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 - Page 151: 6

Answer

$y' = 2xcotx - x^{2}cosec^{2}x + \frac{2}{x^{3}}$

Work Step by Step

$y = x^{2}cotx - \frac{1}{x^{2}}$ $y' = 2xcotx - x^{2}cosec^{2}x + \frac{2}{x^{3}}$
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.