## Calculus: Early Transcendentals (2nd Edition)

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