If $y=\sqrt{\log \left(x^2+1\right)+\sqrt{\log \left(x^2+1\right)+\sqrt{\log \left(x^2+1\right)+\ldots \infty}}},|x| < 1$, then $\frac{d y}{d x}=$

• A

  $\frac{x^2+1}{2 y-1}$
• B

  $\frac{2 x}{2 y-1}$
• C

  $\frac{1}{\left(x^2+1\right)(2 y-1)}$
• Correct answer
  D

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

Hence, the correct answer is option: 'D'.