Derivative of a constant is not zero

From my book

From The First Mean Value Theorem we deduce the means value theorem for derivatives (the law of the mean): $$ f(c)=\frac{1}{b-a} \int_{a}^{b} f(x) d x $$ Let us differentiate the two members of the equation : $f^{\prime}(c)=\frac{1}{b-a} \int_{a}^{b} f^{\prime}(x) d x$ $$ \Rightarrow f^{\prime}(c)=\frac{1}{b-a}[f(x)]_{a}^{b}=\frac{f(b)-f(a)}{b-a} $$

How can we differentiate constants (definite integral is a constant and $f(c)$ is a constant)?! Isn't that 0=0?

The equation you are starting out with is a property of continuous functions. For a continuous funtion $f$, choose some points $a < b$ such that $[a,b]$ is a subset of the domain of definition, and then it is a true statement that there exists a number $c\in [a,b]$ such that $$ f(c) = \frac{1}{b-a}\int_a^b f(t) dt$$ The number $c$ will usually depend in an unknown way on $a, b, $ and $f$ and need not be unique.

As you are correctly noted, the elements of this equation are considered constant, there is nothing which you may want to differentiate here (as you would get $0$ on each side).

What Hans Lundmark is pointing out in his comment is that you may apply this same statement to $f^\prime$, if that is a continuous function. You get exactly the same statement with $f^\prime$ instead of $f$ and a potentially different value of $c$ (which now depends on $a, b$ and $f^\prime$). In that case you may, in fact, apply the fundamental theorem of calculus which then yields, for a continuously differentiable function $f$. $$\exists\, c\in[a,b]:\quad f^\prime(c) = \frac{f(b)-f(c)}{b-a}$$ But this is a rather poor version of that theorem, since it assumes that $f^\prime$ is continuous. One can do better rather easily, see this wikipedia entry.