If $$y = {\tan ^{ - 1}}\sqrt {{{1 - \sin x} \over {1 + \sin x}}} $$, then the value of $${{dy} \over {dx}}$$ at $$x = {\pi \over 6}$$ is

If x² + y² = 4, then $$y{{dy} \over {dx}} + x = $$

If $$y = {A \over x} + B{x^2}$$, then $${x^2}{{{d^2}y} \over {d{x^2}}}$$ =

Let $$f(x) = ta{n^{ - 1}}x$$. Then $$f'(x) + f''(x)=0$$, when x is equal to