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

$1-y^2$

The correct answer is Option (3) → $1-y^2$

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

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

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

$=e^{2x}+e^{-2x}-2-e^{2x}-e^{-2x}-2$

$=\frac{-4}{(e^x-e^{-x})^2}=1-y^2$