Without trigonometric functions, I want to find $f$ which is differentiable at $x_0$ but $f'$ is not continuous at $x_0$ 
Without using the fuctions $\cos$ and $\sin$, I want to find a function $f$ which is differentiable at a point $x_0$ but $f'$ is not continuous at $x_0$.

I used this function which is differentiable at $0$ but $f'$ is not continuous at $0$:
$$f(x) = \left\{\begin{array}{ll}
x^2\sin(\frac{1}{x})& \text{si } x\neq 0\\
\\
0 & \text{si } x=0.
\end{array}\right.$$
But I like to construct a function without $\cos$ and $\sin$.
Thanks a lot.
 A: (Too long for a comment)
It may be not the most exciting example but my first thought was to build $\sin$-ish function. Namely let $\phi\in C^1(\mathbb{R})$ such that $\|\phi\|_{C(\mathbb{R})}<\infty$, $\phi(x)\to 0$ as $x\to 0$, $\phi(x)/x\to 1$ as $x\to 0$ and $\phi'$ has no limit at infinity (so let $\phi$ oscillate). Then $f$ defined as $$f(x)=x^2\phi(1/x) \quad \& \quad f(0)=0$$
has the property that we are interested in. Simply $$\frac{h^2\phi(1/h)}{h}\to 0$$
as $h\to 0$ (so $f'$ exist at zero). But since $f'(x)=2x\phi(1/x)-\phi'(1/x)$ except zero and $f'(0)=0$ the function $f'$ is not continous at $0$ which results from the calculations $$\lim_{h\to 0}2h\phi(1/h)-\phi'(1/h) =-\lim_{h\to 0}\phi'(1/h) =\text{don't exist}.$$
Well maybe some technical stuff is missing but this boring example shows more or less whats is the problem. Maybe some assumptions are not necessary, maybe more exciting examples are there but if you find sufficient $\phi$ you may build your own example.
A: $|x|$ is continuous at 0, but it's derivative is not.
