$\mathrm{If} \mathrm{y}=\frac{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}} \mathrm{then} \frac{\mathrm{dy}}{\mathrm{dx}}=$

Given $y=\frac{e^x-e^{-x}}{e^x+e^{-x}}$

Use quotient rule, Differentiate both sides with  respect to $x$

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right)\\ & =& \frac{\left(e^x+e^{-x}\right)\left(e^x+e^{-x}\right)-\left(e^x-e^{-x}\right)\left(e^x-e^{-x}\right)}{{\left(e^x+e^{-x}\right)}^2}\\ & =& \frac{4}{{\left(e^x+e^{-x}\right)}^2}\\ & =& \boxed{\mathbf{1}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}\end{array}$