$$\text { For all real } x \text {, the minimum value of } \frac{1-x+x^2}{1+x+x^2} \text { is }$$

The set of all points, where the derivative of the functions $$\mathrm{f}(x)=\frac{x}{1+|x|}$$ exists, is

If $$y=[(x+1)(2 x+1)(3 x+1) \ldots(\mathrm{n} x+1)]^{\frac{3}{2}}$$, then $$\frac{\mathrm{d} y}{\mathrm{~d} x}$$ at $$x=0$$ is

If $$y=\log _{\sin x} \tan x$$, then $$\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=\frac{\pi}{4}}$$ has the value