If $u = \tan^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right)$ and $v = 2 \tan^{-1} x$,then $\frac{du}{dv}$ is equal to

  • A
    $4$
  • B
    $1$
  • C
    $1/4$
  • D
    $-1/4$

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