If $y=\operatorname{Sec}^{-1} x$, then $\frac{d^2 y}{d x^2}=$

• Correct answer
  A

  $\frac{1-2 x^2}{x|x|\left(x^2-1\right)^{\frac{3}{2}}}$
• B

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

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

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

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