Proof of how to calculate the coefficient of variation of a nonlinear function I'm struggling to find the proof of Eq. D.14b shown in the figure below:

The variable $b$ represents the “Least Squares” best-fit to the slope between $r_e$ and $r_t$ which both depend on $X_1$ to $X_j$
The variable $\delta$ is the error term, i.e. for each pair $(r_{e,i},r_{t,i})$ the value of value $\delta_i = r_{e,i}/(b*r_{t,i})$
Any pointers are appreciated.
 A: From the context we may deduce that $X_1, \ldots, X_j,$ and $\delta$ are independent positive random variables.  Let's use a more uniform notation, then, and consider any finite set $Z_1, \ldots, Z_n$ of independent random variables and let $Z=Z_1Z_2\cdots Z_n$ be their product.
Independence implies the (raw) moments $\mu^\prime_k$ multiply, because expectations of products of independent random variables are the products of their expectations and the independence of the $Z_i$ implies the $Z_i^k$ are independent, too.  That is, for any $k\ge 0$ for which all the following expressions exist,
$$\mu^\prime_k(Z) = E[Z^k] = E[Z_1^kZ_2^k\cdots Z_n^k] = E[Z_1^k]\cdots E[Z_n^k] = \mu^\prime_k(Z_1)\cdots \mu^\prime_k(Z_n).$$
This multiplicative law of moments is justified by the simplest of algebraic manipulations except for the middle equality, which required the independence assumption.
Consider the ratio $\lambda(Z)=\mu^\prime_2(Z) / (\mu^\prime(Z))^2.$  For later use, notice this ratio is invariant under positively scaling $Z:$ that is, for any positive number $b,$
$$\lambda(bZ) = \frac{\mu^\prime_2(bZ)}{\mu^\prime(bZ)^2} = \frac{b^2\mu^\prime_2(Z)}{b^2\mu^\prime(Z)^2} = \lambda(Z).$$
Applying the multiplicative law to numerator and denominator separately, we see
$$\lambda(bZ) = \lambda(Z)=\lambda(Z_1)\lambda(Z_2)\cdots \lambda(Z_n).\tag{*}$$
Finally, notice that the squared coefficient of variation $V$ of any variable $Y$ can be expressed as
$$V(Y)^2 = \left(\frac{\sigma(Y)}{E[Y]}\right)^2 = \frac{\operatorname{Var}(Y)}{E[Y]^2} = \frac{\mu^\prime_2(Y) - (\mu^\prime_1(Y))^2}{\mu^\prime_1(Y)^2} = \frac{\mu^\prime_2(Y)}{\mu^\prime_1(Y)^2} - 1 = \lambda(Y) - 1.$$
Again, this is just simple algebra and applying the definitions of $V,$ $\lambda,$ $\mu^\prime,$ and $\operatorname{Var}.$
Squaring both sides of $(*)$ and replacing every expression of the form $\lambda(\ )^2$ by $V(\ )^2+1$ yields
$$V(bZ)^2 + 1 = (V(Z_1)^2 + 1)(V(Z_2)^2 + 1) \cdots (V(Z_n)^2 + 1).$$
Adding $-1$ back to both sides produces equation $(D.14\text b),$ QED.
