## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 7: Transcendental Functions - Section 7.6 - Inverse Trigonometric Functions - Exercises 7.6 - Page 421: 29

#### Answer

$$\frac{{dy}}{{dt}} = - \frac{1}{{\sqrt {1 - {t^2}} }}$$

#### Work Step by Step

\eqalign{ & y = {\sec ^{ - 1}}\frac{1}{t},\,\,\,\,0 < t < 1 \cr & y = {\sec ^{ - 1}}{t^{ - 1}} \cr & {\text{find the derivative of }}y{\text{ with respect to }}t \cr & \frac{{dy}}{{dt}} = \frac{{d\left( {{{\sec }^{ - 1}}{t^{ - 1}}} \right)}}{{dt}} \cr & {\text{we can use the formula }}\cr & \frac{{d\left( {{{\sec }^{ - 1}}u} \right)}}{{dt}} = \frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }}\frac{{du}}{{dt}}\,\,\,\,\left| u \right| > 1.\,\,\,\left( {{\text{see table 7}}{\text{.3}}} \right). \cr & {\text{here }}u = {t^{ - 1}},\,\,{\text{then}} \cr & \frac{{dy}}{{dt}} = \frac{1}{{\left| {{t^{ - 1}}} \right|\sqrt {{{\left( {{t^{ - 1}}} \right)}^2} - 1} }}\frac{{d\left( {{t^{ - 1}}} \right)}}{{dt}} \cr & \frac{{dy}}{{dt}} = \frac{1}{{\left| {1/t} \right|\sqrt {{t^{ - 2}} - 1} }}\left( { - {t^{ - 2}}} \right) \cr & {\text{simplifying}} \cr & \frac{{dy}}{{dt}} = \frac{{\left| t \right|}}{{\sqrt {\frac{{1 - {t^2}}}{{{t^2}}}} }}\left( { - \frac{1}{{{t^2}}}} \right) \cr & \frac{{dy}}{{dt}} = \frac{{t\left| t \right|}}{{\sqrt {1 - {t^2}} }}\left( { - \frac{1}{{{t^2}}}} \right) \cr & \,0 < t < 1 \cr & \frac{{dy}}{{dt}} = \frac{{{t^2}}}{{\sqrt {1 - {t^2}} }}\left( { - \frac{1}{{{t^2}}}} \right) \cr & \frac{{dy}}{{dt}} = - \frac{1}{{\sqrt {1 - {t^2}} }} \cr}

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