## Calculus: Early Transcendentals (2nd Edition)

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