If for x in left(0, frac{1}{4}right),the deri | vedclass

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(D) Let $y = 	an^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}ight)$.We can rewrite the argument of $	an^{-1}$ as follows:$y = 	an^{-1}\left(\frac{2(3x\sqr

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

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(D) Let $y = 	an^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}ight)$.We can rewrite the argument of $	an^{-1}$ as follows:$y = 	an^{-1}\left(\frac{2(3x\sqrt{x})}{1-(3x\sqrt{x})^2}ight)$.Using the formula $2	an^{-1}(	heta) = 	an^{-1}\left(\frac{2	heta}{1-	heta^2}ight)$,where $	heta = 3x\sqrt{x} = 3x^{3/2}$,we get:$y = 2	an^{-1}(3x^{3/2})$.Differentiating with respect to $x$ using the chain rule:$\frac{dy}{dx} = 2 \cdot \frac{1}{1+(3x^{3/2})^2} \cdot \frac{d}{dx}(3x^{3/2})$.$\frac{dy}{dx} = \frac{2}{1+9x^3} \cdot (3 \cdot \frac{3}{2} x^{1/2})$.$\frac{dy}{dx} = \frac{2}{1+9x^3} \cdot \frac{9}{2} \sqrt{x} = \frac{9\sqrt{x}}{1+9x^3}$.Given $\frac{dy}{dx} = \sqrt{x} \cdot g(x)$,we have $\sqrt{x} \cdot g(x) = \frac{9\sqrt{x}}{1+9x^3}$.Therefore,$g(x) = \frac{9}{1+9x^3}$.

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

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