Question

Differentiate $\frac{e^x+e^{-x}}{e^x-e^{-x}}$ with respect to $x$.

Answer 1

quotient rule: $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$

by chain rule: $\frac{d}{dx}{e}^{ax}=e^{ax}\cdot \frac{d}{dx}\left(ax\right)=a.e^{ax}$

$\frac{d}{dx}{e}^{-x}=-e^{-x}$

$\frac{dY}{dx}$

$=\frac{(e^x+e^{-x})\frac{d}{dx}(e^x-e^{-x})-\left(e^x{-e}^{-x}\right)\frac{d}{dx}(e^x+e^{-x})}{{(e^x+e^{-x})}^2}$

$=\frac{{(e^x+e^{-x})}^2-{(e^x-e^{-x})}^2}{{(e^x+e^{-x})}^2}$

$=\frac{4}{{(e^x+e^{-x})}^2}$

$Y=\tan hx$

$\frac{dY}{dx}=\sec h^{2}x=\frac{4}{{(e^x+e^{-x})}^2}$