Editing Implicit function theorem (section)

== Two variables case ==

Let <math>f:\R^2 \to \R</math> be a continuously differentiable function defining the [[implicit equation]] of a [[curve]] <math> f(x,y) = 0 </math>. Let <math>(x_0, y_0)</math> be a point on the curve, that is, a point such that <math>f(x_0, y_0)=0</math>. In this simple case, the implicit function theorem can be stated as follows:
{{math theorem|math_statement=If {{tmath|f(x,y)}} is a function that is continuously differentiable in a neighbourhood of the point {{tmath|(x_0,y_0)}}, and 
<math display=inline>\frac{\partial f}{ \partial y} (x_0, y_0) \neq 0,</math> then there exists a unique differentiable function {{tmath|\varphi}} such that {{tmath|1=y_0=\varphi(x_0)}} and {{tmath|1=f(x, \varphi(x))=0}} in a neighbourhood of {{tmath|x_0}}.}}

'''Proof.''' By differentiating the equation {{tmath|1=f(x, \varphi(x))=0}}, one gets
<math display=block>\frac{\partial f}{ \partial x}(x, \varphi(x))+\varphi'(x)\, \frac{\partial f}{ \partial y}(x, \varphi(x))=0. </math>
and thus 
<math display=block>\varphi'(x)=-\frac{\frac{\partial f}{ \partial x}(x, \varphi(x))}{\frac{\partial f}{ \partial y}(x, \varphi(x))}.</math>
This gives an [[ordinary differential equation]] for {{tmath|\varphi}}, with the initial condition {{tmath|1=\varphi(x_0) = y_0}}.

Since <math display=inline>\frac{\partial f}{ \partial y} (x_0, y_0) \neq 0,</math> the right-hand side of the differential equation is continuous, upper bounded and lower bounded on some closed interval around {{tmath|x_0}}. It is therefore [[Lipschitz continuous]],{{dubious|date=April 2025}} and the [[Cauchy-Lipschitz theorem]] applies for proving the existence of a unique solution.