Conventions regarding partial derivatives Look at this expression: 
$$\frac{\partial}{\partial t} (V-\mathbf{v}\cdot\mathbf{A}).$$
This expression occurs in Griffiths EM book (4th ed, p.444). $V=V(\mathbf{r},t)$is the scalar potential, $\mathbf{A} = \mathbf{A}(\mathbf{r},t)$ is the vector potential, and $\mathbf{v} = \mathbf{v}(t)$ is the velocity of the particle in context.
After this expression, the author points out that the derivative operator does not act on $\mathbf{v}$. Now, does he, by that, just warn me of missing the standard convention in how the derivative symbol acts? or does he mean that he is just deviating from the convention in this very expression, in order to make it nicely looking?
 A: The expression provided in the question is,
$$\frac{\partial}{\partial t} \left( V - \vec{v} \cdot \vec{A}\right)$$
Strictly speaking, if $\vec{v}=\vec{v}(t)$, then the differentiation operator must act on the function. However, it depends how you choose to view $\vec{v}$. Consider as an example a generic Lagrangian,
$$\mathcal{L}=f(x(t),t)$$
To obtain the equations of motion associated to the system, we need a particular derivative with respect to the differentiated function $\dot{x}(t)$, i.e.
$$\frac{\partial}{\partial (\partial_t x(t))}\mathcal{L}(x(t),t)$$
The expression is not to be viewed completely literally. You're not suppose to express $\mathcal{L}$ in terms of the function $\partial_t x(t)$, and then differentiate. Instead, you treat it as an independent variable in its own right; so one should take certain expressions in physics with a pinch of salt, as the saying goes. If we thought of $\vec{v}$ as an independent variable, then we'd not subject it to the $\partial_t$.
A: This is another manifestation of the ambiguity in the notation for the partial derivative. If you take $\partial_t C(V,\vec{A},\vec{v})$, with $C(V,\vec{A},\vec{v}) \equiv V(\vec{r},t) - \vec{v}(t)\cdot \vec{A}(\vec{r},t)$, $\partial_t C|_{(V,\vec{A},\vec{v})} =0$, $\partial_t C |_{(\vec{A}, \vec{v})} = \partial_t V$, $\partial_t C|_{(V,\vec{v})} = -\vec{v}\cdot \partial_t \vec{A}$, $\partial_t C|_{(V,\vec{A})} = -\dot{\vec{v}} \cdot \vec{A}$, $\partial_t C|_{\vec{v}} = \partial_t V - \vec{v} \cdot \partial_t \vec{A}$. In naive notation, all of the above are denoted by $\partial_t C$, hence the confusion.
The confusion about this goes so far that I've seen a professor in a field theory class define $\bar{\partial}_\mu$ as the derivative that acts only on the explicit dependencies, because physicists (me included) often abuse the notation and write $\partial_\mu\phi$ when they mean $\partial_\mu\phi(x)$ in the lagrangian. Apparently, this is the reason Goldstein uses the notation $d/dx^\mu$ instead of $\partial/\partial x^\mu$ in the chapter on classical field theory.
