# Thread: Differentiation Simples

1. ## Differentiation Simples

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

2. Originally Posted by davefulton
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
$\displaystyle \dfrac{\partial}{\partial \eta}(\ln(\rho))=\dfrac{d \ln (\rho)}{d \rho} \cdot \dfrac{\partial \rho}{\partial\eta}$

3. Thank you. You have answered many of my questions in the past. You really do live up to your name. Any idea on the second part?