Question

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

Answer 1

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

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

$\Rightarrow \frac{(1-9x^3{)}^2}{(1-9x^3)+(6x\sqrt{x}{)}^2}=\sqrt{x}.g\left(x\right)$

$\Rightarrow \frac{(1-9x^3{)}^2}{(1+9x^3{)}^2}=\sqrt{x}.g\left(x\right)$

$\Rightarrow \frac{(1-9x^3{)}^2+(1+9x^3{)}^2}{(1-9x^3{)}^2-(1+9x^3)}$

$\Rightarrow \frac{2[1+(9x^3{)}^2]-18x^3+18x^3}{1-1-1+(9x^3{)}^2-(9x^3{)}^2-\times 9x^3-2\times 9x^3}$

$\Rightarrow \frac{2(1+(9x^3{)}^2}{-9x^3}$

$\Rightarrow \left[\frac{1+(9x^3{)}^2}{-9x^3}\right]=\sqrt{x}.g\left(x\right)$

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