Answer
$\displaystyle \frac{dp}{dq}=\frac{10p^{2}q+p}{2p-q-10pq^{2}}$
Work Step by Step
We treat p as p(q)
$\displaystyle \frac{d}{dq}[p^{2}]$= .. chain rule...$=2p\displaystyle \frac{dp}{dq}$
$\displaystyle \frac{d}{dq}[pq]$= ... product...$=\displaystyle \frac{dp}{dq}\cdot q+p(1)$
$\displaystyle \frac{d}{dq}[5(pq)^{2}]$= ... chain rule...$=5\displaystyle \cdot 2(pq)\cdot\frac{d}{dq}[pq]=$
$=5\displaystyle \cdot 2(pq)\cdot(\frac{dp}{dq}\cdot q+p)$
So, after differentiating both sides,
we solve for $\displaystyle \frac{dp}{dq}$:
$2p\displaystyle \frac{dp}{dq}-(\frac{dp}{dq}\cdot q+p)=10(pq)\cdot(\frac{dp}{dq}\cdot q+p)$
$2p\displaystyle \frac{dp}{dq}-\frac{dp}{dq}\cdot q-p=10pq^{2}\cdot\frac{dp}{dq}\cdot q+10p^{2}q$
$\displaystyle \frac{dp}{dq}(2p-q-10pq^{2})=10p^{2}q+p$
$\displaystyle \frac{dp}{dq}=\frac{10p^{2}q+p}{2p-q-10pq^{2}}$
