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

\(y=\sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}\)
Differentiating w.r.t. \(x\), we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x} =\frac{\mathrm{d}}{\mathrm{~d} x}\left(\sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}\right) \)
\( =\frac{1}{2 \sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}} \cdot \frac{\mathrm{~d}}{\mathrm{~d} x}\left(\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}\right) \)
\( =\frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}}\)
\(\frac{\left(1+\sin ^{-1} x\right) \cdot \frac{d}{d x}\left(1-\sin ^{-1} x\right)-\left(1-\sin ^{-1} x\right) \frac{d}{d x}\left(1+\sin ^{-1} x\right)}{\left(1+\sin ^{-1} x\right)^2} \)
\( =\frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}} \cdot \frac{\left(1+\sin ^{-1} x\right) \times \frac{-1}{\sqrt{1-x^2}}-\left(1-\sin ^{-1} x\right) \cdot \frac{1}{\sqrt{1-x^2}}}{\left(1+\sin ^{-1} x\right)^2}\)
\(=\frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}} \times \frac{\frac{1}{\sqrt{1-x^2}}\left[-1-\sin ^{-1} x-1+\sin ^{-1} x\right]}{\left(1+\sin ^{-1} x\right)^2} \)
\( \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{(-1) \sqrt{1+\sin ^{-1} x}}{\sqrt{1-\sin ^{-1} x} \cdot\left(\sqrt{1-x^2}\right)\left(1+\sin ^{-1} x\right)^2} \)
\( \left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)_{\mathrm{at} x=0}=\frac{-1 \sqrt{1+\sin ^{-1} 0}}{\left(\sqrt{1-\sin ^1 0}\right)\left(\sqrt{1-0^2}\right)\left(1+\sin ^{-1} 0\right)^2} \)
\( =\frac{-1}{(1)(1)(1+0)} \)
\( \therefore \quad\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{\mathrm{at} x=0}=-1\)