#### Answer

$$h''(1)= \frac{1}{8 } $$

#### Work Step by Step

Given $$ h(w)=\frac{1}{\sqrt{w}+1}$$
Since
\begin{align*}
h'(w)&= \frac{d}{dw}\left(\left(\sqrt{w}+1\right)^{-1}\right)\\
&= \frac{-1}{2\sqrt{w}\left(\sqrt{w}+1\right)^2}\\
h''(w)&= \frac{3w+4\sqrt{w}+1}{4w\sqrt{w}\left(\sqrt{w}+1\right)^4}
\end{align*}
Then $$h''(1)= \frac{1}{8 } $$