Editing Inverse function theorem (section)

===Implicit function theorem===
The inverse function theorem can be used to solve a system of equations
:<math>\begin{align}
&f_1(x) = y_1 \\
&\quad \vdots\\
&f_n(x) = y_n,\end{align}</math>
i.e., expressing <math>y_1, \dots, y_n</math> as functions of <math>x = (x_1, \dots, x_n)</math>, provided the Jacobian matrix is invertible. The [[implicit function theorem]] allows to solve a more general system of equations:
:<math>\begin{align}
&f_1(x, y) = 0 \\
&\quad \vdots\\
&f_n(x, y) = 0\end{align}</math>
for <math>y</math> in terms of <math>x</math>. Though more general, the theorem is actually a consequence of the inverse function theorem. First, the precise statement of the implicit function theorem is as follows:<ref>{{harvnb|Spivak|1965|loc=Theorem 2-12.}}</ref>
*given a map <math>f : \mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^m</math>, if <math>f(a, b) = 0</math>, <math>f</math> is continuously differentiable in a neighborhood of <math>(a, b)</math> and the derivative of <math>y \mapsto f(a, y)</math> at <math>b</math> is invertible, then there exists a differentiable map <math>g : U \to V</math> for some neighborhoods <math>U, V</math> of <math>a, b</math> such that <math>f(x, g(x)) = 0</math>. Moreover, if <math>f(x, y) = 0, x \in U, y \in V</math>, then <math>y = g(x)</math>; i.e., <math>g(x)</math> is a unique solution.
To see this, consider the map <math>F(x, y) = (x, f(x, y))</math>. By the inverse function theorem, <math>F : U \times V \to W</math> has the inverse <math>G</math> for some neighborhoods <math>U, V, W</math>. We then have:
:<math>(x, y) = F(G_1(x, y), G_2(x, y)) = (G_1(x, y), f(G_1(x, y), G_2(x, y))),</math>
implying <math>x = G_1(x, y)</math> and <math>y = f(x, G_2(x, y)).</math> Thus <math>g(x) = G_2(x, 0)</math> has the required property. <math>\square</math>