Proof: If a function $f: (a,b) \rightarrow \mathbb{R}$ is differentiable in $x_{0} \in (a,b)$ and has a local maximum in $x_{0}$, we have that $f'(x_{0})=0$
I'll be honest and say I can't stand tasks like that at all, especially if they are in an exam like this one (taken from an old exam).
I'd like to know the easiest way of proofing this, I don't mind if it's written in maths language or in human language, just so I can see how it's done. I have no problem calculating extremums but explaining the theory was never easy for me. I even got a solution in front of me but it's very very complicated (done by $\varepsilon$-$\delta-proof).$
Edit: If someone wants I can share the solution of the professor but comparing it with your answer it's too much time consuming and complicated. Let me know if you want see it anyway and thank you very much for helping me! :-)
$x_0$ is a local maximum, implies that there exists $c>0$ such that for every $x\in [x_0-c,x_0+c]$, $f(x)\leq f(x_0)$. This implies that $lim_{x\rightarrow x_0, x\leq x_0}{{f(x)-f(x_0)}\over{x-x_0}}\geq 0$ and $lim_{x\rightarrow x_0, x\geq x_0}{{f(x)-f(x_0)}\over{x-x_0}}\leq 0$ since these limits is $f'(x_0)$, you deduce that $f'(x_0)=0$.