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: 25


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

Work Step by Step

Given that $y$ = $sec^{-1}(2s+1)$ on substituting ${2s+1}$ = $u$ 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 $\frac{dy}{ds}$ = $\frac{1}{|(2s+1)|\sqrt ({({2s+1})^2-1})}$ $\times$ $\frac{d {(2s+1)}}{ds}$ or $\frac{dy}{ds}$ = $\frac{1}{|2s+1|\sqrt( {(4 {s^2})+4s})}$ $\times$ $2$ The final answer is: $\frac{1}{|2s+1|\sqrt( {( {s^2})+s})}$
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