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If $y=\sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$, then $\frac{d y}{d x}$ is equal to
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$\frac{-x}{\sqrt{1-x^2}}$
$\begin{aligned} & y=\sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right) \\ & \Rightarrow y=\sin \left(2 \tan ^{-1}\left(\cot \frac{\theta}{2}\right)\right)[\text { let } x=\cos \theta] \\ & \Rightarrow y=\sin \left(2 \tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right\}\right) \\ & \Rightarrow y=\sin \left(2\left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right) \\ & \Rightarrow y=\sin (\pi-\theta)=\sin \theta=\sqrt{1-x^2}\end{aligned}$
$\Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{1-x^2}} \times(-2 x)=\frac{-x}{\sqrt{1-x^2}}$
$\Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{1-x^2}} \times(-2 x)=\frac{-x}{\sqrt{1-x^2}}$
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