Combine a square root with inverse cosine

For $0<x<1$, let $u=\sqrt{x}$. Then $u^{\prime}=1/(2\sqrt{x})$ and $d(\cos^{-1}u)/dx=-u^{\prime}/\sqrt{1-u^2}=-1/[2\sqrt{x}\sqrt{1-x}]=-1/[2\sqrt{x(1-x)}]$, matching the printed answer. The original real function is defined on $[0,1]$; the displayed finite derivative holds on $(0,1)$.

The chain rule gives $2x/[1+(x^2)^2]=2x/(1+x^4)$, the printed answer, for all real $x$.