If $y=\log \left( x+\sqrt{{{x}^{2}}-1} \right)$, then $\dfrac{dy}{dx}=?$
$\dfrac{-1}{\sqrt{{{x}^{2}}-1}}$
$\dfrac{1}{\sqrt{{{x}^{2}}-1}}$
$\dfrac{1}{(1-{{x}^{2}})}$
Solution
$y = log(x + \sqrt{x^2 - 1})$
$\dfrac{dy}{dx} = \dfrac{[1 + \dfrac{1}{2} (x^2 - 1)^{-1/2}.2x]}{x + \sqrt{}x^2 - 1}$
$\dfrac{dy}{dx} = \left[\dfrac{1 + \dfrac{x}{\sqrt{x^2 - 1}}}{x + \sqrt{x^2 - 1}}\right] = \dfrac{1}{\sqrt{x^2 - 1}}$