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 - Page 264: 17


tanh(ln x) = $\frac{x^{2} - 1}{x^{2} +1}$

Work Step by Step

Start with the definition of $\tanh{(x)} = \frac{\sinh{(\theta)}}{\cosh{(\theta)}}$; in this case $\theta = \ln{x}$ $tanh(ln x) = \frac{sinh(ln x)}{cosh(ln x)}$ Switch for the hyperbolical sine and cosine definition: $\sinh(\theta) = \frac{e^{\theta}-e^{-\theta}}{2}$ $\cosh{(\theta)} = \frac{e^{\theta}+e^{-\theta}}{2}$ Therefore: $tanh(ln x) = \frac{(e^{ln x} - e^{-ln x})}{(e^{ln x} + e^{-lnx})}$ $tanh(ln x) =\frac{e^{ln x} - e^{-lnx}}{e^{lnx} + e^{-ln x}}$ Recall that: $\theta^{\log_{\theta}\alpha} = \alpha$ $e^{\ln \theta} = \theta$ Therefore: $tanh(ln x) =\frac{x - x^{-1}}{x + x^{-1}}$ Multiply by $\frac{x}{x}$ to eliminate the $x^{-1}$ $tanh(ln x) =\frac{x - x^{-1}}{x + x^{-1}}$ $\times$ $\frac{x}{x}$ = $\frac{x^{2} -1}{x^{2} + 1}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.