Answer
\[{y^,} = \frac{{2x{e^x} - {e^x}}}{{\,{{\left( {2x + 1} \right)}^2}}}\]
Work Step by Step
\[\begin{gathered}
y = \frac{{{e^x}}}{{2x + 1}} \hfill \\
Differentiate\,\,using\,\,the\,\,quotient\,\,rule \hfill \\
\,\,{\left[ {\frac{u}{v}} \right]^,} = \frac{{v{u^,} - u{v^,}}}{{{v^2}}} \hfill \\
Then \hfill \\
{y^,} = \frac{{\,\left( {2x + 1} \right)\,{{\left( {{e^x}} \right)}^,} - {e^x}\,{{\left( {2x + 1} \right)}^,}}}{{\,{{\left( {2x + 1} \right)}^2}}} \hfill \\
Compute\,\,the\,\,derivatives \hfill \\
{y^,} = \frac{{\,\left( {2x + 1} \right)\,\left( {{e^x}} \right) - {e^x}\,\,{{\left( 2 \right)}^,}}}{{\,{{\left( {2x + 1} \right)}^2}}} \hfill \\
Multiply \hfill \\
{y^,} = \frac{{2x{e^x} + {e^x} - 2{e^x}}}{{\,{{\left( {2x + 1} \right)}^2}}} \hfill \\
{y^,} = \frac{{2x{e^x} - {e^x}}}{{\,{{\left( {2x + 1} \right)}^2}}} \hfill \\
\end{gathered} \]