Thomas' Calculus 13th Edition

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

Chapter 3: Derivatives - Practice Exercises - Page 177: 11


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

Work Step by Step

$$\eqalign{ & y = 2{\tan ^2}x - {\sec ^2}x \cr & y = 2{\left( {\tan x} \right)^2} - {\left( {\sec x} \right)^2} \cr & {\text{differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = 2\frac{d}{{dx}}\left[ {{{\left( {\tan x} \right)}^2}} \right] - \frac{d}{{dx}}\left[ {{{\left( {\sec x} \right)}^2}} \right] \cr & {\text{use the chain rule}} \cr & \frac{{dy}}{{dx}} = 2\left( 2 \right)\left( {\tan x} \right)\frac{d}{{dx}}\left[ {\tan x} \right] - 2\left( {\sec x} \right)\frac{d}{{dx}}\left[ {\sec x} \right] \cr & {\text{find derivatives and simplify}} \cr & \frac{{dy}}{{dx}} = 4\left( {\tan x} \right)\left( {{{\sec }^2}x} \right) - 2\left( {\sec x} \right)\left( {\sec x\tan x} \right) \cr & \frac{{dy}}{{dx}} = 4\tan x{\sec ^2}x - 2\tan x{\sec ^2}x \cr & \frac{{dy}}{{dx}} = 2\tan x{\sec ^2}x \cr} $$
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