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: 27

Answer

$p' = sec^{2}q$

Work Step by Step

$p = 5+\frac{1}{cotq}$ $p' = 0+ \frac{cotq(0) + cosec^{2}q }{cot^{2}q}$ $p' = \frac{ cosec^{2}q }{cot^{2}q}$ $p' = \frac{1}{cos^{2}q}$ $p' = sec^{2}q$
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