Editing Chain rule (section)

=== Third proof ===
[[Constantin Carathéodory]]'s alternative definition of the differentiability of a function can be used to give an elegant proof of the chain rule.<ref>{{cite journal|first=Stephen|last=Kuhn|title=The Derivative á la Carathéodory|journal=[[The American Mathematical Monthly]]|year=1991|volume=98|issue=1|pages=40–44|doi=10.2307/2324035|jstor=2324035}}</ref>

Under this definition, a function {{mvar|f}} is differentiable at a point {{mvar|a}} if and only if there is a function {{mvar|q}}, continuous at {{mvar|a}} and such that {{math|1=''f''(''x'') − ''f''(''a'') = ''q''(''x'')(''x'' − ''a'')}}. There is at most one such function, and if {{mvar|f}} is differentiable at {{mvar|a}} then {{math|1=''f'' ′(''a'') = ''q''(''a'')}}.

Given the assumptions of the chain rule and the fact that differentiable functions and compositions of continuous functions are continuous, we have that there exist functions {{mvar|q}}, continuous at {{math|''g''(''a'')}}, and {{mvar|r}}, continuous at {{mvar|a}}, and such that,
<math display="block">f(g(x))-f(g(a))=q(g(x))(g(x)-g(a))</math>
and
<math display="block">g(x)-g(a)=r(x)(x-a).</math>
Therefore,
<math display="block">f(g(x))-f(g(a))=q(g(x))r(x)(x-a),</math>
but the function given by {{math|1=''h''(''x'') = ''q''(''g''(''x''))''r''(''x'')}} is continuous at {{mvar|a}}, and we get, for this {{mvar|a}}
<math display="block">(f(g(a)))'=q(g(a))r(a)=f'(g(a))g'(a).</math>
A similar approach works for continuously differentiable (vector-)functions of many variables. This method of factoring also allows a unified approach to stronger forms of differentiability, when the derivative is required to be [[Lipschitz continuity|Lipschitz continuous]], [[Hölder condition|Hölder continuous]], etc. Differentiation itself can be viewed as the [[polynomial remainder theorem]] (the little [[Étienne Bézout|Bézout]] theorem, or factor theorem), generalized to an appropriate class of functions.{{ citation needed|date=February 2016}}