I know the proof of the statement "If partial derivatives exist in a n-ball $B(\vec{a})$ and are continuous at $\vec{a}$, then the function is differentiable at $\vec{a}$ " . But I was wondering whether the converse holds.

I was thinking along these lines that, differentiability at a point implies the existence of tangent plane, and that $\vec{a}$ lies on that plane. This would mean that the partial derivatives do exist, but I'm not able to visualize or proceed any further.

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    $\begingroup$ Hint: Do you know any examples of differentiable functions (of one variable) with discontinuous derivative? $\endgroup$ May 25, 2021 at 13:39
  • $\begingroup$ @moishe Yes I do $f(x) = \begin{cases} x^2 \sin(1/x) &\mbox{if } x \neq 0 \\ 0 & \mbox{if } x=0. \end{cases}$, but I cannot visualize the whole thing in higher dimensions, let's say for 3D as to how it involves the tangent plane. $\endgroup$ May 25, 2021 at 13:41
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    $\begingroup$ Then you have answered your own question: The converse to that theorem does not hold. No need for a visualization. $\endgroup$ May 25, 2021 at 13:43

1 Answer 1


I don't have repututation for comment so I will post as answer. The converse of your claim is not true even in dimension 1: differentiability in one point does not imply diffenentiability in any neighborhood of that point, see for example:

\begin{cases}x^2 \, \,\,\text{ if} &x \in \mathbb{Q} \\ 0 &\text{otherwise} \end{cases} It is differentiable at the origin and not continuous in any other point. The same example works also in higher dimension.


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