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: 29


$p' = sec^{2}q$

Work Step by Step

$\displaystyle p = \frac{sinq + cosq}{cosq}$ $\displaystyle p' = \frac{cosq(cosq -sinq) + (sinq+cosq)sinq}{cos^{2}q}$ $\displaystyle p' = \frac{cos^{2}q - sinq\ cosq + sin^{2}q + sinq\ cosq}{cos^{2}q}$ $\displaystyle p' = \frac{1}{cos^{2}q}$ $\displaystyle p' = sec^{2}q$
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.