Deriving $f\frac{\partial v}{\partial z}$:

Take $\frac{\partial}{\partial z}$ of the $x$-momentum equation (Eq. ):

$$\frac{\partial}{\partial z}\left(-fv\right) = \frac{\partial}{\partial z}\left(-\frac{1}{\rho_0}\frac{\partial p}{\partial x}\right)$$

Again, assuming $f$ is independent of $z$:

$$-f\frac{\partial v}{\partial z} = -\frac{1}{\rho_0}\frac{\partial}{\partial x}\left(\frac{\partial p}{\partial z}\right)$$

Substitute the hydrostatic relation (Eq. ) for $\frac{\partial p}{\partial z}$:

$$-f\frac{\partial v}{\partial z} = -\frac{1}{\rho_0}\frac{\partial... ...tial x}\left(-\rho g\right) = \frac{g}{\rho_0}\frac{\partial \rho}{\partial x}$$

Substitute $\rho = \rho_0 + \rho'$ (where $\frac{\partial \rho}{\partial x} = \frac{\partial \rho'}{\partial x}$):

$$-f\frac{\partial v}{\partial z} = \frac{g}{\rho_0}\frac{\partial \rho'}{\partial x}$$

Multiplying by $-1$:

$$\mathbf{f\frac{\partial v}{\partial z} = -\frac{g}{\rho_0} \frac{\partial \rho'}{\partial x}}$$

This completes the derivation of the thermal-wind relation.