University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 272: 73


$\sec^2 (5x-1)$

Work Step by Step

Simplify: $\dfrac{d(\dfrac{1}{5}\tan(5x-1)+C)}{dx}$ Thus, we have $\dfrac{d(\dfrac{1}{5}\tan(5x-1)+C)}{dx}=(\dfrac{1}{5})\dfrac{5}{\cos^2(5x-1)}=\sec^2 (5x-1)$
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.