Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 13 - The Trigonometric Functions - Chapter Review - Review Exercises - Page 701: 51


$$\frac{{dy}}{{dx}} = 10{\sec ^2}5x$$

Work Step by Step

$$\eqalign{ & y = 2\tan 5x \cr & {\text{differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {2\tan 5x} \right] \cr & {\text{use multiple constant rule}} \cr & \frac{{dy}}{{dx}} = 2\frac{d}{{dx}}\left[ {\tan 5x} \right] \cr & {\text{using the chain rule for }}{D_x}\left( {\tan u} \right) = {\sec ^2}u \cdot {D_x}\left( u \right).{\text{ then}} \cr & \frac{{dy}}{{dx}} = 2\left( {{{\sec }^2}5x} \right)\frac{d}{{dx}}\left[ {5x} \right] \cr & \frac{{dy}}{{dx}} = 2\left( {{{\sec }^2}5x} \right)\left( 5 \right) \cr & {\text{simplifying}} \cr & \frac{{dy}}{{dx}} = 10{\sec ^2}5x \cr} $$
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