Taking partial derivatives with dependent variables

Question

When dealing with multivariate calculus e.g. a $f$ s.t.

$f = f(x,y,z,w)$

and $x,y$ are really not independent variables but depend on $z$ and $w$. See this for a good example. we can calculate partial derivative of $f$ w.r.t. $x$ viewed in two different ways:

1. $x$ considered independent of $z, w$
2. we take into account the dependence of $x$ on $z$ and $w$

The linked answer shows the two cases.

My question is, is doing 1. ever correct? (e.g., when solving a problem in physics). With 1, I can get any result I want. I mean I can have a complicated expression and I can invent new variables for things I don't know how to differentiate etc. and get a simple expression for $\partial f / \partial x$.

To restate the question in another form, when we have a function of $n$ multiple variables, does it ever make sense to disguise or view it as a function of $m > n$ variables with some variables dependent on others but the dependence effectively ignored in taking partial derivatives?

Answer 1

I am honestly not sure what’s tricking you up, but I’m going to try anyway.

When we deal with the functions and say that, say, $f=f(x,y,z)$, it already implies the independence of the variables used.

Whenever you say your variable $x$ is not independent but rather a function of the two other variables, $x=x(w, z)$, you implicitly make the case of, in fact, a different function which doesn’t contain that $x$ per se. A simple example would do the explanation.

Imagine you have a function: $$ f(x,y,z) = x^3 + 2y - 3z $$

Here, you can take all the partial derivatives you want, i. e., with respect to any variables, and that is rather clear.

Imagine now, that you say that $x$ is now a dependent variable and, in fact, a function of the two others, namely $x=y^2-z^3$.

Once you define that new function $f=f(y^2-z^3, y, z)$, it is not at all a function of $x$ anymore, and it makes no use to try and define an operator $\frac{\partial}{\partial x}$.