Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.9 - Linearization and Differentials - Exercises 3.9 - Page 175: 26

Answer

$$dy = 2x\sec \left( {{x^2} - 1} \right)\tan \left( {{x^2} - 1} \right)dx$$

Work Step by Step

$$\eqalign{ & y = \sec \left( {{x^2} - 1} \right) \cr & {\text{Calculate }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\sec \left( {{x^2} - 1} \right)} \right] \cr & {\text{Recall that }}\frac{d}{{dx}}\left[ {\sec u} \right] = \sec u\tan u\frac{{du}}{{dx}} \cr & \frac{{dy}}{{dx}} = \sec \left( {{x^2} - 1} \right)\tan \left( {{x^2} - 1} \right)\frac{d}{{dx}}\left[ {{x^2} - 1} \right] \cr & \frac{{dy}}{{dx}} = \sec \left( {{x^2} - 1} \right)\tan \left( {{x^2} - 1} \right)\left( {2x} \right) \cr & \frac{{dy}}{{dx}} = 2x\sec \left( {{x^2} - 1} \right)\tan \left( {{x^2} - 1} \right) \cr & {\text{Write in differential form}} \cr & dy = 2x\sec \left( {{x^2} - 1} \right)\tan \left( {{x^2} - 1} \right)dx \cr} $$
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