Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.7 Hyperbolic Functions - 6.7 Exercises - Page 490: 38


$f'(t)=\dfrac{2\cosh t}{(1-\sinh t)^2}$

Work Step by Step

Given: $f(t)=\dfrac{1+\sinh t}{1-\sinh t}$ Now,$f'(t)=\dfrac{d}{dt}[\dfrac{1+\sinh t}{1-\sinh t}]$ Apply the quotient rule, we get $f'(t)=\dfrac{\cosh t(1-\sinh t)-(1+\sinh t)(-\cosh t)}{(1-\sinh t)^2}$ or, $f'(t)=\dfrac{\cosh t-\cosh t\sinh t+\cosh t+\cosh t\sinh t}{(1-\sinh t)^2}$ or, $f'(t)=\dfrac{2\cosh t}{(1-\sinh t)^2}$
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