Calculus: Early Transcendentals (2nd Edition)

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