Answer
$$ p^{\prime} =\frac{q^{3} \cos q-q^{2} \sin q-q \cos q-\sin q}{\left(q^{2}-1\right)^{2}} $$
Work Step by Step
Given $$ p=\frac{q \sin q}{q^{2}-1}$$
So, we have
\begin{aligned} p^{\prime}&=\frac{dp}{d q}=\frac{(q \sin q)^{\prime} \cdot\left(q^{2}-1\right)-q \sin q \cdot\left(q^{2}-1\right)^{\prime}}{\left(q^{2}-1\right)^{2}}\\
&=\frac{\left(q^{\prime} \cdot \sin q+q \cdot \sin ^{\prime} q\right) \cdot\left(q^{2}-1\right)-q \sin q \cdot 2 q}{\left(q^{2}-1\right)^{2}}\\
&=\frac{(\sin q+q \cos q) \cdot\left(q^{2}-1\right)-2 q^{2} \sin q}{\left(q^{2}-1\right)^{2}}\\
&=\frac{q^{2} \sin q-\sin q+q^{3} \cos q-q \cos q-2 q^{2} \sin q}{\left(q^{2}-1\right)^{2}}\\
&=\frac{q^{3} \cos q-q^{2} \sin q-q \cos q-\sin q}{\left(q^{2}-1\right)^{2}} \end{aligned}
