જો $u=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ અને $v=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$ હોય,તો $\frac{d u}{d v}$ શોધો.

  • A
    $2$
  • B
    $\frac{1-x^2}{1+x^2}$
  • C
    $1$
  • D
    $\frac{1}{2}$

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