If for $x \in\left(0, \frac{1}{4} ight)$, 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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If for $x \in\left(0, \frac{1}{4}ight)$, 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

Let $y=	an ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^3}ight)$ $\begin{aligned} & \quad=	an ^{-1}\left(\frac{3 x \sqrt{x}+3 x \sqrt{x}}{1-3 x \sqrt{x} \cdot 3 x \sqrt{x}}ight) \ & \quad=	an ^{-1}(3 x \sqrt{x})+	an ^{-1} 3 x \sqrt{x} \ & y=2 	an ^{-1}(3 x \cdot \sqrt{x}) \end{aligned}$ Differentiating w.r.t. $x$, we get $\begin{aligned} \frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{2}{1+(3 x \sqrt{x})^2} \cdot \frac{\mathrm{~d}}{\mathrm{~d} x}(3 x \cdot \sqrt{x}) \ & =\frac{2}{1+9 x^3} \cdot \frac{9}{2} \sqrt{x} \ \frac{\mathrm{~d} y}{\mathrm{~d} x} & =\frac{9 \sqrt{x}}{1+9 x^3} \end{aligned}$ $	herefore \quad$ Comparing with $\frac{\mathrm{d} y}{\mathrm{~d} x}=\sqrt{x} \cdot \mathrm{~g}(x)$ $	herefore \quad \mathrm{g}(x)=\frac{9}{1+9 x^3}$

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

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