If for $x \in \left( 0, \frac{1}{4} \right)$,the derivative of $\tan^{-1} \left( \frac{6x\sqrt{x}}{1 - 9x^3} \right)$ is $\sqrt{x} \cdot g(x)$,then $g(x)$ equals:

  • A
    $\frac{3}{1 + 9x^3}$
  • B
    $\frac{9}{1 + 9x^3}$
  • C
    $\frac{3x\sqrt{x}}{1 - 9x^3}$
  • D
    $\frac{3}{1 - 9x^3}$

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