Differentiate w.r.t x

Let $y=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}, u=e^{x}+e^{-x}, v=e^{x}-e^{-x}$

Formula :

$\frac{\mathrm{d}\left(\mathrm{e}^{\mathrm{x}}\right)}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}$

According to the quotient rule of differentiation

If $\mathrm{y}=\frac{u}{v}$

$\mathrm{dy} / \mathrm{dx}=\frac{\mathrm{v} \times \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \times \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^{2}}$

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

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

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

$\left(a^{2}-b^{2}=(a-b)(a+b)\right)$

$=\frac{\left(2 \mathrm{e}^{\mathrm{x}}\right)\left(-2 \mathrm{e}^{-\mathrm{x}}\right)}{\left(\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}\right)^{2}}$

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