If $y = log (x + \sqrt{x^2+a^2})$ then $\frac{dy}{dx}=$

$\frac{1}{\sqrt{x^2+a^2}}$

The correct answer is Option (1) → $\frac{1}{\sqrt{x^2+a^2}}$

$y = \log (x + \sqrt{x^2+a^2})$

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

$=\frac{(\sqrt{x^2+a^2}+x}{(x+\sqrt{x^2+a^2})}×\frac{1}{\sqrt{x^2+a^2}}$

$=\frac{1}{\sqrt{x^2+a^2}}$