I'm having a hard time trying to understand a theorem of multivariable calculus.


  • Statement A = "The partial derivatives of $f$ are continuous in an open set containing (a,b)"
  • Statement B = "$f$ is differentiable at (a,b)"

I was wondering why the theorem: A$\implies$B was only true in that way. Why is A$\iff$B false ?

How can a function be differentiable without continuous partial derivatives ? Do you have an example of such a function ?

Thanks a lot for your help and happy new year ;)

Diego, student from Belgium


An example for such a function is: $f(x,y)=\begin{cases} (x^2+y^2)sin(\frac{1}{\sqrt{x^2+y^2}}) & if\;(x,y)\neq (0,0)\\ 0 & if\;(x,y)=(0,0). \end{cases}$

It is differentiable in $(0,0)$, but the partial derivatives are not continous in $(0,0)$.

More explicitely: The partial derivative $\frac{\partial f}{\partial x}$ is: (analogous for $y$)

$\frac{\partial f}{\partial x}(0,0)=\lim\limits_{h \rightarrow 0}{\frac{f(h,0)-f(0,0)}{h}}=\lim\limits_{h \rightarrow 0}{hsin(\frac{1}{|h|})}=0$.

$\frac{\partial f}{\partial x}(x,y)=2xsin(\frac{1}{\sqrt{x^2+y^2}})-\frac{xcos(\frac{1}{\sqrt{x^2+y^2}})}{\sqrt{x^2+y^2}}$.

If you consider this along the x-axis (i. e. y=0), you get $\frac{\partial f}{\partial x}(x,0)=2xsin(\frac{1}{|x|})-sgn(x)cos(\frac{1}{|x|})$, which is oscillating around $(0,0)$ (hence we don't have $\lim\limits_{x \rightarrow 0}\frac{\partial f}{\partial x}(x,0)=0)$, hence the partial derivative $\frac{\partial f}{\partial x}(x,y)$ is not continous.

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  • 1
    $\begingroup$ +1. $f(x)=x^2\sin(\frac{1}{x}),x\neq 0,f(0)=0$ works in one dimension: $\lim\sup_{x\rightarrow 0}f'(x)=1\neq \lim\inf_{x\rightarrow 0}f'(x)=-1$ $\endgroup$ – Peter Melech Jan 7 '19 at 13:13
  • $\begingroup$ You're right, but since Diego asked for multivariable and spoke of $(a,b)$, I thought he'd need an example in two dimensions. ^^ $\endgroup$ – Student7 Jan 7 '19 at 13:15
  • $\begingroup$ Of course! Your answer is excellent, my comment was just supplementary! $\endgroup$ – Peter Melech Jan 7 '19 at 13:18
  • $\begingroup$ Thank you for your answer ;) Have a very nice day ! $\endgroup$ – Diego H Jan 7 '19 at 13:36
  • $\begingroup$ Thank you, i wish you happy new year and a nice day, too. ^^ $\endgroup$ – Student7 Jan 7 '19 at 13:41

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