# Differential of square $d(w^2)$or square of differential$(dw)^2$!?

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?

-
It seems your question is why in the expression $dl^2 = dx^2 + \cdots + dw^2$ we have that $dw^2 = (dw)^2$. The answer would be: it just is. Conventional notation demands that $dw^2$ be read with the parenthesis as in $(dw)^2$ and not as $d(w^2)$. – Willie Wong Dec 7 '12 at 12:06
To be more precise, the expression really defines a covariant 2-tensor $$\mathrm{d}l \otimes \mathrm{d}l = \mathrm{d}x\otimes\mathrm{d}x + \cdots + \mathrm{d}w \otimes \mathrm{d}w$$ In context it is usually clear how to interpret $dw^2$; if you are newly starting to read stuff in the subject, it may take a little bit of getting used to. – Willie Wong Dec 7 '12 at 12:12

In general, $dx^2=2dx$, but you've been confused by notation enlightening, where the parenthesis are omitted because there should be no possible confusion. Imagine that page (or any work on general relativity) with every parenthesis explicitly written. My eyes'd start to bleed.
Typo: $d (x^2) = 2x dx$, not $2 dx$. – David Speyer Dec 3 '12 at 15:57