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(A) Let $y = 	an ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^3}ight)$.We can rewrite the expression inside the inverse tangent as $y = 	an ^{-1}\left(\fr

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

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(A) Let $y = 	an ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^3}ight)$.We can rewrite the expression inside the inverse tangent as $y = 	an ^{-1}\left(\frac{2(3 x^{3/2})}{1-(3 x^{3/2})^2}ight)$.Using the identity $2 	an ^{-1}(	heta) = 	an ^{-1}\left(\frac{2	heta}{1-	heta^2}ight)$,we get $y = 2 	an ^{-1}(3 x^{3/2})$.Now,differentiate $y$ with respect to $x$ using the chain rule:$\frac{dy}{dx} = 2 \cdot \frac{1}{1+(3 x^{3/2})^2} \cdot \frac{d}{dx}(3 x^{3/2})$.$\frac{dy}{dx} = 2 \cdot \frac{1}{1+9 x^3} \cdot (3 \cdot \frac{3}{2} x^{1/2})$.$\frac{dy}{dx} = 2 \cdot \frac{1}{1+9 x^3} \cdot \frac{9}{2} \sqrt{x}$.$\frac{dy}{dx} = \frac{9}{1+9 x^3} \cdot \sqrt{x}$.Comparing this with $\sqrt{x} \cdot g(x)$,we find $g(x) = \frac{9}{1+9 x^3}$.

For $x \in \left(0, \frac{1}{4}\right)$,if 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

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