How does this derivative notation work? Elasticity of substitution = $\dfrac{\mathrm d \ln(k/l)\;\;\;}{\mathrm d \ln(f_l/f_k)}$
This is type of notation I haven't really worked with before. Is this read as "The change in the natural log of k over l with respect to the change in the natural log of the partial derivative of f with respect to l over the partial derivative of f with respect to k"?
The production function I am working with specifically is $f(k, l) = k^{1/3}l^{2/3}$ and I know it should equal 1 (via another elasticity of substitution equation). But what and how does this equation I posted work?
Thanks :)
 A: The statement is assuming $f_l/f_k$ is a function of $k/l$ (it is in your example; see below), and then the quantity you're interested in is the slope in a log-log plot (or rather 1/slope).
If the slope in a log-log plot is constant, then there's a power relationship (where the power equals the slope.)
For example, if $g(x) = A x^C$, then $\ln g = \ln A + C\, \ln x$, so $\frac{d \ln g}{d \ln x} = C$, or equivalently, for your situation, $\frac{d \ln x}{d \ln g} = 1/C$.

Your example:
$f_l = \frac{2}{3} k^{1/3} l^{-1/3}$
$f_k = \frac{1}{3} k^{-2/3} l^{2/3}$
So $f_l/f_k = 2\, k/l$
Thus $\ln(f_l/f_k) = \ln 2 + \ln(k/l)$
Thus your expression is one.
A: Your reading sounds correct. One way to look at it would be using analogy to the standard $\frac{df}{dx}$. Try to write ln$(k/l)$ as a function of ln$(f_k/f_l)$, then substitute $x$ or some other variable for ln$(f_k/f_l)$.
A: You may read it however you like. What you wrote is correct. Suppose for simplicity that $f = f(x)$ and $g = g(x)$. Then
$$\frac{d \log f}{d \log g} = \frac{d \log f}{(g'/g) dx} = \frac{f'}{f} \frac{g}{g'}.$$
You can generalize this and rewrite the elasticity of substitution.
