Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.7 - Hyperbolic Functions - Exercises 7.7 - Page 430: 8

Answer

$e^{-3x}$

Work Step by Step

Use hyperbolic function formula as follows: $ \cosh x= \dfrac{e^x+e^{-x}}{2}$ and $ \sinh x= \dfrac{e^x-e^{-x}}{2}$ Need to solve: $\cosh (3x) - \sinh (3x)$ This implies: $\cosh (3x) - \sinh (3x)=\dfrac{e^{3x}+e^{-3x}}{2}-\dfrac{e^{3x}-e^{-3x}}{2}$ Hence, $\cosh (3x) - \sinh (3x)=e^{-3x}$
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