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If $y=\sin \left[2 \tan ^{-1}\left\{\sqrt{\frac{1-x}{1+x}}\right\}\right]$, find $\frac{d y}{d x}$


$y=\sin \left[2 \tan ^{-1}\left\{\sqrt{\frac{1-x}{1+x}}\right\}\right]$

Put $x=\cos 2 \theta$

$y=\sin \left[2 \tan ^{-1}\left\{\sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}\right\}\right]$

Using $2 \cos ^{2} \theta-1=\cos 2 \theta$ and $1-2 \sin ^{2} \theta=\cos 2 \theta$

$y=\sin \left[2 \tan ^{-1}\left\{\sqrt{\frac{2 \sin ^{2} \theta}{2 \cos ^{2} \theta}}\right\}\right]$

$y=\sin \left[2 \tan ^{-1}(\tan \theta)\right]$

$y=\sin (2 \theta)$

$y=\sin \left[\frac{2}{2} \times \cos ^{-1} x\right]$

Using $\cos ^{-1} x=\sin ^{-1} \sqrt{1-x^{2}}$

$y=\sin \left[\sin ^{-1} \sqrt{1-x^{2}}\right]$


Differentiating w.r.t $x$ we get


$\frac{d y}{d x}=\frac{1}{2 \sqrt{1-x^{2}}} \frac{d}{d x}\left(1-x^{2}\right)$

$\frac{d y}{d x}=-\frac{2 x}{2 \sqrt{1-x^{2}}}$

$\frac{d y}{d x}=-\frac{x}{\sqrt{1-x^{2}}}$

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