Answer
$$g''(1)= \frac{-1}{8} $$
Work Step by Step
Given $$g(s)=\frac{\sqrt{s}}{s+1}$$
Since
\begin{align*}
g'(s) &= \frac{\frac{d}{ds}\left(\sqrt{s}\right)\left(s+1\right)-\frac{d}{ds}\left(s+1\right)\sqrt{s}}{\left(s+1\right)^2}\\
&= \frac{-s+1}{2\sqrt{s}\left(s+1\right)^2}\\
g''(s)&=\frac{1}{2}\cdot \frac{\frac{d}{ds}\left(-s+1\right)\sqrt{s}\left(s+1\right)^2-\frac{d}{ds}\left(\sqrt{s}\left(s+1\right)^2\right)\left(-s+1\right)}{\left(\sqrt{s}\left(s+1\right)^2\right)^2}\\
&= \frac{3s^2-6s-1}{4s\sqrt{s}\left(s+1\right)^3}
\end{align*}
Then
$$g''(1)= \frac{-1}{8} $$
