Question

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

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

Answer 1

Answer: (D)

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

Answer 2

$f \left(x\right) = 2 \tan^{-1}\left(3x\sqrt{x}\right)$
For x $\in\left(0, \frac{1}{4}\right)$
$f'\left(x\right) = \frac{9\sqrt{x}}{1+9x^{3}}$
$g\left(x\right) = \frac{9}{1+9x^{3}}$