Answer
$$ G'(s)=\frac{1}{2\sqrt s(1+ s)}.$$
Work Step by Step
Recall that $(\tan^{-1} x)'=\dfrac{1}{1+x^2}$
Since $ G(s)=\tan^{-1}(\sqrt s)$, then the derivative, using the chain rule, is given by
$$ G'(s)=\frac{1}{1+(\sqrt s)^2}(\sqrt s)'=\frac{1}{2\sqrt s(1+ s)}.$$
