In my old book in calculus, it says that a sufficient condition for the function $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ to be differentiable at an internal point z, is that the partial derivatives exist at z and are continuous there.

Is this also a necessairy condition? Or is there a function defined in a domain areound z, where the function is differentiable at z, but there is no domain around z where the partial derivatives are continuous?

  • $\begingroup$ It's not necessary. Consider $$f(x,y) = \begin{cases} 0 &, (x,y)\in \mathbb{Q}^2\\ x^2+y^2 &,\text{ otherwise}.\end{cases}$$ It's not partially differentiable everywhere (and not continuous anywhere except $0$). $\endgroup$ – Daniel Fischer Oct 9 '14 at 14:09
  • $\begingroup$ @DanielFischer thanks $\endgroup$ – user119615 Oct 9 '14 at 14:15

It is not a necessary condition, even for functions of one variable, as the following example shows: $$ f(x)=\begin{cases}x^2\,\sin\frac1x &x \ne0,\\ 0 & x=0. \end{cases} $$ $f$ is differentiable on $(-\infty,\infty)$, but $f'$ is discontinuous at $x=0$.


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