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: 36


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

Work Step by Step

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