function has partial derivatives but is not differentiable Can you write me an example of function which has partial derivatives but is not differentiable?
How could I create and prove the function like that?
 A: Zero on the coordinate axes. One on the compementary of the coordinate axes. Isn't even continuous in (0,0).
A: $f(x,y)=\sqrt[3]{x^3+y^3}$ is not differentiabile in (0,0), but their partial derivates exists on that point.
$$f_x(0,0)=\lim_{\Delta x\to0}\frac{f(\Delta x,0)-f(0,0)}{\Delta x}=1; \\ f_y(0,0)=\lim_{\Delta y\to0}\frac{f(0,\Delta y)-f(0,0)}{\Delta y}=1.$$
But $\Delta f(0,0)$ cannot be expressed as $f_x (0,0)\Delta x+f_y (0,0)\Delta y+o(\sqrt{\Delta x^2+\Delta y^2})$ (this is the definition of differentiability of the function at a point).
Let us prove this:
$$\Delta f(0,0)=f(\Delta x,\Delta y)-f(0,0)=\sqrt[3]{\Delta x^3+\Delta y^3}= \\ =f_x (0,0)\Delta x+f_y(0,0)\Delta y+\varepsilon(\Delta x, \Delta y)\cdot\sqrt{\Delta x^2+\Delta y^2}$$
where $$\varepsilon(\Delta x,\Delta y)=\frac{\sqrt[3]{\Delta x^3+\Delta y^3}-\Delta x-\Delta y}{\sqrt{\Delta x^2+\Delta y^2}}.$$
Since for $\Delta x= \Delta y=\frac{1}{n},$ $\lim_{(\Delta x,\Delta y)\to (0,0))}\varepsilon(\Delta x,\Delta y)$ is equivalent with $\lim_{n\to\infty}\varepsilon(\frac{1}{n},\frac{1}{n})=\frac{\sqrt[3]{2}-2}{\sqrt{2}}\neq 0,$ we conclude that the function is not differentiable at $(0,0).$
A: Hold a rod parallel to the floor and turn slowly around.  As you turn, raise the rod (keeping it parallel to the floor) such that the rod achieves its maximum height at $45^{\circ}$, then lower it until its minimum height is achieved at $135^{\circ}$, raise it again until you reach $225^{\circ}$, lower it to $315^{\circ}$ and lower it again to $360^{\circ}$.
