If $y=\log _{e}\left(x+\sqrt{1+x^2}\right)$ then $\frac{d^2 y}{d x^2}$ is equal to:

$\frac{-x}{\left(1+x^2\right)^{\frac{3}{2}}}$

The correct answer is Option (4) → $\frac{-x}{\left(1+x^2\right)^{\frac{3}{2}}}$

$y=\log_e\left(x+\sqrt{1+x^2}\right)$

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

$\frac{d^2y}{dx^2}=\frac{d}{dx}\left((1+x^2)^{-1/2}\right)$

$= -\frac{1}{2}(1+x^2)^{-3/2}\cdot 2x$

$= -\frac{x}{(1+x^2)^{3/2}}$

$\frac{d^2y}{dx^2} = -\frac{x}{(1+x^2)^{3/2}}$