Question

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

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

Answer 1

The correct option is A $\frac{9}{1+9x^3}$

Since
$2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$
So
${\tan }^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right)=2{\tan }^{-1}\left(3x^{\frac{3}{2}}\right)=y$
$\therefore \frac{dy}{dx}=\frac{9\sqrt{x}}{1+9x^3}=\sqrt{x}.g\left(x\right)$
$\Rightarrow g\left(x\right)=\frac{9}{1+9x^3}$