0
$\begingroup$

For the following function

$$ f(x,y) = \begin{cases} \frac{xy}{x^2+y^2} & \text{if $(x,y)\neq (0,0)$} \\ 0 & \text{if $(x,y)=(0,0)$} \\ \end{cases} $$

I know that the partials $f_x$ and $f_y$ both exist at the point $(0,0)$, namely $f_x(0,0)=0$ and $f_y(0,0)=0$. I also know that this function $f(x,y)$ is not continuous at the origin and hence it is also not differentiable at the origin.

Thus, I want to conclude that the partials must not be continuous at the origin $(0,0)$, via the the contrapositive of the differentiability theorem, which states that if all the partial derivatives of a function both exist and are continuous at a point, then that function is differentiable at that point.

However, the source for this problem says that this function $f(x,y)$ is an example of a function whose partials both exist and are continuous at $(0,0)$, but where the function is also not differentiable at $(0,0)$.

So I am confused as to whether this source contains an error or my logic surrounding the differentiability theorem is erroneous.

In review, my question is basically if the partials $f_x$ and $f_y$ for the above given $f(x,y)$ are actually continuous at the origin $(0,0)$ or whether they are discontinuous at the origin. Thanks in advance.

If it helps I will attach an image of the source of this problem:

This is the source of my confusion

$\endgroup$
1
  • $\begingroup$ I think they wanted to say : an example of function that IS NOT continuous at $0$, but both partial derivative exist at $0$. In particular, it's not differentiable at $0$ (since not continuous) $\endgroup$
    – Surb
    Aug 21 '20 at 7:34
1
$\begingroup$

The source contains an error. The partial derivative w.r.t $x$ is $\frac {y^{3}-x^{2}y} {(x^{2}+y^{2})^{2}}$ which does not even have a limit along the $y-$ axis.

$\endgroup$
1
  • $\begingroup$ That is what I was thinking, but i just wanted to make sure the error was not on my end so thanks for clarifying. $\endgroup$
    – r.a.h.
    Aug 21 '20 at 7:39

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.