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.