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: 43

Answer

$$\frac{{dy}}{{dx}} = 30{\sec ^2}x{\left( {1 + 2\tan x} \right)^{14}}$$

Work Step by Step

$$\eqalign{ & y = {\left( {1 + 2\tan x} \right)^{15}} \cr & {\text{Differentiating}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\left( {1 + 2\tan x} \right)}^{15}}} \right] \cr & {\text{Use the chain rule}}{\text{,}} \cr & \frac{{dy}}{{dx}} = 15{\left( {1 + 2\tan x} \right)^{15 - 1}}\frac{d}{{dx}}\left[ {1 + 2\tan x} \right] \cr & \frac{{dy}}{{dx}} = 15{\left( {1 + 2\tan x} \right)^{14}}\left( {0 + 2{{\sec }^2}x} \right) \cr & \frac{{dy}}{{dx}} = 30{\sec ^2}x{\left( {1 + 2\tan x} \right)^{14}} \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.