I can't see how this is done... what's the generic approach to differentiating in log space like this?
Here's what I tried (the specific case I'm interested is has $p=\alpha=1, \delta=3$). I can write $\mathrm{log}\rho$ as
\begin{equation} \log\rho = \delta\log a - \alpha\log r (\delta - \alpha/p)\log(a^p+r^p)\\ = 3\log a - \log r - 2\log(a+r) \end{equation} I attempted a change of variables as
\begin{equation} u = \log r; \frac{dr}{du} = r \end{equation}
which gives
\begin{equation} \frac{d\log\rho}{d\log r} = \frac{d\log \rho}{dr}\frac{dr}{du}\\ = r\frac{d}{dr}\left[3\log a - \log r - 2\log(a+r)\right] \end{equation}
distributing the operator:
\begin{equation} = r\left[0 - 1 - \frac{d}{dr}2\log(a+r)\right] \end{equation}
Chain rule again for the last derivative...
\begin{equation} u' = a + r; \frac{du}{dr} = 1; \frac{d\log u}{u} = \frac{1}{u}\\ \implies r\left[-1-2\frac{\frac{d}{dr}(a+r)}{(a+r)}\right] = -r-2\frac{r}{r+a} \end{equation}
This does equal what is above, for the choices of the parameters I showed above. Is there a more elegant way? Solving it in general seems like it would take twice as long.
By implicit differentiation we have that
$$\log\rho = 3\log a - \log r - 2\log(a+r)\implies d\log\rho=\left(-\frac1r-\frac{2}{a+r}\right)dr$$
$$d \log r=\frac1r dr$$
then
$$\frac{d\log\rho}{d \log r}=-1-\frac{2r}{a+r}=-\frac{a+3r}{a+r}$$