$f(x,y) = (1- \cos(\frac{x^2}{y})) \sqrt{x^2+y^2}$

Question

Let $f(x,y) = (1- \cos(\frac{x^2}{y})) \sqrt{x^2+y^2}$ for $y \ne 0$

How can I prove that f is not differentiable in $(0,0)$. Please some help.

Answer 1

The function is not defined at $(0,0)$, but we can examine the directional derivatives if we define an extension. Consider the extension with $f(0,0)=0$, since the limit of the function is $0$ as $(x,y)\rightarrow (0,0)$ along any path with $y\neq0$.

The directional derivative for an arbitrary direction vector $(a,b)$ is

$$\lim_{h \rightarrow 0}\,\frac{f(ha,hb)-f(0,0)}{h} = \lim_{h \rightarrow 0}\frac1{h}\left[1-\cos\left(\frac{h^2a^2}{hb}\right)\right]\sqrt{h^2a^2+h^2b^2} \\=\lim_{h \rightarrow 0}\left[1-\cos\left(h\frac{a^2}{b}\right)\right]\sqrt{a^2+b^2}\\\ = \begin{cases} 0 \,\,\,\,\,\,\,\,\text{if}\,\,b\neq0 \\ \text{undefined}\,\,\text{if}\,\,b=0 \end{cases}. $$

So the directional derivative at $(0,0)$ is $0$ in any direction that is not parallel to the x-axis. Otherwise it is undefined.