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I'm studying the function:

$$ f(x,y) = \begin{cases} x+y & \text{if } x=0 \text{ or } y=0 \\ 1& \text{if } xy\neq 0 \end{cases}$$

The partial derivatives exist in $(0,0)$, and I can prove that directional derivative doesn't exist in $(0,0)$. Now, Is this function differentiable? Why?

Thanks for your help.

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Yes, if you study the directional derivative with the vector $(a,b)$, such that $a\neq 0$, and $b\neq 0$, do'nt exists. And you are right, but another form to justify? Thanks –  Hiperion Sep 12 '11 at 0:09

1 Answer 1

up vote 3 down vote accepted

No, the function is not differentiable. Here are two proofs of this:

(1) If a function is differentiable at a point, then it is continuous there. Your function is not continuous at $(0,0)$, and therefore not differentiable there.

(2) If a function is differentiable at a point, then all directional derivatives should exist. Your function does not have any directional derivatives in directions other than $\mathbf{e_1} = (1,0)$ and $\mathbf{e_2} = (0,1)$, so the function must not be differentiable.

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Yes, you are right. Thank you very much! –  Hiperion Sep 12 '11 at 0:13

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