No, and it's easy to construct one. Namely, how about $\displaystyle \displaystyle f:[0,1]\to\mathbb{R}$ given by contiuously extending $\displaystyle g: (0,1]\to\mathbb{R}:x\mapsto x^2\sin\left(\frac{1}{x}\right)$. You can easily prove, it's a classic exercise, that $\displaystyle f$ is differentiable on $\displaystyle [0,1]$ but that $\displaystyle f'$ is discontinuous at $\displaystyle 0$. The problem with constructing such a function, the reason why all the obvious examples don't work, is that if $\displaystyle j$ is a function which is the derivative of another function (i.e. $\displaystyle j=k'$ for some $\displaystyle k:[a,b]\to\mathbb{R}$) then $\displaystyle j$ preserves connectedness. Or, said in a less highfalutin tongue, $\displaystyle j$ will have the intermediate value property, that if $\displaystyle x\in(j(c),j(d))$ for some $\displaystyle c,d\in[a,b]$ then $\displaystyle x=j(e)$ for some $\displaystyle e\in[a,b]$. This is roughly what

Darboux's theorem says.