Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.11 - Hyperbolic Functions - 3.11 Exercises: 35

Answer

$$G'(t)=\frac{t^2+1}{2t^2}$$

Work Step by Step

$G'(t)=\frac{d}{dt}\sinh {(\ln t)}$ Using the chain rule: $G'(t)=\frac{d\sinh {(\ln t)}}{d\ln t} \times \frac{d\ln t}{dt}$ $=\cosh {(\ln t)} \times \frac{1}{t}$ $=\frac{\cosh {(\ln t)}}{t}$ Recall that: $$\cosh x=\frac{e^x+e^{-x}}{2}$$ Thus: $G'(t)=\frac{\frac{e^{\ln t}+e^{-\ln t}}{2}}{t}$ $=\frac{t+\frac{1}{t}}{2t}$ $=\frac{\frac{t^2+1}{t}}{2t}$ $=\frac{t^2+1}{2t^2}$
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