Question

If $\frac{1}{2}\left(e^y-e^{-y}\right)=x$ , prove that
$\frac{dy}{dx}=\frac{1}{\sqrt{x^2+1}}$

Answer 1

$\frac{e^y-e^{-y}}{2}=x\to 1$

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

${(e^y+e^{-y})}^2={(e^y-e^{-y})}^2+4$

$=4x^2+4$

$\therefore e^y+e^{-y}=2\sqrt{x^2+1}$

Differentiate $1$ w.r.t. x

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

$\frac{2\sqrt{x^2+1}}{2}\left(\frac{dy}{dx}\right)=1$

$\therefore \frac{dy}{dx}=\frac{1}{\sqrt{x^2+1}}$