Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.3 Derivatives Of Inverse Functions; Derivatives And Integrals Involving Exponential Functions - Exercises Set 6.3 - Page 432: 21

Answer

$$y' = {e^{x\tan x}}\left( {x{{\sec }^2}x + \tan x} \right)$$

Work Step by Step

$$\eqalign{ & y = {e^{x\tan x}} \cr & {\text{differentiate}} \cr & y' = \left( {{e^{x\tan x}}} \right)' \cr & {\text{use the chain rule}} \cr & y' = {e^{x\tan x}}\left( {x\tan x} \right)' \cr & {\text{product rule}} \cr & y' = {e^{x\tan x}}\left( {x\left( {\tan x} \right)' + \tan x\left( x \right)'} \right) \cr & y' = {e^{x\tan x}}\left( {x{{\sec }^2}x + \tan x\left( 1 \right)} \right) \cr & {\text{Simplify}} \cr & y' = {e^{x\tan x}}\left( {x{{\sec }^2}x + \tan x} \right) \cr} $$
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.