If $y=\log \left(x+\sqrt{1+x^2}\right)$ then the value of $y_2(0)$ is :

-1

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

$\Rightarrow y_1=\frac{1}{x+\sqrt{x^2+1}}\left(1+\frac{1}{2 \sqrt{x^2+1}}(2 x)\right)$

$\Rightarrow y_1=\frac{1}{\sqrt{x^2+1}} \Rightarrow y_2=\frac{-x}{\left(1+x^2\right)^{3 / 2}}$

$\Rightarrow y_2(0)=0$

Hence (1) is correct answer.