# Continous partial derivatives $\implies$ differentiable. What about the reciprocal?

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

Let:

• 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.

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