{R}^2$$ and that the derivatives $$\partial_x f(0,0)=\partial_y f(0,0)=0$$.

• Differentiability: first of all, for any point $$(x,y)\ne (0,0)$$ the partial derivatives are continuous at $$(x,y)$$ and they exist in a neigbourhood of $$(x,y)$$. The tricky part is for $$(x,y)=(0,0)$$. There I know only the value of the derivatives at $$(0,0)$$ but I have no other information, therefore I have to study the following limit:

$$\lim_{(h,k) \rightarrow (0,0)} \frac{f(h,k)}{\sqrt{h^2+k^2}}$$

I note that $$|\frac{h^3 k}{(h^2 + k^4) \sqrt{h^2+k^2}}| = |\frac{h^2 \cdot h k}{(h^2 + k^4) \sqrt{h^2+k^2}}|$$ and since $$|\frac{h^2}{h^2 + k^4}|<1$$ I have $$|\frac{h^2 \cdot h k}{(h^2 + k^4) \sqrt{h^2+k^2}}|<|\frac{hk}{\sqrt{h^2+k^2}}| < |\frac{hk}{|h|}| =_{\text{h} \ne 0} |k|$$ (if $$h=0$$ then the limit is identically $$0$$)

Now, as $$(h,k) \rightarrow (0,0)$$ the r.h.s goes to $$0$$ and hence the limit is $$0$$ and the function is differentiable.

Therefore the function is differentiable on the whole $$\mathbb{R}^2$$, right?

is everything correct

• everything is correct, but why don't you write an $\varepsilon,\delta$ proof? – LinAlg Nov 3 '20 at 12:51
• @LinAlg I wasn't able. Could you show it to me, if it's not a problem? – Vefhug Nov 3 '20 at 18:31

Here is a proof that $$\lim_{(x,y) \rightarrow (0,0)} \frac{f(x,y) - f(0,0)}{\sqrt{x^2+y^2}}=0.$$
Let $$\varepsilon>0$$. Take $$\delta=\varepsilon$$, let $$(x,y) \in \mathbb{R}^2\backslash \{(0,0)\}$$ such that $$||(x,y) - (0,0)|| \leq \delta$$.
\begin{align} \left|\frac{f(x,y) - f(0,0)}{\sqrt{x^2+y^2}} - 0\right| &= \left|\frac{x^3y}{(x^2 + y^4) \sqrt{x^2+y^2}}\right| \\ &= \left| \frac{x^2}{x^2 + y^4} \right| \left| \frac{x}{\sqrt{x^2+y^2}} \right| \left| y \right| \\ &\leq \left| y \right| \leq \delta = \varepsilon. \end{align}
So, $$f$$ is differentiable at $$(0,0)$$.