Answer
$\frac{dP}{dV}=-\frac{nRT}{(V-nb)^2}+\frac{2an^2}{V^3}$
Work Step by Step
Step 1. Let $X=V-nb$ and $Y=V$; the equation becomes $P=nRTX^{-1}-an^2Y^{-2}$
Step 2. Using the Power Rule for negative integers given in Exercise 64, we have
$\frac{dP}{dV}=-nRTX^{-2}\frac{dX}{dV}+2an^2Y^{-3}\frac{dY}{dV}$
Step 3. As $\frac{dX}{dV}=\frac{dY}{dV}=1$, we have
$\frac{dP}{dV}=-\frac{nRT}{(V-nb)^2}+\frac{2an^2}{V^3}$
