From the first page of this pdf on measurements of bandstructures it is written that:
In the chain rule above, written again below for clarity:
$$\frac{\partial k_x}{\partial t}\frac{\partial E}{\partial k_x}+\frac{\partial k_y}{\partial t}\frac{\partial E}{\partial k_y}+\frac{\partial k_z}{\partial t}\frac{\partial E}{\partial k_z}\equiv\frac{dE}{dt}$$
I am questioning whether the $\frac{dE}{dt}$ should actually be $\frac{\partial E}{\partial t}$
The reason I say this is because $E\equiv E({\bf{k}}) \equiv E(k_x, k_y, k_z)$ and $k_{x,y,z}\equiv k_{x,y,z}(t)$
I'm using the mnemonic method of tree diagrams outlined in the popular Youtube video by Krista King which for my case the tree diagram looks like:
The problem is that according to Youtube video link above, I am supposed to be writing $$\frac{\partial k_x}{\partial t}\frac{\partial E}{\partial k_x}+\frac{\partial k_y}{\partial t}\frac{\partial E}{\partial k_y}+\frac{\partial k_z}{\partial t}\frac{\partial E}{\partial k_z}\equiv\color{red}{\frac{\partial E}{\partial t}}$$
Before asking this question I searched this site for an answer and found this similar question but does not quite answer my question here.
So, to summarize, which one is it? $$\frac{\partial E}{\partial t}\quad \text{or}\quad \frac{dE}{dt}\,?$$
Update:
Many thanks to comments and answers so far, it seems to be the case that it is correct to write $$\frac{dE}{dt}$$
Shown below is an image taken from the video in question:
So why is the Youtube video saying otherwise?
Best Answer
If you want to be super precise, neither of them is correct. Originally, you have a function $E:\Bbb{R}^3\to\Bbb{R}$, meaning it eats a $3$-tuple of real numbers and it spits out a real number (based on physical context, you would interpret this as energy)
Next, you're given three functions $k_x:\Bbb{R}\to\Bbb{R}$, $k_y:\Bbb{R}\to\Bbb{R}$ and $k_z:\Bbb{R}\to\Bbb{R}$. Finally, you're defining a completely new function, $\mathcal{E}:\Bbb{R}\to\Bbb{R}$ as \begin{align} \mathcal{E}(t)&:= E(k_x(t), k_y(t),k_z(t)). \end{align}
Note that $\mathcal{E}$ is NOT the same thing as $E$. Saying that $\mathcal{E}$ is the same as $E$ is in my opinion (atleast for people just learning the basics) is as bad of a lie as saying that the moon and the earth are the same thing just because they are round. It's completely absurd to mix up completely different things. Yes, there is some similarities, afterall I am using the function $E$ to define $\mathcal{E}$, but that does not mean they are the same thing.
Now, here are various levels of precision with which you can write the chain rule:
The first is in my opinion the most unambiguous way of writing things, because it is literally impossible to misinterpret it. The second is about as good as you can get with Leibniz's notation. The third is as far as I'm willing to take the omissions in notation (because note that derivatives are functions and functions should be evaluated at correct points of their domains, so not specifying this correctly can easily lead to all sorts of basic errors). Worse than all these three is writing (something which physics texts LOVE writing) \begin{align} \dfrac{dE}{dt}&=\dfrac{\partial E}{\partial k_x}\cdot\dfrac{d k_x}{dt}+ \dfrac{\partial E}{\partial k_y}\cdot\dfrac{d k_y}{dt}+ \dfrac{\partial E}{\partial k_z}\cdot\dfrac{d k_z}{dt} \end{align} because the $E$ on the LHS of this equation has a completely different meaning to the $E$ on the RHS. The one on the LHS should really be $\mathcal{E}$.
Finally, the absolute worst, and just plain nonsense is
\begin{align} \dfrac{\partial E}{\partial t}&=\dfrac{\partial E}{\partial k_x}\cdot\dfrac{d k_x}{dt}+ \dfrac{\partial E}{\partial k_y}\cdot\dfrac{d k_y}{dt}+ \dfrac{\partial E}{\partial k_z}\cdot\dfrac{d k_z}{dt}; \end{align} it should be $d$ on the LHS, and $E$ on the LHS should be $\mathcal{E}$.