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:

Let $$f:\left( { - 1,1} \right) \to R$$ be a differentiable function with $$f\left( 0 \right) = - 1$$ and $$f'\left( 0 \right) = 1$$. Let $$g\left( x \right) = {\left[ {f\left( {2f\left( x \right) + 2} \right)} \right]^2}$$. Then $$g'\left( 0 \right) = $$

Let $$y$$ be an implicit function of $$x$$ defined by $${x^{2x}} - 2{x^x}\cot \,y - 1 = 0$$. Then $$y'(1)$$ equals