Soapbox: Engaged.

Implicit differentiation is a slightly sophisticated technique. Because of this, it is often taught in a simpler, hand-waveier way.

The usual notion of the derivative is really an operation you apply to a function. A function is, in its idealized form, nothing more than a rule that maps inputs in its domain to outputs in its codomain. In particular, a function has no notion of a "dependent variable".

If I know that f is a differentiable function, I don't know if it's "a function of x" or "a function of y". It's a meaningless question. When I say f(x) = x² + 1, the x is just a means to an end. I could just as easily have written f(y) = y² + 1 or f(t) = t² + 1. All three define the same function, and nothing distinguishes them after I have defined them. In fact, I could just as easily say it in words: "let f be the function which squares its input". This defines the exact same f, even though it is not defined in terms of a variable at all!

That said, we usually slur our words. Newton and Leibniz did not have the sophisticated notion of a function at their disposal. (I believe the modern notion of a function was an invention during Euler's reign). When I say "the function x^2", really what I would want is what computer scientists call a lambda expression. That is, I really want to talk about this thing I might write down as x ↦ x^2, which denotes the function which takes a variable (which I happen to call x here) and squares it. Using this notation, I could equivalently write f as either f(x) = x² or as f = (x ↦ x^2). (I mean, think of it like this: f(x) = ... tells you what f evaluated at x is... but what is f itself?).

So then what does Leibniz notation even mean?? How does d/dx x² really work? Well if we use D to mean the derivative operator (defined as Df(x) = f'(x)), then d/dx x² really means D(x ↦ x^2). We can think of it as a fancy syntax which implicitly creates a function out of whatever variable appears in the bottom.

In general, Leibniz notation is more damning than this. You also have to account for when a variable denotes a function applied with some input implicitly. This is what happens when we deal with the chain rule. We say dy/dt = dy/dx dx/dt, but really, x is denoting a funciton evaluated with an implicit argument t, and y is denoting a function evaluated with an implicit argument x(t). When you spell this out in full, you get (y o x)'(t) = y'(x(t)) x'(t). (But it looks like a mess because we are taught the lie in school).

What implicit differentiation is really about is an application of the chain rule, combined with implicit arguments. I'll leave it to others and your Google-fu to find the details. I just wanted to stand on this soapbox for a while.

Good bye!