Question

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

• A. $\frac{3\mathrm{x}\sqrt{\mathrm{x}}}{1-9{\mathrm{x}}^3}$
• B. $\frac{3\mathrm{x}}{1-9{\mathrm{x}}^3}$
• C. $\frac{3}{1+9{\mathrm{x}}^3}$
• D. $\frac{9}{1+9{\mathrm{x}}^3}$

Answer 1

The correct option is D

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

Explanation for the correct answer:

Step 1: Finding the derivative

Let, $\mathrm{y}={\tan }^{-1}\left(\frac{6\mathrm{x}\sqrt{\mathrm{x}}}{1-9{\mathrm{x}}^3}\right)$

Differentiate the given function with respect to $x$

$\begin{array}{rcl}\frac{\mathrm{dy}}{\mathrm{dx}}& =& \frac{1}{1+{\displaystyle \frac{36{\mathrm{x}}^2.\mathrm{x}}{{(1-9{\mathrm{x}}^3)}^2}}}.\frac{(1-9{\mathrm{x}}^3)\left(6\times {\displaystyle \frac{3}{2}}{\mathrm{x}}^{{\displaystyle \frac{1}{2}}}\right)\left(6{\mathrm{x}}^{{\displaystyle \frac{3}{2}}}27{\mathrm{x}}^2\right)}{{(1-9{\mathrm{x}}^3)}^2}\left[\because \frac{\mathrm{d}}{\mathrm{dx}}\left({\tan }^{-1}\mathrm{x}\right)=\frac{1}{1+{\mathrm{x}}^2}\right]\\ & =& \frac{{\left(1-9{\mathrm{x}}^3\right)}^2}{{\left(1-9{\mathrm{x}}^3\right)}^2+36{\mathrm{x}}^3}.\frac{\left(1-9{\mathrm{x}}^3\right)\left(9{\mathrm{x}}^{{\displaystyle \frac{1}{2}}}\right)6\sqrt{\mathrm{x}}.\mathrm{x}\left(27{\mathrm{x}}^2\right)}{{(1-9{\mathrm{x}}^3)}^2}\left[\because {\mathrm{x}}^{\frac{1}{2}}=\sqrt{\mathrm{x}},{\mathrm{x}}^{\frac{3}{2}}=\mathrm{x}\sqrt{\mathrm{x}}\right]\\ & =& \frac{\sqrt{\mathrm{x}}\left(9\right(1-9{\mathrm{x}}^3)+6\mathrm{x}(27{\mathrm{x}}^2)}{{(1-9{\mathrm{x}}^3)}^2+36{\mathrm{x}}^3}\\ & =& \frac{\sqrt{\mathrm{x}}(9-81{\mathrm{x}}^3+162{\mathrm{x}}^3)}{\left(1+81{\mathrm{x}}^6-18{\mathrm{x}}^3\right)+36{\mathrm{x}}^3}\\ & =& \frac{\sqrt{\mathrm{x}}(9+81{\mathrm{x}}^3)}{1+81{\mathrm{x}}^6+18{\mathrm{x}}^3}\end{array}$

Step 2: Equating the derivative to $\begin{array}{rcl}& & \sqrt{\mathrm{x}}.\mathrm{g}\left(\mathrm{x}\right)\end{array}$

$\begin{array}{rcl}\therefore \mathrm{g}\left(\mathrm{x}\right)\times \sqrt{x}& =& \frac{\sqrt{\mathrm{x}}(9+81{\mathrm{x}}^3)}{1+81{\mathrm{x}}^6+18{\mathrm{x}}^3}\\ & \Rightarrow & g\left(x\right)=\frac{9(1+9{\mathrm{x}}^3)}{{(1+9{\mathrm{x}}^3)}^2}\\ & \Rightarrow & \mathrm{g}\left(\mathrm{x}\right)=\frac{9}{(1+9{\mathrm{x}}^3)}\end{array}$

Hence, the correct answer is an option (D).