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: 26


$\frac{1}{|s|\sqrt( {( 25{s^2})-1})}$

Work Step by Step

Given that $y$ = $sec^{-1}(5s)$ on substituting ${5s}$ = $u$ on applying chain rule derivative $\frac{dy}{ds}$ = $\frac{{dsec^{-1}{u}}}{du}$ $\times$ $\frac{du}{ds}$ or $\frac{dy}{ds}$ = $\frac{1}{{|u|}\sqrt ({u^2}-1)}$ $\times$ $\frac{du}{ds}$ so $\frac{dy}{ds}$ = $\frac{1}{|(5s)|\sqrt ({({5s})^2-1})}$ $\times$ $\frac{d {(5s)}}{ds}$ or $\frac{dy}{ds}$ = $\frac{1}{|5s|\sqrt( {(25{s^2})-1})}$ $\times$ $5$ The final answer is: $\frac{1}{|s|\sqrt( {( 25{s^2})-1})}$
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