Answer
$p' = -cot^{2}q - sinq -1=-sinq-cosec^2q$
Work Step by Step
$p = (1+cosecq)cosq$
$p' = (-cosecq\ cotq)cosq - (1+cosecq)sinq$
$p' = -cot^{2}q - sinq -1$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
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: 28
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
