Answer
\[{y^,} = - 32{e^{2x + 1}}\]
Work Step by Step
\[\begin{gathered}
y = - 16{e^{2x + 1}} \hfill \\
Find\,\,the\,\,derivative \hfill \\
{y^,} = \,{\left( { - 16{e^{2x + 1}}} \right)^,} \hfill \\
{y^,} = \, - 16\,{\left( {{e^{2x + 1}}} \right)^,} \hfill \\
Use\,\,the\,\,formula \hfill \\
\frac{d}{{dx}}\,\,\left[ {{e^{g\,\left( x \right)}}} \right] = {e^{g\,\left( x \right)}}{g^,}\,\left( x \right) \hfill \\
Then \hfill \\
{y^,} = \, - 16\,\left( {{e^{2x + 1}}} \right)\,{\left( {2x + 1} \right)^,} \hfill \\
{y^,} = \, - 16\,\left( {{e^{2x + 1}}} \right)\,\left( 2 \right) \hfill \\
Multiplying\, \hfill \\
{y^,} = - 32{e^{2x + 1}} \hfill \\
\hfill \\
\end{gathered} \]
