derivate of a piecewise function $f(x)$ at$ x=0$. There is a piecewise function $f(x)$
$$f(x)= \begin{cases} 1 ,\ \ \text{if}\ \  x \geq \ 0 \\ 0,\ \  \text{if}\ \  x<0 \end{cases}$$ 
what is the derivative of the $f(x)$ at $x=0$?
Is it $0$? Or since it is not continuous, the derivative does not exists? 

Hint: Prove that if a given function $f$ is differentiable at a point $a$, then $f$ is continuous at $a$.
Sketch: If $f$ is differentiable at $a$ then we may write (Theorem) $f(a+h) = f(a) + hf'(a) + \frac{r(h)}{h}h$ where $lim_{h \to 0 } \frac{r(h)}{h} = 0 $, then $lim_{h\to 0} f(a+h) = f(a)$, that is, $f$ is continuous at $a$. 
Show that your function is not continuous at $0$. Then use the contrapositive $p \Rightarrow q \equiv\  \sim q \Rightarrow\ \sim p$. 
