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

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(B) Let $F(x) = 	an^{-1} \left( \frac{6x\sqrt{x}}{1 - 9x^3} ight)$.We can rewrite the argument as $\frac{2(3x^{3/2})}{1 - (3x^{3/2})^2}$.Using the

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

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(B) Let $F(x) = 	an^{-1} \left( \frac{6x\sqrt{x}}{1 - 9x^3} ight)$.We can rewrite the argument as $\frac{2(3x^{3/2})}{1 - (3x^{3/2})^2}$.Using the identity $2	an^{-1}(	heta) = 	an^{-1} \left( \frac{2	heta}{1 - 	heta^2} ight)$,we get $F(x) = 2	an^{-1}(3x^{3/2})$.Now,differentiate $F(x)$ with respect to $x$:$F'(x) = 2 \cdot \frac{1}{1 + (3x^{3/2})^2} \cdot \frac{d}{dx}(3x^{3/2})$.$F'(x) = 2 \cdot \frac{1}{1 + 9x^3} \cdot (3 \cdot \frac{3}{2} x^{1/2})$.$F'(x) = 2 \cdot \frac{1}{1 + 9x^3} \cdot \frac{9}{2} \sqrt{x} = \frac{9\sqrt{x}}{1 + 9x^3}$.Given $F'(x) = \sqrt{x} \cdot g(x)$,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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