#### Answer

\[ = \,{\left( {z + 4} \right)^2}\,\left( {3\tan z + \,\left( {z + 4} \right){{\sec }^2}z} \right)\]

#### Work Step by Step

\[\begin{gathered}
y = \,{\left( {z + 4} \right)^3}\tan z \hfill \\
\hfill \\
differentiate \hfill \\
\hfill \\
{y^,} = \frac{d}{{dz}}\,\left( {\,{{\left( {z + 4} \right)}^3}\tan z} \right) \hfill \\
\hfill \\
use\,\,the\,\,Product\,\,rule \hfill \\
\hfill \\
= \,{\left( {\,{{\left( {z + 4} \right)}^3}} \right)^,}\tan z + \,{\left( {z + 4} \right)^3}\,{\left( {\tan z} \right)^,} \hfill \\
\hfill \\
use\,\,the\,\,Chain\,\,rule \hfill \\
\hfill \\
= 3\,{\left( {z + 4} \right)^2}\tan z + \,{\left( {z + 4} \right)^3}{\sec ^2}z \hfill \\
\hfill \\
factor \hfill \\
\hfill \\
= \,{\left( {z + 4} \right)^2}\,\left( {3\tan z + \,\left( {z + 4} \right){{\sec }^2}z} \right) \hfill \\
\hfill \\
\end{gathered} \]