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On page 142 of Lackmann's Midlatitude 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!

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