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

$$\frac{dp}{dq}=\frac{6q-4p}{3p^2+4q}$$
$$p^3+4pq-3q^2=2$$ To find $dp/dq$, we would use the methods of implicit differentiation. First, differentiate both sides of the equation with respect to $q$: $$3p^2\frac{dp}{dq}+4\Big(q\frac{dp}{dq}+p\frac{dq}{dq}\Big)-\frac{d(3q^2)}{dq}=\frac{d(2)}{dq}$$ $$3p^2\frac{dp}{dq}+4\Big(q\frac{dp}{dq}+p\Big)-6q=0$$ $$3p^2\frac{dp}{dq}+4q\frac{dp}{dq}+4p-6q=0$$ $$\frac{dp}{dq}(3p^2+4q)+4p-6q=0$$ Next, we need to separate elements with $dp/dq$ and those without $dp/dq$ into 2 sides of the equation: $$\frac{dp}{dq}(3p^2+4q)=6q-4p$$ Finally, calculate for $dp/dq$: $$\frac{dp}{dq}=\frac{6q-4p}{3p^2+4q}$$