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