0
$\begingroup$

$f\begin{pmatrix}x\\y\end{pmatrix}:=\begin{cases}\frac{x^2y}{x^6+2y^2}& (x,y)^T\neq(0,0)^T\\0&(x,y)^T=(0,0)^T\end{cases}$

I want to find out if $f$ is continuously partially differentiable at $(0,0)^T$.

The partial derivates at $(0,0)^T$ are both equal to $0$. So if $f$ were differentiable at $(0,0)^T$ we would have

$$\lim_{x,y\to 0}\frac{f(x,y)}{\sqrt{x^2+y^2}}=0$$ Using polar coordinates I end up with

$$\lim_{r\to 0}\frac{r^3 \cos^2 \phi \sin \phi}{(r^6 \cos^6 \phi +2 r^2 \sin^2\phi)r}$$

Which does't converge to $0$. So this means $f$ is not differentiable at $(0,0)^T$ and therefore not continuously partially differentiable at $(0,0)^T$.

Is that correct?

$\endgroup$
2
$\begingroup$

When you wrote$$\lim_{x,y\to 0}\frac{f(x,y)}{\sqrt{x^2+y^2}},$$you meant$$\lim_{x,y\to 0}\frac{f(x,y)}{\sqrt{x^2+y^2}}=0.$$Otherwise, it is correct.

$\endgroup$
  • $\begingroup$ Thanks! When checking differentiability at $(0,0)$ can I simply do it like I did it or do I have to first check if the function is continuous at $(0,0)$? $\endgroup$ – conrad Nov 10 '18 at 19:30
  • $\begingroup$ No, you don't have to check continuity. $\endgroup$ – José Carlos Santos Nov 10 '18 at 19:32
  • $\begingroup$ @user610431 In fact, if you know a function is not continous at a point, you automatically also know that it's not differentiable at that point $\endgroup$ – Ovi Nov 10 '18 at 20:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.