JEE Main 2017Session, Shift 1MathematicsMediumLimit, Continuity and Differentiability420× since 2002If for x∈(0,14)x \in \left( {0,{1 \over 4}} \right)x∈(0,41), the derivatives of tan−1(6xx1−9x3){\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)tan−1(1−9x36xx) is x.g(x)\sqrt x .g\left( x \right)x.g(x), then g(x)g\left( x \right)g(x) equalsAOption A. 3xx1−9x3{{{3x\sqrt x } \over {1 - 9{x^3}}}}1−9x33xxBOption B. 3x1−9x3{{{3x} \over {1 - 9{x^3}}}}1−9x33xCOption C. 31+9x3{{3 \over {1 + 9{x^3}}}}1+9x33DOption D. 91+9x3{{9 \over {1 + 9{x^3}}}}1+9x39CheckSkip · show solution