I'm struggling to derive the 1D material derivative myself from a book called 'A 1D Introduction to Continuum Mechanics'. It says if $f$ is a measurable quantity of a 1D continuum material, $f$ can be defined in the Eulerian and Lagrangian systems as follows:
\begin{align} f^E(x,t) &= f^L(\xi^E(x,t),t) \\ f^L(\xi,t) &= f^E(x^L(\xi,t),t)\end{align}
Where $x$ is the position on the x-axis, t is time, and $\xi$ is the initial position of the particle at $t=0$. The books instruction is to differentiate $f^L(\xi,t) = f^E(x^L(\xi,t),t)$ with respect to time and use the chain rule to observe:
$$\frac{\partial f^L}{\partial t} = \frac{\partial f^E}{\partial t} + v^E \frac{\partial f^E}{\partial x}~,$$ and thus, $$\frac{D f^E}{D t} = \frac{\partial f^E}{\partial t} + v^E \frac{\partial f^E}{\partial x}$$
Additionally the books states $v^E = v^L$, and I think $x^E = x^L$. I've only been able to partially solve the problem:
$$\frac{\partial}{\partial t} f^L(\xi,t) = \frac{\partial}{\partial t} f^E(x^L(\xi,t),t) \\ \frac{\partial f^L}{\partial t} = \frac{\partial f^E}{\partial x^L} \frac{\partial x^L}{\partial t} + \frac{\partial x^L(\xi,t)}{\partial \xi} \frac{\partial \xi}{\partial t} \\ \frac{\partial f^L}{\partial t} = \frac{\partial f^E}{\partial x^L} (v^{(E)}) + (?) \frac{\partial \xi}{\partial t}$$