Here is the Implicit Function Theorem statement:
"Let $g : R^k \times R^n \to R^n$ be a continously differentiable function s.t. $g(x_0, y_0) = c$ and $D_yg(x_0,y_0) : R^n \to R^n$ is an isomorphism. Then $\exists_{x_0 \in U}$ and $\exists_{y_0 \in V}$, $\exists_{\phi : U \to V}$ s.t. 1) $\left \{ (x,y) : g(x,y) = c \right \}$, $U \times V = $graph($\phi$) and 2) $\phi$ is differentiable and $D\phi(x) = -\left [ D_yg(x,\phi(x)) \right ]^{-1} \left [ D_xg(x,\phi(x))\right ]$"
I am confused on the meaning of condition $1)$, the mapping $\phi : U \to V$, and how we prove that the functions $g : R^k \times R^n \to R^n$ and $\phi : U \to V$ is differentiable.
Looking at an IFT example in http://bama.ua.edu/~mjvanessen/IFT%20Examples.pdf , I was trying to piece the statement of the theorem to the problem (#1). In the first problem we are looking for a function $\phi(x)$ s.t. $g(x,\phi(x)) = 0$, but how does the mapping $\phi : U \to V$ work in the problem. Can someone show me this mapping in the problem we are working on?
Then, how do we show that $g()$ is continuously differentiable and $\phi()$ is differentiable? If someone could explain in detail this would help me understand this theorem a lot better. Thanks.
$\phi$ gives the solution of $g(x,y)=0$. Namely, $y=\phi(x)$.
$g$ continuously differentiable is the hypothesis. Usually $g$ will be some known function and checking that is continuously differentiable will be trivial. The differentiability of $\phi$ is the conclusion. $\phi$ is differentiable because of the theorem and you don't need show it again. And $\phi$ will be "unknown" (no explicit formula except in some cases).
EDIT: try with the example $x^2+y^2=1$. This curve is not the graph of a function $y=y(x)$, but some chunks are graphs of some function(s).