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
    Commented Aug 21, 2020 at 7:34

1 Answer 1

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.
    Commented Aug 21, 2020 at 7:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .