Answer
$$s^{\prime}(x)=\frac{-s^{2}\left(2 \sqrt{x+s}+x^{2}\right)}{x^{2}\left(s^{2}+2 \sqrt{x+s}\right)} $$
Work Step by Step
Given $$\sqrt{x+s}=\frac{1}{x}+\frac{1}{s}$$
Differentiate both sides with respect to $x$,
\begin{align*}
\frac{d}{dx}\sqrt{x+s}&= \frac{d}{dx}\frac{1}{x}+ \frac{d}{dx}\frac{1}{s}\\
\frac{1+s'(x)}{2\sqrt{x+s} }&= \frac{-1}{x^2}-\frac{s'(x)}{s^2}\\
\frac{1 }{2\sqrt{x+s} }+\frac{s'(x)}{2\sqrt{x+s} }&= \frac{-1}{x^2}-\frac{s'(x)}{s^2}\\
\frac{1}{2 \sqrt{x+s}} s^{\prime}(x)+\frac{1}{s^{2}} s^{\prime}(x)&=\frac{-1}{x^{2}}-\frac{1}{2 \sqrt{x+s}}\\
\frac{s^{2}+2 \sqrt{x+s}}{2 s^{2} \sqrt{x+8}} s^{\prime}(x)&=\frac{-2 \sqrt{x+s}-x^{2}}{2 x^{2} \sqrt{x+s}}
\end{align*}
Then
$$s^{\prime}(x)=\frac{-s^{2}\left(2 \sqrt{x+s}+x^{2}\right)}{x^{2}\left(s^{2}+2 \sqrt{x+s}\right)} $$
