A Distinction Between Different Types of Partial Derivatives I recently noticed a subtle distinction between different types of partial derivatives: 


*

*those that involve differentiation with respect to a parameter (that is, an   independent coordinate)

*those that involve differentiation with respect to a variable that is a function of a parameter or another variable.  
I don't recall this distinction being made in my calculus courses, and I'd like to check if my reasoning is valid and if there is a name for this observation.
Consider two functions $f(x,y)$ and $g(z,t)$, where $x$, $y$, and $t$ are independent parameters but $z(t)$.  For example, $f(x,y)$ could represent that magnitude of a $2D$ magnetic field while $g(z,t)$ might represent the energy of a particle moving in a time-dependent potential.
In the first case, the partial derivative $\partial{f}/\partial{x}$ is also a total derivative since
$$
\frac{df}{dx} = \frac{\partial{f}}{\partial{x}} + \frac{\partial{f}}{\partial{y}}\frac{dy}{dx} = \frac{\partial{f}}{\partial{x}}
$$
since $y$ is independent of $x$.
In the second case, the partial derivative $\partial{g}/\partial{t}$ is more of a "pure" partial derivative since it only describes the explicit dependence of $g(z,t)$ on $t$, with the complete dependence being described by the total derivative
$$
\frac{dg}{dt} = \frac{\partial{g}}{\partial{t}} + \frac{\partial{g}}{\partial{z}}\frac{dz}{dt}
$$
Clearly, the first case is just a special instance of the more general second case, but the first case exhibits some useful properties that are not found in general in the second case.  Most importantly in my mind, since $\partial{f}/\partial{x}$ is essentially a total derivative, any partial differential equation for $f(x,y)$ involving only derivatives in $x$ is in fact just an ordinary differential equation.  The same is not true for a partial differential equation for $g(z,t)$ involving only derivatives in $t$, unless the explicit form of $z(t)$ is known a priori.
As a simple example, suppose $f(x,y)$ is unknown but the form of $f_x(x,y)$ is given.  We then have
$$
\frac{\partial{f}}{\partial{x}} = \frac{df}{dx} = f_x(x,y)
\rightarrow
f(x,y) = \int f_x(x,y) dx
$$
I don't believe that this simple integration of a differential equation is possible in general in the second case.  This is certainly a substantive difference between the two type of partial derivatives, but I don't believe it was discussed in my undergraduate curriculum.
 A: I'm going to elaborate a bit on my comment above. First of all, I am going to be more pedantic in my notation to emphasize what's going on. @wj32 is quite correct that the Leibniz notation can lead to confusion.
OK, so we're given a differentiable function $f\colon\mathbb R^2\to\mathbb R$, and we compose it with a function $\mathbf g\colon\mathbb R\to\mathbb R^2$. For the moment, let's think of the latter as $\mathbf g(t) = (x(t),y(t))$. Write $F(t) =(f\circ\mathbf g)(t)$. Then the chain rule tells us that
$$F'(t)=(f\circ\mathbf g)'(t) = \nabla f(\mathbf g(t)) \cdot \mathbf g'(t) = \frac{\partial f}{\partial x}x'(t) + \frac{\partial f}{\partial y}y'(t) \,.$$
You would think of $F'$ as a "total derivative," more so if $f$ also were a function of $t$ as well. We can certainly incorporate this by letting $x(t)=t$ (or by introducing a third variable, as necessary).
But no one will argue that if $x(t)=t$ and $y$ is independent of $t$ (i.e., $y(t)=y_0$), then we get $y'(t)=0$ and
$$F'(t)= \frac{\partial f}{\partial x}(t,y_0)\cdot 1 + \frac{\partial f}{\partial y}(t,y_0)\cdot 0 = \frac{\partial f}{\partial x}(t,y_0)\,.$$
This is precisely the setting of your first equation, and there's none of this discussion of derivatives' being undefined. At this stage, call $t=x$ if you want.
I will elaborate on my remark about the notation used by physical chemists, economists, and others. If we have a function $f$ of $(p,V,T)$ (say the entropy of a gas), but there is some equation constraining $p$, $V$, and $T$ (say, the ideal gas law $pV = T$, where I've set $nR = 1$ for simplicity), what should $\dfrac{\partial f}{\partial T}$ mean? Ordinarily, this means that we differentiate $f$ with respect to $T$, holding $p$ and $V$ constant. Of course, we can't do this. So the chemist writes $\Big(\dfrac{\partial f}{\partial T}\Big)_V$ or $\Big(\dfrac{\partial f}{\partial T}\Big)_p$ to nail it down. These are fitting in the realm of "total derivatives" (although no chemist calls it such, that I know of) and we will have
$$\Big(\dfrac{\partial f}{\partial T}\Big)_V = \dfrac{\partial f}{\partial T} + \dfrac{\partial f}{\partial p}\Big(\dfrac{\partial p}{\partial T}\Big)_V\,.$$
A: You are perceptive to notice that there is a difference, but you have it backwards due to a few misconceptions.
In the case of $f(x,y)$, where $x$ and $y$ are independent, the total derivative $\frac{d}{dx}f(x,y)$ is undefined, because when $x$ and $y$ are independent, $\frac{dy}{dx}$ is undefined.
In the case of $g(z,t)$, where $z$ is a function of $t$, the total derivative $\frac{d}{dt}g(z,t)$ is defined. Since there is a known function between $z$ and $t$, $\frac{dz}{dt}$ is defined.
So what you are noticing is the difference between a function with a well-defined total derivative, and one with an undefined total derivative.
