University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.9 - Inverse Trigonometric Functions - Exercises - Page 181: 29


$-\frac{|t|}{{t}\sqrt( { {1}-{t^2}})}$

Work Step by Step

Given that $y$ = $sec^{-1}{\frac{1}{t}}$ where t lies between $0$ and $1$ on substituting ${\frac{1}{t}}$ = $u$ on applying chain rule derivative $\frac{dy}{dt}$ = $\frac{{dsec^{-1}{u}}}{du}$ $\times$ $\frac{du}{dt}$ or $\frac{dy}{dt}$ = $\frac{1}{{|u|}\sqrt ({u^2}-1)}$ $\times$ $\frac{du}{dt}$ so $\frac{dy}{dt}$ = $\frac{1}{|(\frac{1}{t})|\sqrt ({({\frac{1}{t}})^2-1})}$ $\times$ $\frac{d {(\frac{1}{t})}}{dt}$ or $\frac{dy}{dt}$ = $\frac{1}{|\frac{1}{t}|\sqrt( {(\frac{1}{t^2})-1})}$ $\times$ $-\frac{1}{t^2}$ The final answer is: $-\frac{|t|}{{t}\sqrt( { {1}-{t^2}})}$
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