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I have some trouble with the following exercise. It is asking to examine/check the various notions of differentiability for a function $ \mathbb R^2 \to \mathbb R.$ I am a little be confused. So, any help/hint would be very helpful.

Exercise: For any point $ (x,y) \in \mathbb R^2 $ examine whether the function

$$ f(x,y)= \left\{ \begin{array}{ll} xy & x \geq 0, \quad y \geq 0 \\ 0, & \textrm{otherwise} \end{array} \right. $$

is continuous, partial differentiable, differentiable, and continuous differentiable. Moreover, wherever they exist, give the partial derivatives, the derivative and the directional derivative for every direction.

Here are some thoughts.

We add the cases $ x=0 $ or $y=0$ to the second formula, so $f(x,y) =xy $ when $ x, y$ are both strictly positive.

Now

$$ f(x,y)= \left\{ \begin{array}{ll} xy & x > 0, \quad y >0 \\ 0, & \textrm{otherwise} \end{array} . \right. $$

When

$$ (x,y) \in (0,\infty) \times (-\infty,0) \cup (-\infty,0) \times (0,\infty) \cup (-\infty,0) \times (-\infty,0) $$

we obtain that $ f \equiv 0.$ Since this above set is a union of open sets we conclude that $f$ is continuously differentiable on $ \mathbb R^2 \setminus \{ ( (0,\infty) \times (0,\infty) ) \}.$

Question: How could I work the rest of the problem?

I would really appreciate any hint/idea, that could help me to solve it. I have done similat exercises, I am now stuck because of the shape of $ f.$

Thank you in advance.

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    Have you tried taking the partials of $f$ (from the definition, as limits of difference quotients) along the non-negative axes? (It's clear that $f$ is smooth -- in fact real-analytic -- everywhere else.) It may help to sketch $g(x) = f(x, c)$ for a generic $c > 0$. – 2017-02-04
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    @AndrewD.Hwang: Thank you very much for your comment. You are right. I got stuck without any reason. If I understand it correctly, we should obtain that the partial derivatives do not exist for $x,y > 0.$ Am I right ? For the directional derivative should I got by def. and considering the cases $ x,y > 0$ and $x=y=0, $ and other one ? Can I conclude something for the directional derivative using what I know about the partial derivatives ? Thank you in advance for any comment. – 2017-02-04
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    Specifically, at each point $(x, 0)$ with $x > 0$, focus on the partial with respect to $y$, and at each point $(0, y)$ with $y > 0$, focus on the partial with respect to $x$. As I read it, if at some point at least one partial does not exist, you're not asked about directional derivatives. The origin is going to require separate consideration. <> I expect part of the goal is to get practice organizing a complex argument into systematic, easy-to-follow cases. – 2017-02-04

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