If \(y=\sqrt{\frac{1-\sin ^{-1}(x)}{1+\sin ^{-1}(x)}}\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) at \(x=0\) and \(y=1\) is

\(
y=\sqrt{\frac{1-\sin ^{-1}(x)}{1+\sin ^{-1}(x)}}
\)
Taking log on both sides, we get
\(
\log y=\frac{1}{2}\left[\log \left(1-\sin ^{-1} x\right)-\log \left(1+\sin ^{-1} x\right)\right]
\)
Differentiating w. r. t. \(x\), we get
\(\frac{1}{y} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{2}\left[\frac{1}{1-\sin ^{-1} x} \cdot \frac{-1}{\sqrt{1-x^2}}-\frac{1}{1+\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}\right] \)
\( \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-y}{2 \sqrt{1-x^2}}\left(\frac{1}{1-\sin ^{-1} x}+\frac{1}{1+\sin ^{-1} x}\right) \)
\( \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{(0,1)}=\frac{1}{2(1)}\left(\frac{1}{1-0}+\frac{1}{1+0}\right)=-1\)