## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises - Page 192: 63

#### 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

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

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