For $x \in\left(0, \frac{1}{4} ight)$, if the derivative of $ an ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^3} ight)$ is $\sqrt{x} \cdot g(x)$, then $g(x)$ equals

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For $x \in\left(0, \frac{1}{4}ight)$, if the derivative of $	an ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^3}ight)$ is $\sqrt{x} \cdot g(x)$, then $g(x)$ equals

Aicharya · Accepted Answer

$\begin{aligned} & y=	an ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^3}ight)=	an ^{-1}\left\{\frac{2\left(3 x^{\frac{3}{2}}ight)}{1-\left(3 x^{\frac{3}{2}}ight)^2}ight\}=2 	an ^{-1}\left(3 x^{\frac{3}{2}}ight) \ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=2 	imes \frac{1}{1+9 x^3} 	imes 3 	imes \frac{3}{2} \sqrt{x}=\frac{9}{1+9 x^3} \cdot \sqrt{x} \ & \Rightarrow g(x)=\frac{9}{1+9 x^3}\end{aligned}$

For $x \in\left(0, \frac{1}{4}\right)$, if the derivative of $\tan ^{-1}\left(\frac{6 x \sqrt{x}}{1-9…

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