Taylor expansion multivariable: mutual independence of variables required?

Question

I know that Taylor expansion, which is normally shown and proved in the case of one variable, can be extended to functions of more than one variable.

For example, consider a function of two variables:

z = f (x,y)

My question is: for the multivariable Taylor expansion to hold, is it required that the two independent variables (x and y in this case) are THEMSELVES mutually independent?

Put in another way: supposing a relation y = g(x) exists (which may be unknown!), is it still possible to perform the Taylor expansion of the multivariable function z = f(x,y), along with all the logical reasoning associated to this operation? (e.g. linear approximation close to a reference point (x0, y0)).

I intuitively have an idea of the answer, but I would like a confirmation and possibly a proof for it.

Answer 1

If $f$ is a smooth function of two variables, $f(s,t)$ has a Taylor expansion, let's say $$ f(s,t) = a_{00} + a_{10} (s - s_0) + a_{01} (t - t_0) + a_{20} (s-s_0)^2 + a_{11} (s-s_0)(t-t_0) + \ldots $$
Then you can specialize this to $$f(x, g(x)) = a_{00} + a_{10} (x - s_0) + a_{01} (g(x) - t_0) + a_{20} (x - s_0)^2 + a_{11} (x - s_0)(g(x) - t_0) + \ldots $$ If $g$ is a smooth function with $g(s_0) = t_0$, you can substitute the Taylor expansion of $g$ around $s_0$ to get a Taylor series for $f(x,g(x))$ in powers of $x - s_0$.

Answer 2

Yes, we usually assume that the variables are independent. Even if there's a relationship between them, it is not known, so they are still considered independent. However if there's a relationship between them, then the function $f(x,y)$ is really a function of only one independent variable. In this case all the partial derivatives with respect to the variable assumed dependent will vanish.