Calculus: Early Transcendentals (2nd Edition)

$= \frac{{8x{e^{8x}}}}{{\,{{\left( {x + 1} \right)}^9}}}$
$\begin{gathered} y = \,{\left( {\frac{{{e^x}}}{{x + 1}}} \right)^8} \hfill \\ \hfill \\ differentiate\,\,both\,\,sides \hfill \\ {y^,} = \frac{d}{{dx}}{\left( {\frac{{{e^x}}}{{x + 1}}} \right)^8} \hfill \\ \hfill \\ use\,\,the\,\,chain\,\,rule\,\,for\,\,powers \hfill \\ \hfill \\ \frac{d}{{dx}}\left( {g{{\left( x \right)}^n}} \right) = ng{\left( x \right)^{n - 1}}g'\left( x \right) \hfill \\ \hfill \\ Therefore, \hfill \\ \hfill \\ {y^,} = 8\,{\left( {\frac{{{e^x}}}{{x + 1}}} \right)^7} \cdot \,{\left( {\frac{{{e^x}}}{{x + 1}}} \right)^,} \hfill \\ \hfill \\ use\,\,quotient\,\,rule \hfill \\ \hfill \\ {y^,} = 8\,{\left( {\frac{{{e^x}}}{{x + 1}}} \right)^7} \cdot \frac{{{e^x}\,\left( {x + 1} \right) - {e^x}}}{{\,{{\left( {x + 1} \right)}^2}}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ {y^,} = 8\,{\left( {\frac{{{e^x}}}{{x + 1}}} \right)^7} \cdot \frac{{x{e^x}}}{{\,{{\left( {x + 1} \right)}^2}}} \hfill \\ \hfill \\ = \frac{{8x{e^{8x}}}}{{\,{{\left( {x + 1} \right)}^9}}} \hfill \\ \hfill \\ \end{gathered}$