this is probably a very common question, but i couldn't find the answer on my books.
is every darboux function the derivative of a function? even the nowhere continuous ones?



No. Unfortunately, there is no such simple criterion for a derivative. At least none is known. Ref: Bruckner, Andrew, Differentiation of real functions. Second edition. CRM Monograph Series, 5. American Mathematical Society, Providence, RI, 1994. xii+195 pp. ISBN: 0821869906, MR1274044 (94m:26001) 


Let $\phi(x)=x\chi_{[1,1]}(x)+\text{sgn}(x)\chi_{[1,1]^c}(x)$, $f$ be the Conway base 13 function, then $\phi\circ f$ is a Darboux function and is nowhere continuous. Moreover, $\phi\circ f$ is Borel measurable and bounded (in particular, it is $L^1$). But $\phi\circ f$ is not a derivative because any derivative is a limit of continuous functions, hence its discontinuity points form a set of first category (in particular cannot be the whole interval). See this post. In sum, $\phi\circ f$ is a bounded Borel measurable Darboux function but is not a derivative. 

