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 194: 96

Answer

\[\begin{align} & \mathbf{a}.\text{ }f'\left( g\left( x \right) \right)g''\left( x \right)+{{\left( g'\left( x \right) \right)}^{2}}f''\left( g\left( x \right) \right) \\ & \mathbf{b}.=\left( 36{{x}^{2}}+10 \right)\cos \left( 3{{x}^{4}}+5{{x}^{2}}+2 \right) \\ & \text{ }-{{\left( 12{{x}^{3}}+10x \right)}^{2}}\sin \left( 3{{x}^{4}}+5{{x}^{2}}+2 \right) \\ \end{align}\]

Work Step by Step

\[\begin{align} & \mathbf{a}\text{.} \\ & \frac{d}{dx}\left[ f\left( g\left( x \right) \right) \right] \\ & \text{By the chain rule} \\ & \frac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=f'\left( g\left( x \right) \right)g'\left( x \right) \\ & \text{Calculate the second derivative} \\ & \frac{{{d}^{2}}}{d{{x}^{2}}}\left[ f\left( g\left( x \right) \right) \right]=\frac{d}{dx}\left[ f'\left( g\left( x \right) \right)g'\left( x \right) \right] \\ & \text{Use the product rule} \\ & \frac{{{d}^{2}}}{d{{x}^{2}}}\left[ f\left( g\left( x \right) \right) \right]=f'\left( g\left( x \right) \right)\frac{d}{dx}\left[ g'\left( x \right) \right]+g'\left( x \right)\frac{d}{dx}\left[ f'\left( g\left( x \right) \right) \right] \\ & =f'\left( g\left( x \right) \right)g''\left( x \right)+g'\left( x \right)f''\left( g\left( x \right) \right)\frac{d}{dx}\left[ g\left( x \right) \right] \\ & =f'\left( g\left( x \right) \right)g''\left( x \right)+g'\left( x \right)f''\left( g\left( x \right) \right)g'\left( x \right) \\ & =f'\left( g\left( x \right) \right)g''\left( x \right)+{{\left( g'\left( x \right) \right)}^{2}}f''\left( g\left( x \right) \right) \\ & \\ & \mathbf{b}.\frac{{{d}^{2}}}{d{{x}^{2}}}\left( \sin \left( 3{{x}^{4}}+5{{x}^{2}}+2 \right) \right) \\ & \text{Let }g\left( x \right)=3{{x}^{4}}+5{{x}^{2}}+2\text{ and }f\left( x \right)=\sin x,\text{ then} \\ & g'\left( x \right)=12{{x}^{3}}+10x \\ & g''\left( x \right)=36{{x}^{2}}+10 \\ & \\ & f\left( g\left( x \right) \right)=\sin \left( 3{{x}^{4}}+5{{x}^{2}}+2 \right) \\ & \frac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=\frac{d}{dx}\left[ \sin \left( 3{{x}^{4}}+5{{x}^{2}}+2 \right) \right] \\ & \text{The second derivative is} \\ & \underbrace{f'\left( g\left( x \right) \right)g''\left( x \right)+{{\left( g'\left( x \right) \right)}^{2}}f''\left( g\left( x \right) \right)}_{\Downarrow } \\ & =\cos \left( 3{{x}^{4}}+5{{x}^{2}}+2 \right)\left( 36{{x}^{2}}+10 \right) \\ & +{{\left( 12{{x}^{3}}+10x \right)}^{2}}\left( -\sin \left( 3{{x}^{4}}+5{{x}^{2}}+2 \right) \right) \\ & \text{Simplifying} \\ & =\left( 36{{x}^{2}}+10 \right)\cos \left( 3{{x}^{4}}+5{{x}^{2}}+2 \right) \\ & -{{\left( 12{{x}^{3}}+10x \right)}^{2}}\sin \left( 3{{x}^{4}}+5{{x}^{2}}+2 \right) \\ \end{align}\]
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