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?

Unfortunately, no.
$$
x(t)=\int_0^t(t-s)^q\,\mathrm{d}s
$$
Then
$$
x'(t)=\overbrace{\vphantom{\int_0^t}0^q}^{1/0}+q\,\,\overbrace{\int_0^t(t-s)^{q-1}\,\mathrm{d}s}^{\text{diverges}}
$$
makes no sense. It's better to compute
$$
\begin{align}
x(t)
&=\left.\frac{-1}{q+1}(t-s)^{q+1}\right]_0^t\\
&=\frac{t^{q+1}}{q+1}
\end{align}
$$
and differentiate.
Or as egreg suggests, change variables before differentiating the integral:
$$
\begin{align}
x(t)
&=\int_0^t(t-s)^q\,\mathrm{d}s\\
&=\int_0^ts^q\,\mathrm{d}s\\
\end{align}
$$
