Given that,
$$
y=\tan ^{-1}\left\{\frac{x}{1+\sqrt{1-x^2}}\right\}+\sin \left\{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right\}
$$
Let $x=\cos 2 \theta$
$$
\begin{aligned}
& y=\tan ^{-1}\left\{\frac{\cos 2 \theta}{1+\sqrt{\sin ^2 2 \theta}}\right\}+\sin \left\{2 \tan ^{-1} \sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}\right\} \\
& \Rightarrow y=\tan ^{-1}\left\{\frac{\cos 2 \theta}{1+\sin 2 \theta}\right\} \\
& +\sin \left\{2 \tan ^{-1} \sqrt{\frac{2 \sin ^2 \theta}{2 \cos ^2 \theta}}\right\} \\
& \Rightarrow y=\tan ^{-1}\left\{\frac{\cos ^2 \theta-\sin ^2 \theta}{(\cos \theta+\sin \theta)^2}\right\} \\
& +\sin \left\{2 \tan ^{-1}(\tan \theta)\right\} \\
& \Rightarrow y=\tan ^{-1}\left\{\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}\right\}+\sin 2 \theta \\
& y=\tan ^{-1}\left\{\frac{\cos \theta\left(1-\frac{\sin \theta}{\cos \theta}\right)}{\cos \theta\left(1+\frac{\sin \theta}{\cos \theta}\right)}\right\}+\sin 2 \theta \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad y=\tan ^{-1}\left\{\frac{1-\tan \theta}{1+\tan \theta}\right\}+\sin 2 \theta \\
& \Rightarrow \quad y=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\theta\right)\right)+\sin 2 \theta \\
& \Rightarrow \quad y=\frac{\pi}{4}-\theta+\sqrt{1-\cos ^2} 2 \theta \\
& \Rightarrow \quad \frac{d y}{d x}=\frac{1}{2} \cdot \frac{1}{\sqrt{1-r^2}}+\frac{1}{2 \sqrt{1-r^2}}(2 x)
\end{aligned}
$$
Now, on differentiating w.r.t. $x$, we get
$$
\begin{aligned}
& \frac{d y}{d x}=\frac{1}{2} \cdot \frac{1}{\sqrt{1-x^2}}+\frac{1}{2 \sqrt{1-x^2}}(-2 \mathrm{x}) \\
& \therefore \quad \frac{d y}{d x}=\frac{1-2 x}{2 \sqrt{1-x^2}}
\end{aligned}
$$