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

4

Let $u=\sec ^{-1}\left(\frac{1}{2 x^2-1}\right), v=\sqrt{1-x^2}$

Put $x=\cos \theta \Rightarrow u=2 \theta, v=\sin \theta$

$\Rightarrow \frac{d u}{d \theta}=2, \frac{d v}{d \theta}=\cos \theta \Rightarrow \frac{d u}{d v}=\frac{2}{\cos \theta}=\frac{2}{x}$

$\left.\Rightarrow \frac{d u}{d v}\right|_{x=\frac{1}{2}}=4$

Hence (2) is correct answer.