I know that given a certain function $f(x)$, there are some values of $x$ that could be in its domain but not in its derivative's. However, I believe that all the values that belong to the domain of $f'(x)$, have to belong to the domain of the original function as well.
Is this true? If yes, how can it be proved?
It depends on how you define the domain of $f'(x)$. For me the definition of the domain of $f'(x)$ is $$\mbox{dom}(f')=\left\{x\in\mbox{dom}(f): \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\ \ \ \mbox{exists and it is finite}\right\}$$ so $\mbox{dom}(f')\subseteq\mbox{dom}(f)$ by definition.
Note that $x$ must be an element of $\mbox{dom}(f)$ because you have to evaluate $f$ in $x$ to build the limit that define the derivative. (Hope it's clear, I don't speak english so well.)
An example. Consider $f(x)=\ln(x)$. Its domain is $\mbox{dom}(f)=(0, +\infty)$ and its derivative is $f'(x)=\frac{1}{x}$.
The "natural domain" of $y=\frac{1}{x}$ is $\mathbb{R}\setminus\{0\}$ while $\mbox{dom}(f')=(0, +\infty)$.
Please read this question : Differentiable only on interior points
In fact in addition to Ixion's answer, $f'$ could be defined in $x_0$ if $x_0\in\operatorname{dom}(f)\cap\overline{\operatorname {dom}(f)\setminus\{x_0\}}$ because limits can only make sense in limit points.
How would you define $f'(-1)$ in this case ? $\begin{cases} f(-1)=5 \\ f(x)=x & x\ge 0\end{cases}$
But the initial inclusion $\operatorname{dom}(f')\subset\operatorname{dom}(f)$ holds, yes.
