Question

$y=e^x+e^{-x}$ prove that $\frac{dy}{dx}=\sqrt{y^2-4}$

Answer 1

$\frac{dy}{dx}=e^x-e^{-x}$

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

$y^2-4=e^{2x}+e^{-2x}-2$

$\sqrt{y^2-4}=|e^x-e^{-x}|$

and we know that for $x>0\sqrt{y^2-4}=e^x-e^{-x}=\frac{dy}{dx}$

and for $x<0\sqrt{y^2-4}=-(e^x-e^{-x})=\frac{-dy}{dx}$