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

Question

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