Question

If for $xϵ(0,\frac{1}{4})$, 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. $\frac{3}{1+9x^3}$
• B. $\frac{9}{1+9x^3}$
• C. $\frac{3x\sqrt{x}}{1-9x^3}$
• D. $\frac{3x}{1-9x^3}$

Answer 1

Answer: (B)

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

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

Now differentiating w.r.t. $x$ we get

$\frac{d\left[{\tan }^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right)\right]}{dx}=\frac{9\sqrt{x}}{1+9x^3}$

Using $\frac{d\left({\tan }^{-1}x\right)}{dx}=\frac{1}{1+x^2}$

Hence $g\left(x\right)=\frac{9}{1+9x^3}$

Hence option $B$ is correct.