Except it doesn't.

The partial derivative is, qualitatively, "Treat everything else as fixed and just do a normal derivative on this variable". A total derivative is, qualitatively, "Consider all unknowns as functions of this variable and do the chain rule on each one of them". If you explicitly know the function, ie $$u(x) = \sin^{2}x$$ then you can write down it's x derivative as a function of x manifestly. Your reference is about reparametrisations and how to deal with them. I don't deny the equation, I deny your view of it.

Please stop talking down to me. Especially in the context of my having gone out of the way and having explained the above to Pete from base principle.

You haven't explained it from base principle. In fact I've asked you more than once to consider the definition of the partial derivative, which you seem unwilling to do. I'll do it for you since you seem so unwilling to do it from 'base principle'.

$$f(x,y) = 3x+ \sin^{2}x + \ln y$$

$$\partial_{x}f(X,Y) \equiv \lim_{h \to 0} \frac{f(X+h,Y)-f(X,Y)}{h}$$

$$f(X+h,Y) = 3(X+h) + \sin^{2}(X+h) + \ln Y = 3X + 3h + \ln Y + (\sin X \cos h + \cos X \sin h)^{2} = 3X + 3h + \ln Y + (\sin X \cos h + \cos X \sin h)^{2}$$

Therefore

$$f(X+h,Y)-f(X,Y) = 3h + (\sin X \cos h + \cos X \sin h)^{2} - \sin^{2}X$$

Using $$\sin h \sim h$$ and $$\cos h \sim 1$$ we have

$$f(X+h,Y)-f(X,Y) = 3h + (\sin X + h\cos X)^{2} - \sin^{2}X = 3h + 2h\sin X \cos X h^{2}\cos^{2} X $$

Therefore

$$\partial_{x}f(X,Y) = 3 + 2\sin X \cos X = 3+\sin 2X$$

As stated many times. The equation you keep quoting is what happens if you were to turn around and say "But I know that $$y = y(x) = e^{x}$$!" (for instance), "What if I'd put that in to start with?". Well you can either compute $$f(x,e^{x})$$ and then do $$\partial_{x}f$$ or you can just compute $$d_{x}f(x,y)$$ and then put in the relevant expression for $$\partial_{x}y$$ which appears in the formula. Let's see this in action,

$$g(x,y) = 3x + y$$

$$d_{x}g(X,Y) = 3 + \frac{\partial y}{\partial x}|_{x=X}$$

If we know afterwards that $$y(x) = \sin^{2}x$$ then $$\partial_{x}y = \sin 2x$$ and so $$d_{x}g(X,Y) = 3 + \sin 2X$$. But what if we put it in before hand, $$g(x,\sin^{2}x) \equiv h(x)$$ and so $$\partial_{x}g = \partial_{x}h = \partial_{x}(3x+\sin^{2}x)|_{x=X} = 3+\sin 2X$$

That's how the total derivative comes into it. It's a sort of apriori/posteri substitution. The partial derivative when you're working in terms of functions of x (or functions of theta as was in the original example) is the same as if you changed parametrisations, didn't know the parametrisation, did a total derivative and then converted back. By altering the arguments in the way you are you're confusing yourself. However, as I've just illustrated, when put in the proper context the equation you keep pointing at is entirely consistent with the definition, which you misstated, and in complete agreement with what

*we* have been telling

*you*.

Any disagreement should explain why the definition is wrong, because now I've explicitly done it for you, you have no excuse to ignore that. You should also address how your method contradicts all of Lagrangian/Hamiltonian mechanics, another point you keep skipping.