If $$f\left( x \right) = \left| {\matrix{ {\cos x} & x & 1 \cr {2\sin x} & {{x^2}} & {2x} \cr {\tan x} & x & 1 \cr } } \right|,$$ then $$\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}$$

Let f be a polynomial function such that

f (3x) = f ' (x) . f '' (x), for all x $$ \in $$ R. Then :

If y = $${\left[ {x + \sqrt {{x^2} - 1} } \right]^{15}} + {\left[ {x - \sqrt {{x^2} - 1} } \right]^{15}},$$

then (x² $$-$$ 1) $${{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}}$$ is equal to :

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