Do you use the material derivative in the Eulerian or Lagrangian description of continuum mechanics? When do you use the material derivative, in the Eulerian point of view when looking at a constant position in space, or in the Lagrangian point of view, following a particle?
Since I saw that another name for the material derivative is the "Lagrangian derivative", things have gotten quite confusing.
 A: I think a better way of phrasing your question is, "When is the material derivative useful?"
The material derivative $\frac{D}{Dt}$ is useful in the sense that the physical laws one finds for rigid bodies readily translate to continuum mechanical laws using the material derivative.
For example, conservation of mass is written down as
$$\frac{D(\rho dV)}{Dt} = 0$$
where $\rho dV$ is a differential mass element within the continuum. Similarly, you find that Newton's second law can be written as:
$$\frac{D(\rho \vec{v} dV)}{Dt} = \sum \vec{F}$$
where the $\rho \vec{v} dV$ is now a differential momentum element.
Writing things this way is considered a "Lagrangian approach", because you are writing physical laws in terms of the changes of elements of the continuum.
However, the material derivative can also be written down in terms of quantities defined at specific points of space, directly from the definition $\frac{D\phi}{Dt} = \frac{\partial\phi}{\partial t} + \vec{v}\cdot \nabla \phi$. So one can write physical laws in terms of quantities defined at points in space—the "Eulerian approach"—by taking the "Lagrangian" equations above and applying the definition of the material derivative. For example, conservation of mass becomes:
$$\frac{D(\rho dV)}{Dt} = \frac{\partial \rho}{\partial t} + \vec{v}\cdot\nabla\rho +\rho (\nabla \cdot \vec{v})=0$$
So rather than thinking of the material derivative as being a "Lagrangian" or "Eulerian" object, you can think of it as providing a link between both perspectives of formulating physical law.
A: Material derivative of a fluid property $f$, denoted $Df/Dt$, is the rate of change of $f$ following a given fluid particle. The qualifier "Eulerian" or "Lagrangian" is irrelevant to the interpretation of $Df/Dt$.
The formula for $Df/Dt$ in terms of spatial coordinates and time, however, depends on whether you adopt the Lagragian or Eulerian perspective. In the Lagrangian perspective, $f$ is a function of the particle label $\mathbf{a}$ (which is fixed for a given fluid particle, for e.g. its coordinates at some reference time) and time $t$. In this case, $Df/Dt=\partial f/\partial t$. In the Eulerian perspective, $f$ is a function of laboratory coordinates $\mathbf{x}$ and time $t$. Unlike the particle label $\mathbf{a}$ which is fixed for all times, the lab-coordinate $\mathbf{x}$ of the particle changes as it moves relative to lab reference frame. In this case, $Df/Dt=\partial f/\partial t+\mathbf{u}\cdot\nabla f$, where $\mathbf{u}$ is the flow velocity field measured in the lab reference frame.
To summarize, the meaning of $Df/Dt$ is independent of which perspective you adopt (Lagrangian vs Eulerian) but its formula does depend on it.
