Differentiate w.r.t x
$\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}$
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 $\mathrm{y}=\frac{u}{v}$
$\mathrm{dy} / \mathrm{dx}=\frac{\mathrm{v} \times \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \times \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^{2}}$
$=\frac{\left(e^{x}-e^{-x}\right) \times\left(e^{x}-e^{-x}\right)-\left(e^{x}+e^{-x}\right) \times\left(e^{x}+e^{-x}\right)}{\left(e^{x}-e^{-x}\right)^{2}}$
$=\frac{\left(e^{x}-e^{-x}\right)^{2}-\left(e^{x}+e^{-x}\right)^{2}}{\left(e^{x}-e^{-x}\right)^{2}}$
$=\frac{\left(e^{x}-e^{-x}+e^{x}+e^{-x}\right)\left(e^{x}-e^{-x}-e^{x}-e^{-x}\right)}{\left(e^{x}-e^{-x}\right)^{2}}$
$\left(a^{2}-b^{2}=(a-b)(a+b)\right)$
$=\frac{\left(2 \mathrm{e}^{\mathrm{x}}\right)\left(-2 \mathrm{e}^{-\mathrm{x}}\right)}{\left(\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}\right)^{2}}$
$=\frac{-4}{\left(e^{x}-e^{-x}\right)^{2}}$
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