Editing Implicit function (section)

==Implicit function theorem==
[[Image:Implicit circle.svg|thumb|right|200px|The unit circle can be defined implicitly as the set of points {{math|(''x'', ''y'')}} satisfying {{math|1=''x''<sup>2</sup> + ''y''<sup>2</sup> = 1}}. Around point {{mvar|A}}, {{mvar|y}} can be expressed as an implicit function {{math|''y''(''x'')}}. (Unlike in many cases, here this function can be made explicit as {{math|1=''g''<sub>1</sub>(''x'') = {{sqrt|1 − ''x''<sup>2</sup>}}}}.) No such function exists around point {{mvar|B}}, where the [[tangent space]] is vertical.]]
{{main|Implicit function theorem}}

Let {{math|''R''(''x'', ''y'')}} be a [[differentiable function]] of two variables, and {{math|(''a'', ''b'')}} be a pair of [[real number]]s such that {{math|1=''R''(''a'', ''b'') = 0}}. If {{math|{{sfrac|∂''R''|∂''y''}} ≠ 0}}, then {{math|1=''R''(''x'', ''y'') = 0}} defines an implicit function that is differentiable in some small enough [[neighbourhood (mathematics)|neighbourhood]] of {{open-open|''a'', ''b''}}; in other words, there is a differentiable function {{mvar|f}} that is defined and differentiable in some neighbourhood of {{mvar|a}}, such that {{math|1=''R''(''x'', ''f''(''x'')) = 0}} for {{mvar|x}} in this neighbourhood.

The condition {{math|{{sfrac|∂''R''|∂''y''}} ≠ 0}} means that {{math|(''a'', ''b'')}} is a [[singular point of a curve|regular point]] of the [[implicit curve]] of implicit equation {{math|1=''R''(''x'', ''y'') = 0}} where the [[tangent]] is not vertical.

In a less technical language, implicit functions exist and can be differentiated, if the curve has a non-vertical tangent.<ref name="Stewart1998"/>{{rp|§11.5}}