The derivative of $$(\log x)^x$$ with respect to $$\log x$$ is

$$y=\sqrt{e^{\sqrt{x}}}$$, then $$\frac{d y}{d x}=$$

If $$y=\sin ^{-1}\left[\cos \sqrt{\frac{1+x}{2}}\right]+x^x$$, then $$\frac{d y}{d x}$$ at $$x=1$$ is

If $$x=a(t+\sin t), y=a(1-\cos t)$$, then $$\frac{d y}{d x}=$$