Question: If for $x \in\left(0, \frac{1}{4}\right)$, the derivative of $\tan ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^{3}}\right)$ is $\sqrt{x} \cdot g(x)$, then $g(x)$ equals
$\frac{3}{1+9 x^{3}}$
$\frac{9}{1+9 x^{3}}$
$\frac{3 x \sqrt{x}}{1-9 x^{3}}$
$\frac{3 x}{1-9 x^{3}}$
Correct Option: , 2
Solution:
