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

Question

If for $x \in \left( {0,{1 \over 4}} \right)$, the derivatives of

${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$ is $\sqrt x .g\left( x \right)$, then $g\left( x \right)$ equals

${{{3x\sqrt x } \over {1 - 9{x^3}}}}$

${{{3x} \over {1 - 9{x^3}}}}$

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

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

Solution

Please login to view the detailed solution steps...

Go to DASH