Question

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

Answer 1

$\mathrm{We}\mathrm{have},y=e^x+e^{-x}$
Differentiate it with respect to x,
$$\begin{gathered} \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{d}{dx}\left(e^x+e^{-x}\right) \\ =\frac{d}{dx}{e}^x+\frac{d}{dx}{e}^{-x} \\ =e^x+e^{-x}\frac{d}{dx}\left(-x\right)\left[\mathrm{Using}\mathrm{chain}\mathrm{rule}\right] \\ =e^x+e^{-x}\left(-1\right) \\ =\left(e^x-e^{-x}\right) \\ =\sqrt{{\left(e^x+e^{-x}\right)}^2-4e^x\times e^{-x}}\left[\because \left(a-b\right)=\sqrt{{\left(a+b\right)}^2-4ab}\right] \\ =\sqrt{y^2-4}\left[\because e^x+e^{-x}=y\right] \end{gathered}$$