Is the converse of the statement "If partial derivatives exist and are continuous, then the function is differentiable at that point" true?

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.

• Hint: Do you know any examples of differentiable functions (of one variable) with discontinuous derivative? May 25, 2021 at 13:39
• @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. May 25, 2021 at 13:41
• Then you have answered your own question: The converse to that theorem does not hold. No need for a visualization. May 25, 2021 at 13:43

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