Editing Chain rule (section)

=== First proof ===
One proof of the chain rule begins by defining the derivative of the composite function {{math|''f'' ∘ ''g''}}, where we take the [[Limit of a function|limit]] of the [[difference quotient]] for {{math|''f'' ∘ ''g''}} as {{mvar|x}} approaches {{mvar|a}}:
<math display="block">(f \circ g)'(a) = \lim_{x \to a} \frac{f(g(x)) - f(g(a))}{x - a}.</math>

Assume for the moment that <math>g(x)\!</math> does not equal <math>g(a)</math> for any <math>x</math> near <math>a</math>. Then the previous expression is equal to the product of two factors:
<math display="block">\lim_{x \to a} \frac{f(g(x)) - f(g(a))}{g(x) - g(a)} \cdot \frac{g(x) - g(a)}{x - a}.</math>

If <math>g</math> oscillates near {{mvar|a}}, then it might happen that no matter how close one gets to {{mvar|a}}, there is always an even closer {{mvar|x}} such that {{math|1=''g''(''x'') = ''g''(''a'')}}. For example, this happens near {{math|1=''a'' = 0}} for the [[continuous function]] {{mvar|g}} defined by {{math|1=''g''(''x'') = 0}} for {{math|1=''x'' = 0}} and {{math|1=''g''(''x'') = ''x''<sup>2</sup> sin(1/''x'')}} otherwise. Whenever this happens, the above expression is undefined because it involves [[division by zero]]. To work around this, introduce a function <math>Q</math> as follows:
<math display="block">Q(y) = \begin{cases}
\displaystyle\frac{f(y) - f(g(a))}{y - g(a)}, & y \neq g(a), \\
f'(g(a)), & y = g(a).
\end{cases}</math>
We will show that the difference quotient for {{math|''f'' ∘ ''g''}} is always equal to:
<math display="block">Q(g(x)) \cdot \frac{g(x) - g(a)}{x - a}.</math>

Whenever {{math|''g''(''x'')}} is not equal to {{math|''g''(''a'')}}, this is clear because the factors of {{math|''g''(''x'') − ''g''(''a'')}} cancel. When {{math|''g''(''x'')}} equals {{math|''g''(''a'')}}, then the difference quotient for {{math|''f'' ∘ ''g''}} is zero because {{math|''f''(''g''(''x''))}} equals {{math|''f''(''g''(''a''))}}, and the above product is zero because it equals {{math|''f''′(''g''(''a''))}} times zero. So the above product is always equal to the difference quotient, and to show that the derivative of {{math|''f'' ∘ ''g''}} at {{math|''a''}} exists and to determine its value, we need only show that the limit as {{math|''x''}} goes to {{math|''a''}} of the above product exists and determine its value.

To do this, recall that the limit of a product exists if the limits of its factors exist. When this happens, the limit of the product of these two factors will equal the product of the limits of the factors. The two factors are {{math|''Q''(''g''(''x''))}} and {{math|(''g''(''x'') − ''g''(''a'')) / (''x'' − ''a'')}}. The latter is the difference quotient for {{mvar|g}} at {{mvar|a}}, and because {{mvar|g}} is differentiable at {{mvar|a}} by assumption, its limit as {{mvar|x}} tends to {{mvar|a}} exists and equals {{math|''g''′(''a'')}}.

As for {{math|''Q''(''g''(''x''))}}, notice that {{math|''Q''}} is defined wherever ''{{mvar|f}}'' is. Furthermore, ''{{mvar|f}}'' is differentiable at {{math|''g''(''a'')}} by assumption, so {{math|''Q''}} is continuous at {{math|''g''(''a'')}}, by definition of the derivative. The function {{mvar|g}} is continuous at {{mvar|a}} because it is differentiable at {{mvar|a}}, and therefore {{math|''Q'' ∘ ''g''}} is continuous at {{mvar|a}}. So its limit as ''{{mvar|x}}'' goes to ''{{mvar|a}}'' exists and equals {{math|''Q''(''g''(''a''))}}, which is {{math|''f''′(''g''(''a''))}}.

This shows that the limits of both factors exist and that they equal {{math|''f''′(''g''(''a''))}} and {{math|''g''′(''a'')}}, respectively. Therefore, the derivative of {{math|''f'' ∘ ''g''}} at ''a'' exists and equals {{math|''f''′(''g''(''a''))}}{{math|''g''′(''a'')}}.