Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.6 - Inverse Trigonometric Functions - Exercises 7.6 - Page 421: 22

Answer

$$\frac{{dy}}{{dx}} = \frac{1}{{x\sqrt {{x^2} - 1} }}$$

Work Step by Step

$$\eqalign{ & y = {\cos ^{ - 1}}\left( {1/x} \right) \cr & y = {\cos ^{ - 1}}\left( {{x^{ - 1}}} \right) \cr & {\text{find the derivative of }}y{\text{ with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{{d\left( {{{\cos }^{ - 1}}\left( {{x^{ - 1}}} \right)} \right)}}{{dx}} \cr & {\text{we can use the formula }}\cr & \frac{{d\left( {{{\cos }^{ - 1}}u} \right)}}{{dx}} = - \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dx}}.\,\,\,\left( {{\text{see table 7}}{\text{.3}}} \right). \cr & {\text{here }}u = {x^{ - 1}},\,\,{\text{then}} \cr & \frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {1 - {{\left( {{x^{ - 1}}} \right)}^2}} }}\frac{{d\left( {{x^{ - 1}}} \right)}}{{dx}} \cr & \frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {1 - {x^{ - 2}}} }}\left( { - {x^{ - 2}}} \right) \cr & {\text{simplifying}} \cr & \frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {1 - 1/{x^2}} }}\left( { - \frac{1}{{{x^2}}}} \right) \cr & \frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {\frac{{{x^2} - 1}}{{{x^2}}}} }}\left( { - \frac{1}{{{x^2}}}} \right) \cr & \frac{{dy}}{{dx}} = \frac{1}{{x\sqrt {{x^2} - 1} }} \cr} $$
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