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: 24


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

Work Step by Step

Given that $y$ = $sin^{-1}(1-{t})$ on substituting ${1-t}$ = $u$ on applying chain rule derivative $\frac{dy}{dt}$ = $\frac{{dsin^{-1}{u}}}{du}$ $\times$ $\frac{du}{dt}$ or $\frac{dy}{dt}$ = $\frac{1}{\sqrt (1-{u^2})}$ $\times$ $\frac{du}{dt}$ so $\frac{dy}{dt}$ = $\frac{1}{\sqrt (1-{({1-t})^2})}$ $\times$ $\frac{d {(1-t)}}{dt}$ or $\frac{dy}{dt}$ = $\frac{1}{\sqrt( {2t}-{t^2})}$ $\times$ $-1$ The final answer is: $\frac{-1}{\sqrt( {2t}-{t^2})}$
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