यदि u = \(\rm \sin^{−1}\left(\frac{2 x}{1+x^2}\right)\) और v = \(\rm \tan^{−1}\left(\frac{2x}{1−x^2}\right)\), तब \(\frac{du}{dv}\) है: 

1
\(\frac{1}{2}\)
2
x
3
\(\rm \frac{1−x^2}{1+x^2}\)
4
1

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