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: 8

Answer

$y' = -cosecx(cosec^{2}x +cot^{2}x)$

Work Step by Step

$y = \frac{cosx}{sin^{2}x}$ $y = cotx\ cosecx$ $y' = -cosec^{2}x\ cosecx - cotx\ (cosecx\ cotx)$ $y' = -cosec^{3}x- cot^{2}{x}\ cosecx$ $y' = -cosecx(cosec^{2}x +cot^{2}x)$
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