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

  • 1
    $\begingroup$ everything is correct, but why don't you write an $\varepsilon,\delta$ proof? $\endgroup$ – LinAlg Nov 3 '20 at 12:51
  • $\begingroup$ @LinAlg I wasn't able. Could you show it to me, if it's not a problem? $\endgroup$ – 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 $(0,0)$.

  • 1
    $\begingroup$ Thanks for your time @LinAlg $\endgroup$ – Vefhug Nov 10 '20 at 11:13

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