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: 28

Answer

$p' = -cot^{2}q - sinq -1=-sinq-cosec^2q$

Work Step by Step

$p = (1+cosecq)cosq$ $p' = (-cosecq\ cotq)cosq - (1+cosecq)sinq$ $p' = -cot^{2}q - sinq -1$
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