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