Let y = ${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$ = ${\tan ^{ - 1}}\left[ {{{2.\left( {3{x^{{3 \over 2}}}} \right)} \over {1 - {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}} \right]$ = 2${\tan ^{ - 1}}\left( {3{x^{{3 \over 2}}}} \right)$ $\therefore$ ${{dy} \over {dx}} = 2.{1 \over {1 + {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}.3 \times {3 \over 2}{\left( x \right)^{{1 \over 2}}}$ = ${9 \over {1 + 9{x^3}}}.\sqrt x$ $\therefore$ g(x) = ${9 \over {1 + 9{x^3}}}$