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

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

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

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