#### Answer

\[ = \frac{{{e^t}\,\left( {{t^2} + t + 1} \right)}}{{\,{{\left( {t + 1} \right)}^2}}}\]

#### Work Step by Step

\[\begin{gathered}
y = \frac{{t{e^t}}}{{t + 1}} \hfill \\
\hfill \\
{y^,} = \frac{d}{{dt}}\,\left( {\frac{{t{e^t}}}{{t + 1}}} \right) \hfill \\
\hfill \\
use\,\,quotient\,\,rule \hfill \\
\hfill \\
{y^,} = \frac{{\,{{\left( {t{e^t}} \right)}^,}\,\left( {t + 1} \right) - t{e^t}\,{{\left( {t + 1} \right)}^,}}}{{\,{{\left( {t + 1} \right)}^2}}} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
{y^,} = \frac{{\,\left( {{t^,}{e^t} + t\,{{\left( {{e^t}} \right)}^,}} \right)\,\left( {t + 1} \right) - t{e^t}}}{{\,{{\left( {t + 1} \right)}^2}}} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
{y^,} = \frac{{\,\left( {{e^t} + t{e^t}} \right)\,\left( {t + 1} \right) - t{e^t}}}{{\,{{\left( {t + 1} \right)}^2}}} \hfill \\
\hfill \\
multiply\,\,and\,\,simplify \hfill \\
\hfill \\
\hfill \\
{y^,} = \frac{{t{e^t} + {e^t} + {t^2}{e^t} + t{e^t} - t{e^t}}}{{\,{{\left( {t + 1} \right)}^2}}} \hfill \\
\hfill \\
= \frac{{{e^t}\,\left( {{t^2} + t + 1} \right)}}{{\,{{\left( {t + 1} \right)}^2}}} \hfill \\
\end{gathered} \]