Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.4 Derivatives of Exponential Functions - 4.4 Exercises - Page 232: 19


$\displaystyle \frac{x(e^{x}-e^{-x})-(e^{x}+e^{-x})}{x^{2}}$ or $\displaystyle \frac{e^{x}(x-1)-e^{-x}(x+1)}{x^{2}}$

Work Step by Step

$y=\displaystyle \frac{e^{x}+e^{-x}}{x}$ $\color{blue}{\text{... Use the quotient rule,}\\ \displaystyle \frac{dy}{dx}=\left[\frac{u(x)}{v(x)}\right]^{\prime} =\displaystyle \frac{v(x)\cdot u^{\prime}(x)-u(x)\cdot v^{\prime}(x)}{[v(x)]^{2}}|}$, with $\displaystyle \frac{d}{dx}(e^{x})=e^{x},\ \ \frac{d}{dx}(e^{-x})=-e^{-x},\ \ \ \displaystyle \frac{d}{dx}(x)=1$ $\displaystyle \frac{dy}{dx} =\frac{x(e^{x}-e^{-x})-(e^{x}+e^{-x})}{x^{2}}$ or $\displaystyle \frac{e^{x}(x-1)-e^{-x}(x+1)}{x^{2}}$
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