The value of $${{dy} \over {dx}}$$ at $$x = {\pi \over 2}$$, where y is given by $$y = {x^{\sin x}} + \sqrt x $$ is

The second order derivative of a sin³t with respect to a cos³t at $$t = {\pi \over 4}$$ is

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 $$y{{dy} \over {dx}} + x = $$