The derivative of $\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$, is
Answer & explanation
Correct answer: option 2
Let $y=\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ and $z=\sqrt{1-x^2}$ Then,
$y=\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)=\cos ^{-1}\left(2 x^2-1\right)$
$\Rightarrow y= \begin{cases}2 \cos ^{-1} x &, \text { if } 0 \leq x \leq 1 \\ 2 \pi-2 \cos ^{-1} x & , \text { if }-1 \leq x \leq 0\end{cases}$
∴ $\frac{d y}{d x}= \begin{cases}\frac{-2}{\sqrt{1-x^2}}, & \text { if } 0<x<1 \\ \frac{2}{\sqrt{1-x^2}}, & \text { if }-1<x<0\end{cases}$
$z=\sqrt{1-x^2} \Rightarrow \frac{d z}{d x}=\frac{-x}{\sqrt{1-x^2}}$ for all $x \in(-1,1)$
∴ $\frac{d y}{d z}= \begin{cases}\frac{2}{x}, & \text { if } 0<x<1 \\ \frac{-2}{x}, & \text { if }-1<x<0\end{cases}$
Hence, $\left(\frac{d y}{d z}\right)_{x=1 / 2}=4$