My question concerns exercise 2.4 from Hal Varian's Microeconomic Analysis.
Please note that I am not requesting a solution to the exercise, and only asking how to understand what I am being asked to do here.
2.4. Let $f(x_1, x_2)$ be a production function with two factors and let $w_1$ and $w_2$ be their respective prices. Show that the elasticity of the factor share $(w_2 x_2 / w_1 x_1)$ with respect to $(x_1 / x_2)$ is given by $1/\sigma - 1$.
Here $\sigma$ is defined earlier as the elasticity of substitution $$\frac{\text{TRS}}{x_2 / x_1}\frac{d(x_2 / x_1)}{d \text{TRS}}$$
where TRS is the technical rate of substitution $\frac{-\partial f / \partial x_1}{\partial f / \partial x_2}$.
The text has not yet defined the factor share or its elasticity yet, but in general, I interpret the phrase "the elasticity of $A$ with respect to $B$" as referring to the quantity $\frac{B}{A}\frac{dA}{dB}$. It is not hard to show using the chain rule that this quantity is equal to $d (\ln A) / d (\ln B)$.
Using the first definition, I take $A = w_2 x_2 / w_1 x_1$ and $B = x_1 / x_2$. Then $A = \frac{w_2}{w_1}\frac{1}{B}$, and $$\frac{B}{A}\frac{dA}{dB} = \frac{B}{A}\frac{w_2}{w_1} \frac{-1}{B^2}= -1$$
Using the second definition, I get the same result. We know $$\ln A = \ln (w_2 x_2 / w_1 x_1 ) = \ln (w_2 / w_1) - \ln (x_1 / x_2) = \text{const.} - \ln B $$ Differentiating both sides with respect to the symbol $\ln B$ yields $d (\ln A) / d (\ln B) = -1$.
Is my math wrong, or have I misunderstood the question?