in Curved space it seems $d(w^2)=(dw)^2$ how is it possible!?
$$x^2+y^2+z^2+w^2=\kappa^{-1}R^2,$$ first: $$dw=-w^{-1}(xdx+ydy+zdz),$$
$$\kappa^{-1}R^2-(x^2+y^2+z^2)=w^2,$$ $$dl^2 = dx^2 + dy^2 + dz^2+dw^2,$$
$$dl^2 = dx^2 + dy^2 + dz^2 +\frac{(xdx+ydy+zdz)^2}{\kappa^{-1}R^2-x^2-y^2-z^2}$$
Remark: in general as i know $dx^2=2xdx$ so $dw^2$ should be:
$$d(w^2)=(2w)w^{-1}(xdx+ydy+zdz)=2(xdx+ydy+zdz),$$ but here
$$d(w^2)=(dw)^2=w^{-2}(xdx+ydy+zdz)^2,$$ how is it possible?