Answer
\[{y^,} = \frac{{x\,\left( {2 - x} \right)}}{{{e^x}}}\]
Work Step by Step
\[\begin{gathered}
y = \frac{{{x^2}}}{{{e^x}}} \hfill \\
Find\,\,the\,\,derivative\,\,using\,\,the\,\,quotient\,\,rule \hfill \\
\,\,{\left[ {\frac{u}{v}} \right]^,} = \frac{{v{u^,} - u{v^,}}}{{{v^2}}} \hfill \\
Then \hfill \\
{y^,} = \frac{{{e^x}\,{{\left( {{x^2}} \right)}^,} - {x^2}\,{{\left( {{e^x}} \right)}^,}}}{{\,{{\left( {{e^x}} \right)}^2}}} \hfill \\
{y^,} = \frac{{{e^x}\,\left( {2x} \right) - {x^2}\,\left( {{e^x}} \right)}}{{{e^{2x}}}} \hfill \\
Multiply \hfill \\
{y^,} = \frac{{2x{e^x} - {x^2}{e^x}}}{{{e^{2x}}}} \hfill \\
Factor\,\,the\,\,numerator \hfill \\
{y^,} = \,\frac{{x{e^x}\,\left( {2 - x} \right)}}{{{e^{2x}}}} \hfill \\
Cancel\,\,{e^x} \hfill \\
{y^,} = \frac{{x\,\left( {2 - x} \right)}}{{{e^x}}} \hfill \\
\end{gathered} \]
