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: 55

Answer

\[\frac{dy}{dx}=2x\cdot f'\left( g\left( {{x}^{2}} \right) \right)\cdot g'\left( {{x}^{2}} \right)\]

Work Step by Step

\[\begin{align} & \text{Let }y=f\left( g\left( {{x}^{2}} \right) \right) \\ & \text{Differentiate both sides with respect to }x \\ & \frac{dy}{dx}=\frac{d}{dx}\left[ f\left( g\left( {{x}^{2}} \right) \right) \right] \\ & \text{By the chain rule} \\ & \frac{dy}{dx}=f'\left( g\left( {{x}^{2}} \right) \right)\frac{d}{dx}\left[ g\left( {{x}^{2}} \right) \right] \\ & \frac{dy}{dx}=f'\left( g\left( {{x}^{2}} \right) \right)g'\left( {{x}^{2}} \right)\frac{d}{dx}\left[ {{x}^{2}} \right] \\ & \frac{dy}{dx}=f'\left( g\left( {{x}^{2}} \right) \right)g'\left( {{x}^{2}} \right)\left( 2x \right) \\ & \text{Simplifying} \\ & \frac{dy}{dx}=2x\cdot f'\left( g\left( {{x}^{2}} \right) \right)\cdot g'\left( {{x}^{2}} \right) \\ \end{align}\]
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