If $$y=\tan ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$$, then $$\frac{d y}{d x}=$$

If $$x^y \cdot y^x=16$$, then $\frac{d y}{d x}$ at $(2,2)$$ is

If $$y^2=a x^2+b x+c$$, where $$a, b, c$$ are constants, then $$y^3 \frac{d^2 y}{d x^2}$$ is equal to

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