If $y=\tan ^2\left(\operatorname{Cos}^{-1} \sqrt{\frac{1+x^2}{2}}\right)$, then $\frac{d y}{d x}=$

• A

  $-\frac{4 x}{\left(1-x^2\right)^2}$
• B

  $\frac{4 x}{\left(1+x^2\right)^2}$
• Correct answer
  C

  $-\frac{4 x}{\left(1+x^2\right)^2}$
• D

  $-\frac{4 x}{1+x^2}$

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