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My textbook has this example. It states that the existence of all directional derivatives at a point need not imply differentiability. However, in this previous question I was looking for some clarification-the theorem states that if the partials exist and $f$ is continuous then it is differentiable. Now wouldn't the existence of the directional derivatives imply that the partial derivatives exist? Hence if we have the directional derivatives and $f$ is continuous can we assume that the derivatives exist?

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  • $\begingroup$ I think there is a misstatement in your question. You say that if the partials exist and $f$ is continuous at X, then $f$ is differentiable at $X$. I'm pretty sure you need continuity of the partials, but I don't have a counterexample. $\endgroup$ – saulspatz Feb 11 '18 at 20:09
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You are right on the fact that if all directional derivatives exist than the partial derivatives exist also. But to guaranty the derivability of the function we need also continuity, and the examples in your book are just of functions that are not continuous at $(0,0)$.

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  • $\begingroup$ Thanks. So if we were to have continuity then the existence of directional derivative implies differentiability. $\endgroup$ – conums Feb 11 '18 at 20:07
  • $\begingroup$ Yes, because the existence of directional derivatives ( for all directions) implies the existence of partial derivatives that, essentially, are directional derivative in the directions of the coordinate axis. $\endgroup$ – Emilio Novati Feb 11 '18 at 20:29

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