$$x=\frac{1-t^2}{1+t^2}$$ and $$y=\frac{2 a t}{1+t^2}$$, then $$\frac{d y}{d x}=$$

If y = 2 sin x + 3 cos x and y + A$$\mathrm{\frac{d^2y}{dx^2}}$$ = B, then the values of A, B are respectively

If $$y = {\tan ^{ - 1}}\left\{ {{{a\cos x - b\sin x} \over {b\cos x + a\sin x}}} \right\}$$, then $${{dy} \over {dx}}$$

If $$e^{-y} \cdot y=x$$, then $$\frac{d y}{d x}$$ is