Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises - Page 192: 74

Answer

\[\begin{align} & \text{a}\text{. }4{{x}^{3}}+6{{x}^{2}}+2x \\ & \text{b}\text{. }4{{x}^{3}}+6{{x}^{2}}+2x \\ \end{align}\]

Work Step by Step

\[\begin{align} & \text{a}\text{. }\frac{d}{dx}{{\left( {{x}^{2}}+x \right)}^{2}}\text{ using the chain rule} \\ & \frac{d}{dx}{{\left( {{x}^{2}}+x \right)}^{2}}=2{{\left( {{x}^{2}}+x \right)}^{2-1}}\frac{d}{dx}\left( {{x}^{2}}+x \right) \\ & \text{ }=2\left( {{x}^{2}}+x \right)\left( 2x+1 \right) \\ & \text{Simplifying} \\ & \text{ }=2\left( 2{{x}^{3}}+{{x}^{2}}+2{{x}^{2}}+x \right) \\ & \text{ }=2\left( 2{{x}^{3}}+3{{x}^{2}}+x \right) \\ & \text{ }=4{{x}^{3}}+6{{x}^{2}}+2x \\ & \\ & \text{b}\text{. Expanding }{{\left( {{x}^{2}}+x \right)}^{2}},\text{ recall that }{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\ & {{\left( {{x}^{2}}+x \right)}^{2}}={{x}^{4}}+2{{x}^{3}}+{{x}^{2}} \\ & \text{Differentiating} \\ & \frac{d}{dx}\left( {{x}^{4}}+2{{x}^{3}}+{{x}^{2}} \right)=4{{x}^{3}}+6{{x}^{2}}+2x \\ \end{align}\]
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