यदि $y = \sin^{-1} \left( \frac{2x}{1 + x^2} \right)$,जहाँ $0 < x < 1$ और $0 < y < \frac{\pi}{2}$ है,तो $\frac{dy}{dx} = $

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

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