Find $\large \frac {dy}{dx}$ in the following: $y = sec^{-1} \left(\large\frac{1}{2x^2 - 1}\right), 0 < x < \; { \frac{1}{\sqrt2} } \\ $

Question

$\begin{array}{1 1} \large\frac{dy}{dx}=\frac{-2}{\sqrt{1+x^2}} \\\large\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}} \\ \large\frac{dy}{dx}=\frac{-2}{\sqrt{1-x^2}} \\ \large\frac{dy}{dx}=\frac{2}{\sqrt{1-x^3}}\end{array}$