Answer
\[f\,\left( x \right) = - \frac{1}{{2\ln 2\,\left( {x + 1} \right)}}\]
Work Step by Step
\[\begin{gathered}
f\,\left( x \right) = {\log _2}\frac{8}{{\sqrt {x + 1} }} \hfill \\
\hfill \\
Rewrite \hfill \\
\hfill \\
f\,\left( x \right) = {\log _2}8 - {\log _2}\,{\left( {x + 1} \right)^{\frac{1}{2}}} \hfill \\
\hfill \\
f\,\left( x \right) = {\log _2}8 - \frac{1}{2}{\log _2}\,\left( {x + 1} \right) \hfill \\
\hfill \\
Differentiate,{\text{ use }}\frac{d}{{dx}}\left[ {{{\log }_a}u} \right] = \frac{1}{{\ln a\left( u \right)}}{u^,} \hfill \\
\hfill \\
f\,\left( x \right) = 0 - \frac{1}{2}\,\left( {\frac{1}{{\,\left( {x + 1} \right)\ln 2}}} \right)\,\left( 1 \right) \hfill \\
\hfill \\
then \hfill \\
\hfill \\
f\,\left( x \right) = - \frac{1}{{2\ln 2\,\left( {x + 1} \right)}} \hfill \\
\end{gathered} \]
