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

If $$g$$ is the inverse of a function $$f$$ and $$f'\left( x \right) = {1 \over {1 + {x^5}}},$$ then $$g'\left( x \right)$$ is equal to:

If $$y = \sec \left( {{{\tan }^{ - 1}}x} \right),$$ then $${{{dy} \over {dx}}}$$ at $$x=1$$ is equal to :

$${{{d^2}x} \over {d{y^2}}}$$ equals: