Let $f:\mathbb R^2\to \mathbb R$ be defined by $$f(x,y)=\begin{cases}\frac{x|y|}{\sqrt{x^2+y^2}},& (x,y)\ne(0,0)\\ 0,& (x,y)=(0,0).\end{cases}$$
- For which non-zero vectors $u$ does the directional derivative exist at the point $(0,0)$?
- Do the partial derivatives exist?
- Is $f$ differentiable at $0$?
- Is $f$ continuous at $0$?
According to my calculations, the directional derivative is zero for all non-zero vectors $u$ at $(0,0)$ which implies the partial derivatives are zero at $(0,0)$. I am not sure how to proceed with 3. and 4. I am not allowed to use the existence of partial derivatives and their continuity to check for differentiability since the author introduces this theorem in the next section.
Thanks.
Answer
$f$ is continuous at $(0,0)$ as you can prove the inequality $$ \vert f(x,y) \vert \le \frac{\vert xy \vert}{\sqrt{x^2+y^2}} \le \frac{\sqrt{x^2+y^2}}{2}$$
$f$ is not differentiable at $(0,0)$. If it would be the case its derivative would be equal to the zero matrix as the partial derivatives are equal to $0$. Hence all the directional derivatives would vanish which is not the case. A contradiction proving that $f$ is not differentiable at the origin.
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