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For a two variable function, does the existence of continuous partial derivatives of order 1 with respect to $x$ and $y$ at a point $(x,y)$ imply the existence of the directional derivative in any direction at the point $(x,y)$?

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  • $\begingroup$ Hint: Try defining the derivative with respect to an arbitrary direction, then separate it so it can be derived from the partial derivatives. $\endgroup$ – Mefitico May 2 at 17:42
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Yes, because:

  1. The continuity of the partial derivatives implies that the function is differentiable at $(x,y)$. (This is a standard multi-variable calculus theorem.)
  2. When a function is differentiable, it has directional derivatives in any direction.
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  • $\begingroup$ Is it because the increment theorem is valid if the partial derivatives are continuous? $\endgroup$ – Mathematicaluniverse May 2 at 18:46
  • $\begingroup$ No. It's a conseuence of the mean value theorem. $\endgroup$ – José Carlos Santos May 2 at 18:55
  • $\begingroup$ Could u explain how? $\endgroup$ – Mathematicaluniverse May 2 at 19:28
  • $\begingroup$ It's too long for a comment, but you will find a proof here. $\endgroup$ – José Carlos Santos May 2 at 19:33
  • $\begingroup$ Thank you very much for the answer , the increment theorem is basically the same fact and it's proof is exactly the same proof u told me to check out . It basically states that change in f = fy*change in y + fx * change in x + epsilon1*change in x + epsilon 2 * change in y where epsilon 1 and epsilon 2 tend to 0 as change in x and change in y tend to 0 $\endgroup$ – Mathematicaluniverse May 2 at 19:52

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