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For $x\epsilon(0,\frac{1}{4})$ , if the derivative of $\tan^{-1}(\frac{6x \sqrt{x}}{1-9x^{3}})$ is √x .g(x), then g(x) equals

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

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

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

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

Option A

Let y= $tan^{-1}(\frac{6x\sqrt{x}}{1-9x^{3}})$

= $tan^{-1}(\frac{2.(3x^{\frac{3}{2}})}{1-(3x^{\frac{3}{2}})^{2}}]$

= $2\tan^{-1}(3x^{\frac{3}{2}})$

[ $\because 2 tan^{-1}x= \tan^{-1}\frac{2x}{1-x^{2}}$ ]

$\therefore \frac{\text{d}y}{\text{d}x}=2.\frac{1}{1+(3x^{\frac{3}{2}})^{2}}.3\times \frac{3}{2}x^{\frac{1}{2}}$

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

g(x) = $\frac{9}{1+9x^{3}}$

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