University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.5 - Derivatives of Trigonometric Functions - Exercises: 7

Answer

$y' = sinx +tanx\ secx=sinx(sec^2{x}+1)$

Work Step by Step

$y = sinx\ tanx$ $y' = cosx\ tanx + sinx\ sec^{2}x$ $y' = cosx\ \frac{sinx}{cosx} + sinx\ \frac{1}{cos^{2}x}$ $y' = sinx +tanx\ secx$
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.