The derivative of

${\sec }^{-1}\left(\frac{1}{2{\mathrm{x}}^2-1}\right)$ w.r.t. $\sqrt{1-{\mathrm{x}}^2}$ at $\mathrm{x}=\frac{1}{2}$ is

The given functions are $f\left(x\right)=se{c}^{-1}\left(\frac{1}{2x^2-1}\right)$ and $g\left(x\right)=\sqrt{1-x^2}$

Substitute $x=\cos \theta$

The given functions are $se{c}^{-1}\left(\frac{1}{\cos 2\theta }\right)=2\theta =2{\cos }^{-1}x$ and $\sqrt{1-x^2}=\sin \theta$

The derivative of $2\theta$ with resect to $\sin \theta$

hence, it is $\frac{2}{\cos \theta }=\frac{2}{x}$

Substiute $x=\frac{1}{2}$

it implies that, the derivative of $f\left(x\right)=se{c}^{-1}\left(\frac{1}{2x^2-1}\right)$ with respect to $g\left(x\right)=\sqrt{1-x^2}$ at $x=\frac{1}{2}$ is $\boxed{\mathbf{4}}$