Derivation of a function including singularity point. If\begin{equation}-1<q<0,x(t)= \int_0^t (t-s)^q ds.\end{equation}
(I think)Then \begin{equation}x'(t)=0^q+q\int_0^t (t-s)^{q-1} ds.\end{equation}
Is my answer right?

If you do the substitution $t-s=u$, then $s=t-u$ and $ds=-du$; thus your function becomes
$$
x(t)=-\int_t^0 u^q\,du=\int_0^t u^q\,du
$$
Can you go on from here?
