If $$y=\sqrt{(x-\sin x)+\sqrt{(x-\sin x)+\sqrt{(x-\sin x) \ldots.}}}$$, then $$\frac{\mathrm{d} y}{\mathrm{~d} x}=$$

Let $$f: R \rightarrow R$$ be a function such that $$\mathrm{f}(x)=x^3+x^2 \mathrm{f}^{\prime}(1)+x \mathrm{f}^{\prime \prime}(2)+6, x \in \mathrm{R}$$, then $$\mathrm{f}(2)$$ equals

$$\text { If } y=\left(\sin ^{-1} x\right)^2+\left(\cos ^{-1} x\right)^2, \text { then }\left(1-x^2\right) y_2-x y_1=$$

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