Answer
\[{y^{^,}} = - 2{x^2}{e^{ - 2x}} + 2x{e^{ - 2x}}\]
Work Step by Step
\[\begin{gathered}
y = {x^2}{e^{ - 2x}} \hfill \\
Use\,\,the\,\,product\,\,rule\,\,to\,\,find\,\,the\,\,derivative \hfill \\
{y^,} = {x^2}\,{\left( {{e^{ - 2x}}} \right)^,} + \,\left( {{e^{ - 2x}}} \right)\,\left( {{x^2}} \right) \hfill \\
and \hfill \\
\frac{d}{{dx}}\,\,\left[ {{e^{g\,\left( x \right)}}} \right] = {e^{g\,\left( x \right)}}{g^,}\,\left( x \right) \hfill \\
Then \hfill \\
{y^,} = {x^2}\,\left( {{e^{ - 2x}}} \right)\,\left( { - 2} \right) + \,\,\left( {{e^{ - 2x}}} \right)\left( {2x} \right) \hfill \\
Multiplying\, \hfill \\
{y^{^,}} = - 2{x^2}{e^{ - 2x}} + 2x{e^{ - 2x}} \hfill \\
\end{gathered} \]
