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Answer:$\frac{dy}{dx}=\frac{-x}{\sqrt{1-x^{2}}}$

Hint:

\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constant })=0 ; \\ &\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \end{aligned}

Given:

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

Solution:

Put $x=\cos 2\theta$

\begin{aligned} &y=\sin \left[2 \tan ^{-1} \sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}\right] \\ &y=\sin \left[2 \tan ^{-1} \sqrt{\frac{2 \sin ^{2} \theta}{2 \cos ^{2} \theta}}\right] \\ &\text { Since } 1-\cos 2 \theta=2 \sin ^{2} \theta \\ &1+\cos 2 \theta=2 \cos ^{2} \theta \end{aligned}

\begin{aligned} &1+\cos 2 \theta=2 \cos ^{2} \theta \\ &y=\sin \left[2 \tan ^{-1} \sqrt{\tan ^{2} \theta}\right] \\ &\text { Since } \frac{\sin \theta}{\cos \theta}=\tan \theta \\ &y=\sin \left[2 \tan ^{-1}(\tan \theta)\right] \\ &y=\sin [2 \theta] \end{aligned}                                                                            $\left \{ \tan ^{-1}\left ( \tan \theta \right )=\theta \: if\: \theta \varepsilon \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ] \right \}$

$y=\sin \left(2 \times \frac{1}{2} \cos ^{-1} x\right)$                                                                            $\left [ Since,x=\cos 2\theta ,\theta =\frac{1}{2}\cos ^{-1}x \right ]$

\begin{aligned} &y=\sin \left(\cos ^{-1} x\right) \\ &y=\sin \left(\sin ^{-1} \sqrt{1-x^{2}}\right) \\ &y=\sqrt{1-x^{2}} \end{aligned}                                                                            $\left[\sin ^{-1}(\sin \theta) \Rightarrow 0 \text { if } \theta \in\left[\frac{-\pi}{2},\frac{\mathrm{\pi}}{2}\right]\right]$

Differentiating it with respect to $x$

\begin{aligned} \frac{d y}{d x} &=\frac{d}{d x}\left(\sqrt{1-x^{2}}\right) \\ &=\frac{d}{d x}\left(1-x^{2}\right)^{1 / 2} \end{aligned}

As we know,

\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \\ &\qquad \begin{aligned} \frac{d y}{d x} &=\frac{1}{2}\left(1-x^{2}\right)^{\frac{1}{2}} \frac{d}{\partial x}\left(1-x^{2}\right) \\ d y &=\frac{1}{d x} \\ &=\frac{1}{2 \sqrt{1-x^{2}}}(-2 x) \\ \frac{d y}{d x} &=\frac{-x}{\sqrt{1-x^{2}}} \end{aligned} \end{aligned}

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