On page 142 of Lackmann'sMidlatitude Synoptic Meteorology, Equation 6.7 gives the derivative of the Boussinesq buoyancy$b = \frac{g \theta'}{\theta_0 (z)}$(perturbation potential temperature$\theta'$and base state$\theta_0 (z)$) as
\begin{equation} \frac{db}{dt} = -w \frac{g}{\theta_{00}} \frac{d \theta_0}{dz} \end{equation}
where$\theta_{00}$is the base state potential temperature$\theta_0 (z)$at$z = 0$. While I understand this to be an application of the chain rule
\begin{equation} \frac{db}{dt} = \frac{\partial z}{\partial t} \frac{\partial \theta_0}{\partial z} \frac{\partial b}{\partial \theta_0}, \end{equation}
I am unable to see why
\begin{equation} \frac{\partial b}{\partial \theta_0} = - \frac{g}{\theta_{00}} \end{equation}
when I instead would expect
\begin{equation} \frac{\partial b}{\partial \theta_0} = - \frac{g \theta'}{\theta_0^2 (z)}. \end{equation}
Any help would be much appreciated!
