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If $u=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ and $v=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$ then $\frac{d u}{d v}$ is
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$1$
Here, $u=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$
$\left.u=2 \tan ^{-1} x \quad \because \because 2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right]$
$\begin{aligned} & \Rightarrow \frac{d u}{d x}=2 \times \frac{1}{1+x^2} \\ & \text { and } v=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\end{aligned}$
$\begin{aligned} y & =2 \tan ^{-1} x \quad\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right] \\ \Rightarrow \quad & \quad \frac{d v}{d x}=\frac{2}{1+x^2}\end{aligned}$
From Eqs. (i) and (ii), we get
$\frac{d v}{d x}=1$
$\left.u=2 \tan ^{-1} x \quad \because \because 2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right]$
$\begin{aligned} & \Rightarrow \frac{d u}{d x}=2 \times \frac{1}{1+x^2} \\ & \text { and } v=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\end{aligned}$
$\begin{aligned} y & =2 \tan ^{-1} x \quad\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right] \\ \Rightarrow \quad & \quad \frac{d v}{d x}=\frac{2}{1+x^2}\end{aligned}$
From Eqs. (i) and (ii), we get
$\frac{d v}{d x}=1$
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