## University Calculus: Early Transcendentals (3rd Edition)

$$\frac{dP}{dV}=\frac{-nRT}{(V-nb)^2}+\frac{2an^2}{V^3}$$
$$P=\frac{nRT}{V-nb}-\frac{an^2}{V^2}$$ Find $dP/dV$: $$\frac{dP}{dV}=\frac{d}{dV}\Big(\frac{nRT}{V-nb}-\frac{an^2}{V^2}\Big)$$ Remember that we concern ourvselves with the variable $V$ only. All the others are constants, so we treat them like numbers. $$\frac{dP}{dV}=\frac{d}{dV}\Big(\frac{nRT}{V-nb}\Big)-\frac{d}{dV}\Big(\frac{an^2}{V^2}\Big)$$ $$\frac{dP}{dV}=\frac{(nRT)'(V-nb)-(nRT)(V-nb)'}{(V-nb)^2}-\frac{(an^2)'(V^2)-(an^2)(V^2)'}{V^4}$$ $$\frac{dP}{dV}=\frac{0\times(V-nb)-(nRT)(1-0)}{(V-nb)^2}-\frac{0\times(V^2)-(an^2)(2V)}{V^4}$$ $$\frac{dP}{dV}=\frac{-nRT}{(V-nb)^2}-\Big(\frac{-2an^2V}{V^4}\Big)$$ $$\frac{dP}{dV}=\frac{-nRT}{(V-nb)^2}+\frac{2an^2}{V^3}$$