Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 3 - Section 3.11 - Hyperbolic Functions - 3.11 Exercises - Page 267: 50

Answer

$\frac{{dy}}{{d\theta }} = - \csc \theta $

Work Step by Step

$$\eqalign{ & y = {\operatorname{sech} ^{ - 1}}\left( {\sin \theta } \right),{\text{ 0}} < \theta < \frac{\pi }{2} \cr & {\text{Differentiate}} \cr & \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {{{\operatorname{sech} }^{ - 1}}\left( {\sin \theta } \right)} \right] \cr & {\text{Use Derivatives of Inverse Hyperbolic Functions }} \cr & \frac{d}{{d\theta }}\left[ {{{\operatorname{sech} }^{ - 1}}u} \right] = - \frac{1}{{u\sqrt {1 - {u^2}} }}\frac{{du}}{{d\theta }},{\text{ let }}u = \sin \theta ,{\text{ so}} \cr & \frac{{dy}}{{d\theta }} = - \frac{1}{{\sin \theta \sqrt {1 - {{\left( {\sin \theta } \right)}^2}} }}\frac{d}{{d\theta }}\left[ {\sin \theta } \right] \cr & {\text{Compute the derivative and simplify}} \cr & \frac{{dy}}{{d\theta }} = - \frac{1}{{\sin \theta \sqrt {1 - {{\sin }^2}\theta } }}\left( {\cos \theta } \right) \cr & {\text{Use the identity }}{\sin ^2}\theta + {\cos ^2}\theta = 1 \cr & \frac{{dy}}{{d\theta }} = - \frac{1}{{\sin \theta \sqrt {{{\cos }^2}\theta } }}\left( {\cos \theta } \right) \cr & \frac{{dy}}{{d\theta }} = - \frac{{\cos \theta }}{{\sin \theta \cos \theta }} \cr & \frac{{dy}}{{d\theta }} = - \frac{1}{{\sin \theta }} \cr & \frac{{dy}}{{d\theta }} = - \csc \theta \cr} $$
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