But in the case of an expression with both direct and indirect dependencies on the variable of interest,

*and* other direct dependencies, then there is ambiguity, eg

$$y = g(x) \\

\frac{\partial}{\partial x} f(x, y, z)$$

could potentially mean either

$$\frac{\partial}{\partial x} f(x, y, z)|_z$$ or $$\frac{\partial}{\partial x} f(x, y, z)|_{y,z}$$

So it seems to me that for an expression like that you should be explicit, and that the two meanings are related like this:

$$\frac{\partial}{\partial x} f(x, y, z)|_z = \frac{\partial}{\partial x} f(x, y, z)|_{y,z} \, + \, \frac{\partial}{\partial y} f(x, y, z) \, \frac{dy}{dx}$$

Is that right?

Or am I just hopelessly confused?

I've slept on this and here's how I'd put things.

First of all, if you look up how partial derivation is defined in standard math texts, then normally the thing you take partial derivatives of are

*functions*. In physics we'd normally write something like $$f \,=\, f(x,\, y)$$ to say that $$f$$ is a function of two real variables, and we're letting $$x$$ and $$y$$ denote its first and second parameters, respectively. Then $$f$$'s partial derivatives are noted $$\frac{\partial f}{\partial x}$$ and $$\frac{\partial f}{\partial y}$$ (and preferably

**not** $$\frac{\partial}{\partial x} f(x,\, y)$$). These are new

*functions*, analogous to $$f'$$ in the single variable case. So you can write e.g.

$$\frac{\partial f}{\partial x}(x_{0},\, y_{0})$$

to denote the partial derivative $$\frac{\partial f}{\partial x}$$ evaluated at the point $$(x_{0},\, y_{0})$$. This is analogous to writing something like $$f'(x_{0})$$ in the single variable case. For functions with longer names/definitions, you could write something like this:

$$

\Bigl( \frac{\partial}{\partial y^{i}} \, f \,\circ\, g \Bigr)(\bar{y}_{0}) \,.

$$

This makes it clear that the function you're taking the partial derivative of is $$f \,\circ\, g$$, and then you're evaluating it at the point $$\bar{y}_{0}$$.

When you write something like this:

$$

\frac{\partial}{\partial x} \bigl( x^{2} \,+\, \sin(x + y) \,-\, yz \bigr) \,,

$$

the thing between the brackets is an

*expression*. In line with what temur told you, I'd read it as the definition of an unnamed function

$$

(x,\, y,\, z) \to x^{2} \,+\, \sin(x + y) \,-\, yz

$$

that you're taking the partial derivative of, in this case with respect to $$x$$. So

$$

\frac{\partial}{\partial x} \bigl( x^{2} \,+\, \sin(x + y) \,-\, yz \bigr) \,=\, 2x \,+\, \cos(x + y) \,.

$$

Or more correctly, the result is the

*function* $$(x,\, y,\, z) \to 2x \,+\, \cos(x + y)$$.

With this convention, if $$f = f(x,\, y,\, z)$$ and $$g = g(x)$$, then:

$$\frac{\partial}{\partial x} f(x,\, y,\, z)$$ is synonymous with $$\frac{\partial f}{\partial x}$$,

$$

\frac{\partial}{\partial u} f(u,\, v,\, w) \,=\, \frac{\partial f}{\partial x}

$$ (**not** $$\frac{\partial f}{\partial u$$ - why?),

and

$$

\frac{\partial}{\partial x} f(x,\, g(x),\, z) \,=\, \frac{\partial f}{\partial x} \,+\, \frac{\partial f}{\partial y} \, \frac{\mathrm{d} g}{\mathrm{d} x} \,.

$$

This is all pretty much in line with what temur has just told you, but hopefully it's still helpful.