Answer
\[ = 6\sin \,\left( {{e^{3x + 1}}} \right) \cdot \cos \,\left( {{e^{3x + 1}}} \right) \cdot \,{e^{3x + 1}}\]
Work Step by Step
\[\begin{gathered}
y = {\sin ^2}\,\left( {{e^{3x + 1}}} \right) \hfill \\
\hfill \\
Chain\,\,rule \hfill \\
\hfill \\
f\,{\left( {g\,\left( t \right)} \right)^,} = {f^,}\,\left( {g\,\left( t \right)} \right) \cdot {g^,}\,\left( t \right) \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
{y^,} = 2\sin \,\left( {{e^{3x + 1}}} \right) \cdot \,{\left( {\sin \,\left( {{e^{3x + 1}}} \right)} \right)^,} \hfill \\
\hfill \\
use\,\,\frac{d}{{dx}}\left[ {\sin u} \right] = u'\cos u \hfill \\
\hfill \\
y' = 2\sin \,\left( {{e^{3x + 1}}} \right) \cdot \cos \,\left( {{e^{3x + 1}}} \right) \cdot \,{\left( {{e^{3x + 1}}} \right)^,} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
y' = 2\sin \,\left( {{e^{3x + 1}}} \right) \cdot \cos \,\left( {{e^{3x + 1}}} \right) \cdot \,{e^{3x + 1}} \cdot \,{\left( {3x + 1} \right)^,} \hfill \\
\hfill \\
y' = 2\sin \,\left( {{e^{3x + 1}}} \right) \cdot \cos \,\left( {{e^{3x + 1}}} \right) \cdot \,{e^{3x + 1}} \cdot \,3 \hfill \\
\hfill \\
= 6\sin \,\left( {{e^{3x + 1}}} \right) \cdot \cos \,\left( {{e^{3x + 1}}} \right) \cdot \,{e^{3x + 1}} \hfill \\
\hfill \\
\end{gathered} \]
