Why is this true?

for $\displaystyle \rho(\eta,t)$

$\displaystyle

\dfrac{\partial}{\partial \eta}(\ln(\rho))=\dfrac{1}{\rho}\cdot\dfrac{\parti al \rho}{\partial\eta}

$

Is this needed to derive

$\displaystyle

\dfrac{D\ln(\phi)}{Dt}=\nabla\cdot\bf U.

$

from the continuity equation

$\displaystyle

\dfrac{\partial\rho}{\partial t}+\nabla\cdot(\rho \bf U)=0.

$

where $\displaystyle \phi(\textbf{x},t)=1/\rho(\textbf{x},t)$ is the specific volume