JEE Math Practice 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
- A.$${{{3x\sqrt x } \over {1 - 9{x^3}}}}$$
- B.$${{{3x} \over {1 - 9{x^3}}}}$$
- C.$${{3 \over {1 + 9{x^3}}}}$$
- D.$${{9 \over {1 + 9{x^3}}}}$$
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