Is there a function that is differentiable/continuous everywhere but its derivative is not?
The function
$\displaystyle f(x)=\begin{cases}0\quad x<0\\
x^{2}\quad x\ge 0\end{cases}$
satisfies the following conditions: $\displaystyle f(x)$ is differentiable (and therefore continuous) everywhere. $\displaystyle f'(x)$ is continuous but not differentiable at $\displaystyle 0.$
Does that satisfy your requirements?
no sorry i don't think so, i am look for a function s.t f(x) is differentiable everywhere but it's derivative f'(x) is not continuous everywhere, i think in that example f'(x) is continuous everywhere (this question came up in a tutorial today and we came to no conclusion)
You didn't make it clear whether you needed the derivative to be just non-differentiable, or actually discontinuous (the latter being a stronger condition). I think if you have a discontinuous derivative at a point, then the anti-differentiated function (the original function you differentiated) is going to have to have some sort of corner or cusp. That would mean that the original function is not differentiable at that point.
let $\displaystyle f(x)=\begin{cases} -1\text{ if } x < 0 \\ 1 \text{ if } x \ge 0\end{cases}$
Now define the fucntion
$\displaystyle \displaystyle g(x)=\int_{-5}^{x}f(t)\, dt$
$\displaystyle g(x)$ is everywhere continuous and differentiable but it has derivative $\displaystyle f(x)$ is not
This should do the trick
The "standard example", found in many texts, is $\displaystyle f(x)= x^2 sin(1/x)$ if x is not 0, f(0)= 0. That is differentiable for all x but its derivative is not continuous at x= 0.
Note: while a derivative is not necessarily continuous, it does satisfy the "intermediate value property" so such an example is not going to be simple.