Answer
\[y' = - 15{\sin ^4}\,\left( {\cos 3x} \right) \cdot \cos \,\left( {\cos 3x} \right) \cdot \sin 3x\]
Work Step by Step
\[\begin{gathered}
y = {\sin ^5}\,\left( {\cos 3x} \right) \hfill \\
\hfill \\
Chain\,\,rule \hfill \\
\hfill \\
f\,{\left( {g\,\left( t \right)} \right)^,} = {f^,}\,\left( {g\,\left( t \right)} \right) \cdot {g^,}\,\left( t \right) \hfill \\
\hfill \\
{y^,} = 5{\sin ^4}\,\left( {\cos 3x} \right) \cdot \,{\left( {\sin \,\left( {\cos 3x} \right)} \right)^,} \hfill \\
\hfill \\
use\,\,\frac{d}{{dx}}\left[ {\sin u} \right] = u'\cos u \hfill \\
\hfill \\
y' = 5{\sin ^4}\,\left( {\cos 3x} \right) \cdot \cos \,\left( {\cos 3x} \right) \cdot \,{\left( {\cos 3x} \right)^,} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
y' = 5{\sin ^4}\,\left( {\cos 3x} \right) \cdot \cos \,\left( {\cos 3x} \right) \cdot \,\,\left( { - \sin 3x} \right) \cdot \,{\left( {3x} \right)^,} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
y' = 5{\sin ^4}\,\left( {\cos 3x} \right) \cdot \cos \,\left( {\cos 3x} \right) \cdot \,\sin 3x \cdot 3 \hfill \\
\hfill \\
y' = - 15{\sin ^4}\,\left( {\cos 3x} \right) \cdot \cos \,\left( {\cos 3x} \right) \cdot \sin 3x \hfill \\
\end{gathered} \]
