#### Answer

\[{y^,} = 4\,{\left( {\,\left( {x + 2} \right)\,\left( {{x^2} + 1} \right)} \right)^3} \cdot \,\,\,\left[ {3{x^2} + 4x + 1} \right]\]

#### Work Step by Step

\[\begin{gathered}
y = \,{\left( {\,\left( {x + 2} \right)\,\left( {{x^2} + 1} \right)} \right)^4} \hfill \\
\hfill \\
Chain\,\,rule \hfill \\
\hfill \\
{y^,} = 4\,{\left( {\,\left( {x + 2} \right)\,\left( {{x^2} + 1} \right)} \right)^3} \cdot \,\,{\left( {\,\left( {x + 2} \right)\,\left( {{x^2} + 1} \right)} \right)^,} \hfill \\
\hfill \\
Product\,\,rule \hfill \\
\hfill \\
{y^,} = 4\,{\left( {\,\left( {x + 2} \right)\,\left( {{x^2} + 1} \right)} \right)^3} \cdot \,\,\,\left[ {\,{{\left( {x + 2} \right)}^,} \cdot \,\left( {{x^2} + 2} \right) + \,\left( {x + 2} \right) \cdot \,{{\left( {{x^2} + 1} \right)}^,}} \right] \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
{y^,} = 4\,{\left( {\,\left( {x + 2} \right)\,\left( {{x^2} + 1} \right)} \right)^3} \cdot \,\,\,\left[ {{x^2} + 1 + \,\left( {x + 2} \right) \cdot 2x} \right] \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
{y^,} = 4\,{\left( {\,\left( {x + 2} \right)\,\left( {{x^2} + 1} \right)} \right)^3} \cdot \,\,\,\left[ {{x^2} + 1 + 2{x^2} + 4x} \right] \hfill \\
\hfill \\
{y^,} = 4\,{\left( {\,\left( {x + 2} \right)\,\left( {{x^2} + 1} \right)} \right)^3} \cdot \,\,\,\left[ {3{x^2} + 4x + 1} \right] \hfill \\
\end{gathered} \]