The derivative of cosec^-1$\left(\frac{1}{2x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$, is

-4

Let $y=cosec^{-1}\left(\frac{1}{2 x^2-1}\right)$ and $z=\sqrt{1-x^2}$

We have,

$y=cosec^{-1}\left(\frac{1}{2 x^2-1}\right)$

$\Rightarrow y=\sin ^{-1}\left(2 x^2-1\right)=\frac{\pi}{2}-\cos ^{-1}\left(2 x^2-1\right)$

$\Rightarrow y= \begin{cases}\frac{\pi}{2}-2 \cos ^{-1} x, & \text { if } 0 \leq x \leq 1 \\ -\frac{3 \pi}{2}+2 \cos ^{-1} x, & \text { if }-1 \leq x \leq 0\end{cases}$

$\Rightarrow \frac{d y}{d x}= \begin{cases}\frac{2}{\sqrt{1-x^2}}, & \text { if } 0<x<1 \\ \frac{-2}{\sqrt{1-x^2}}, & \text { if }-1<x<0\end{cases}$

and, $z=\sqrt{1-x^2} \Rightarrow \frac{d z}{d x}=\frac{-x}{\sqrt{1-x^2}}$ for all $x \in(-1,1)$

∴  $\frac{d y}{d z}=\frac{d y / d x}{d z / d x}=\left\{\begin{aligned}-\frac{2}{x}, & \text { if } 0<x<1 \\ \frac{2}{x}, & \text { if }-1<x<0\end{aligned}\right.$

$\Rightarrow\left(\frac{d y}{d z}\right)_{x=1 / 2}=-4$