Question

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

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

Answer 1

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

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

and $x\in (0,\frac{1}{9})$

$y={\tan }^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right)=y={\tan }^{-1}\left(\frac{2\left(3x^{3/2}\right)}{1-{\left(3x^3\right)}^{3/2}}\right)=2{\tan }^{-1}{\left(3x^{3/2}\right)}^{}\left[2{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)\right]$

$\frac{dy}{dx}2{\tan }^{-1}{\left(3x^{3/2}\right)}^{}=2x\frac{1}{1+9x^3}\times 3\times \frac{3}{2}\times x^{1/2}=\frac{9}{1+x^3\sqrt{3}}=\frac{9}{1+9x^3}$