How do I find the derivative of this consumer surplus function with respect to $Q_0$? I know how to do it when I have function values but I'm not sure how to evaluate everything in its original notation.
$CS(Q_0)$ = $\int_{0}^{Q_0} P(Q)\,dQ$ –$P_0Q_0$
Where $P_0$ = $P(Q_0)$
Best Answer
We differentiate the given equation with respect to $Q_{0}$:
$$\frac{d}{dQ_{0}}CS(Q_{0}) = \frac{d}{dQ_{0}}\int_{0}^{Q_{0}}P(Q)\,dQ - \frac{d}{dQ_{0}}P_{0}Q_{0}.$$
Now, by the fundamental theorem of calculus we know that $$\frac{d}{dQ_{0}}\int_{0}^{Q_{0}}P(Q)\,dQ = P(Q_{0}).$$ Also, the product rule tells us that $$-\frac{d}{dQ_{0}}P_{0}Q_{0} = -\frac{dP_{0}}{dQ_{0}}Q_{0} - P_{0}\frac{dQ_{0}}{dQ_{0}}.$$ Because we know that $P_{0} = P(Q_{0})$ we get $$-\frac{d}{dQ_{0}}P_{0}Q_{0} = -\frac{dP(Q_{0})}{dQ_{0}}Q_{0} - P_{0}\frac{dQ_{0}}{dQ_{0}} = -P'(Q_{0})Q_{0} - P_{0}.$$
All together, then, we get $$CS'(Q_{0}) = P(Q_{0}) - P'(Q_{0})Q_{0} - P_{0} = -P'(Q_{0})Q_{0}.$$