Answer
\[{y^,} = \frac{{\,{{\left( {x + 1} \right)}^{10}}}}{{\,{{\left( {2x - 4} \right)}^8}}}\,\left( {\frac{{10}}{{x + 1}} - \frac{8}{{x - 2}}} \right)\]
Work Step by Step
\[\begin{gathered}
f\,\left( x \right) = \frac{{\,{{\left( {x + 1} \right)}^{10}}}}{{\,{{\left( {2x - 4} \right)}^8}}} \hfill \\
\hfill \\
Take\,\,\ln \,\,to\,\,each\,\,side \hfill \\
\hfill \\
\ln \,y = \ln \,\left( {\frac{{\,{{\left( {x + 1} \right)}^{10}}}}{{\,{{\left( {2x - 4} \right)}^8}}}} \right) \hfill \\
\hfill \\
use\,\,\log \,\,properties \hfill \\
\hfill \\
\ln y = 10\ln \,\left( {x + 1} \right) - 8\ln \,\left( {2x - 4} \right) \hfill \\
\hfill \\
Differentiate,{\text{ use }}\frac{d}{{dx}}\left[ {\ln u} \right] = \frac{{{u^,}}}{u} \hfill \\
\hfill \\
\frac{{{y^,}}}{y} = 10\,\left( {\frac{1}{{x + 1}}} \right) - 8\,\left( {\frac{2}{{2x - 4}}} \right) \hfill \\
\hfill \\
Simplify \hfill \\
\hfill \\
{y^,} = \frac{{10}}{{x + 1}} - \frac{8}{{x - 2}} \hfill \\
\hfill \\
Therefore \hfill \\
\hfill \\
{y^,} = \frac{{\,{{\left( {x + 1} \right)}^{10}}}}{{\,{{\left( {2x - 4} \right)}^8}}}\,\left( {\frac{{10}}{{x + 1}} - \frac{8}{{x - 2}}} \right) \hfill \\
\hfill \\
\end{gathered} \]
