Answer
$\frac{dA}{dq}=-\frac{km}{q^2}+\frac{h}{2}$
$\frac{d^2A}{dq^2}=\frac{2km}{q^3}$
Work Step by Step
Step 1. Given $A(q)=\frac{km}{q}+cm+\frac{hq}{2}=kmq^{-1}+cm+\frac{h}{2}q$,
using the Power Rule for negative integers given in Exercise 64, we have
$\frac{dA}{dq}=-kmq^{-2}+0+\frac{h}{2}=-kmq^{-2}+\frac{h}{2}=-\frac{km}{q^2}+\frac{h}{2}$
Step 2. : $\frac{d^2A}{dq^2}=2kmq^{-3}+0=\frac{2km}{q^3}$
