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Solve this following

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

  1. $\frac{3}{1+9 x^{3}}$

  2. $\frac{9}{1+9 x^{3}}$

  3. $\frac{3 x \sqrt{x}}{1-9 x^{3}}$

  4. $\frac{3 x}{1-9 x^{3}}$


Correct Option: , 2

Solution:

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