If $y=\tan ^2\left(\cos ^{-1} \sqrt{\frac{1+x^2}{2}}\right)$, then $\frac{d y}{d x}=$

If $y=x^{\log x}+(\log x)^x, x>1$, then $\left(\frac{d y}{d x}\right)_{x=e}=$

If $y=\sqrt{\log \left(x^2+1\right)+\sqrt{\log \left(x^2+1\right)+\sqrt{\log \left(x^2+1\right)+\ldots+\infty}}, \text {, } 100.00}$, $|x|<1$, then $\frac{d y}{d x}=$

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