If $y = \sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$,then $\frac{d y}{d x}$ is equal to

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

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