How do you prove $\delta (ds^2) = 2 ds \delta(ds)$? How do you prove $\delta (ds^2) = 2 ds \delta(ds)$ ?
To give context, this comes from: Dirac's Theory of General Relativity p19:
http://imgur.com/mrkT5C7
I'm not comfortable with proofs regarding variations of functions. They always look intuitively obvious. This looks like the chain rule. But how would In prove it rigorously?  I see varitaions come up a lot in mathematical physics, but ive never covered them in detail in mathematics.
When I look at my old notes I wrote: $d(x^2)= 2x dx$. Then substituted $ds$ for $x$ and took $dx= \delta (ds)$ That seems a bit sketchy to me now. 
This is from page 19. General Theory of Relativity by Dirac. 
 A: Not sure about rigorousness, but as far as I've understood, the definition of the variation of a functional $F[\rho]$ is
\begin{equation}
\tag{1}\label{1}
\delta F[\rho] \equiv \lim_{\varepsilon \to 0} \frac{F[\rho + \varepsilon\eta] - F[\rho]}{\varepsilon} = \frac{dF[\rho + \varepsilon\eta]}{d\varepsilon}\Bigg|_{\varepsilon = 0}.
\end{equation}
Using the last equality, one may use the chain rule for "normal" derivation to prove the chain rule for the variation of the product of two functionals $F$ and $G$ defined on the same function-space:
\begin{align}
\delta(FG[\rho]) \equiv \delta(F[\rho]\,G[\rho]) &\equiv \frac{d}{d\varepsilon}F[\rho + \varepsilon\eta]\,G[\rho + \varepsilon\eta]\Bigg|_{\varepsilon = 0} \\
&= \frac{d}{d\varepsilon}F[\rho + \varepsilon\eta]\Bigg|_{\varepsilon = 0}\,G[\rho] + F[\rho]\,\frac{d}{d\varepsilon}G[\rho + \varepsilon\eta]\Bigg|_{\varepsilon = 0} \\
&\equiv (\delta F)G + F(\delta G),
\tag{2}\label{2}
\end{align}
where the functional dependence have been suppressed in the last line for the sake of readability.
Your wanted result follows directly from \eqref{2}.
