Answer
$$G'(q)=\frac{-2}{q^2}-\frac{2}{q^3}$$
Work Step by Step
$$G(q)=(1+q^{-1})^2$$
$$G(q)=1+(2\times1\times q^{-1})+(q^{-1})^2$$
$$G(q)=1+2q^{-1}+q^{-2}$$
So, $$G'(q)=\frac{d}{dq}(1)+2\frac{d}{dq}(q^{-1})+\frac{d}{dq}q^{-2}$$
$$G'(q)=0+2\times(-1)\times q^{-2}+(-2)\times q^{-3}$$
$$G'(q)=-2\times\frac{1}{q^2}-2\times\frac{1}{q^3}$$
$$G'(q)=\frac{-2}{q^2}-\frac{2}{q^3}$$