Calculus: Early Transcendentals 9th Edition

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

Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises - Page 225: 74

Answer

$R'\left( t \right) = - \frac{1}{{t\sqrt {{t^2} - 1} }}$

Work Step by Step

$$\eqalign{ & R\left( t \right) = \arcsin \left( {\frac{1}{t}} \right) \cr & {\text{Differentiate}} \cr & R'\left( t \right) = \frac{d}{{dt}}\left[ {\arcsin \left( {\frac{1}{t}} \right)} \right] \cr & {\text{Use the Derivatives of Inverse Trigonometric Functions }} \cr & {\text{and the chain rule}}{\text{. }}\frac{d}{{dt}}\left[ {{{\sin }^{ - 1}}u} \right] = \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dt}},\;u = \frac{1}{t} \cr & R'\left( t \right) = \frac{1}{{\sqrt {1 - {{\left( {\frac{1}{t}} \right)}^2}} }}\frac{d}{{dt}}\left[ {\frac{1}{t}} \right] \cr & {\text{Therefore}}{\text{,}} \cr & R'\left( t \right) = \frac{t}{{\sqrt {{t^2} - 1} }}\left( { - \frac{1}{{{t^2}}}} \right) \cr & R'\left( t \right) = - \frac{1}{{t\sqrt {{t^2} - 1} }} \cr} $$
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