Differentiate w.r.t x:

Let $y=\sqrt{\frac{1+e^{x}}{1-e^{x}}}, u=1+e^{x}, v=1-e^{x}, z=\frac{1+e^{x}}{1-e^{x}}$

Formula :

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

According to the quotient rule of differentiation

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

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

$=\frac{\left(1-e^{x}\right) \times\left(e^{x}\right)-\left(1+e^{x}\right) \times\left(-e^{x}\right)}{\left(1-e^{x}\right)^{2}}$

$=\frac{e^{x}-e^{2 x}+e^{x}+e^{2 x}}{\left(1-e^{x}\right)^{2}}$

$=\frac{2 e^{x}}{\left(1-e^{x}\right)^{2}}$

According to chain rule of differentiation

$\mathrm{dy} / \mathrm{dx}=\frac{\mathrm{dy}}{\mathrm{dz}} \times \frac{\mathrm{dz}}{\mathrm{dx}}$

$=\left[\frac{1}{2} \times\left(\frac{1+\mathrm{e}^{\mathrm{x}}}{1-\mathrm{e}^{\mathrm{x}}}\right)^{\frac{1}{2}-1}\right] \times\left[\frac{2 \mathrm{e}^{\mathrm{x}}}{\left(1-\mathrm{e}^{\mathrm{x}}\right)^{2}}\right]$

$=\left[\frac{\mathrm{e}^{\mathrm{x}}}{1} \times\left(\frac{1+\mathrm{e}^{\mathrm{x}}}{1}\right)^{-\frac{1}{2}}\right] \times\left[\frac{1}{\left(1-\mathrm{e}^{\mathrm{x}}\right)^{2-\frac{1}{2}}}\right]$

$=\left[\mathrm{e}^{\mathrm{x}} \times\left(1+\mathrm{e}^{\mathrm{x}}\right)^{-\frac{1}{2}}\right] \times\left(1-\mathrm{e}^{\mathrm{x}}\right)^{-\frac{3}{2}}$