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
Correct Option: , 2
Solution:
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