Answer
$p' = sec^{2}q$
Work Step by Step
$p = 5+\frac{1}{cotq}$
$p' = 0+ \frac{cotq(0) + cosec^{2}q }{cot^{2}q}$
$p' = \frac{ cosec^{2}q }{cot^{2}q}$
$p' = \frac{1}{cos^{2}q}$
$p' = sec^{2}q$
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: 27
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