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: 35


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

Work Step by Step

Given that $y$ = $cosec^{-1}{{e}^{t}}$, on substituting ${{e}^{t}}$ = $u$ and applying the chain rule derivative, we get: $\frac{dy}{dt}$ = $\frac{{dcosec^{-1}{u}}}{du}$ $\times$ $\frac{du}{dt}$ or $\frac{dy}{dt}$ = $-\frac{1}{{|u|}\sqrt {({u^2}-1)}}$ $\times$ $\frac{du}{dt}$ so $\frac{dy}{dt}$ = $-\frac{1}{|({e}^{t})|\sqrt {({{e}^{2t}}-1})}$ $\times$ $\frac{d {{e}^{t}}}{dt}$ or $\frac{dy}{dt}$ = $-\frac{1}{|{e}^{t}|\sqrt{( {({e}^{2t})-1})}}$ $\times$ ${e}^{t}$ The final answer is: = $-\frac{1}{\sqrt{( {({e}^{2t})-1})}}$
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