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: 33

Answer

a) $y'' = cosecx(cot^{2}x +cosec^{2}x)$ b) $y'' = secx(tan^{2}x + sec^{2}x)$

Work Step by Step

a) $y = cosecx$ $y' = -cosecx\ cotx$ $y'' = cosecx\ cot^{2}x + cosec^{3}x$ $y'' = cosecx(cot^{2}x +cosec^{2}x)$ b) $y = secx$ $y' = secx\ tanx$ $y'' = secx\ tan^{2}x + sec^{3}x$ $y'' = secx(tan^{2}x + sec^{2}x)$
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