Editing Chain rule (section)

=== Second proof ===
Another way of proving the chain rule is to measure the error in the linear approximation determined by the derivative. This proof has the advantage that it generalizes to several variables. It relies on the following equivalent definition of differentiability at a point: A function ''g'' is differentiable at ''a'' if there exists a real number ''g''′(''a'') and a function ''ε''(''h'') that tends to zero as ''h'' tends to zero, and furthermore
<math display="block">g(a + h) - g(a) = g'(a) h + \varepsilon(h) h.</math>
Here the left-hand side represents the true difference between the value of ''g'' at ''a'' and at {{math|''a'' + ''h''}}, whereas the right-hand side represents the approximation determined by the derivative plus an error term.

In the situation of the chain rule, such a function ''ε'' exists because ''g'' is assumed to be differentiable at ''a''. Again by assumption, a similar function also exists for ''f'' at ''g''(''a''). Calling this function ''η'', we have
<math display="block">f(g(a) + k) - f(g(a)) = f'(g(a)) k + \eta(k) k.</math>
The above definition imposes no constraints on ''η''(0), even though it is assumed that ''η''(''k'') tends to zero as ''k'' tends to zero. If we set {{math|1=''η''(0) = 0}}, then ''η'' is continuous at 0.

Proving the theorem requires studying the difference {{math|''f''(''g''(''a'' + ''h'')) − ''f''(''g''(''a''))}} as ''h'' tends to zero. The first step is to substitute for {{math|''g''(''a'' + ''h'')}} using the definition of differentiability of ''g'' at ''a'':
<math display="block">f(g(a + h)) - f(g(a)) = f(g(a) + g'(a) h + \varepsilon(h) h) - f(g(a)).</math>
The next step is to use the definition of differentiability of ''f'' at ''g''(''a''). This requires a term of the form {{math|''f''(''g''(''a'') + ''k'')}} for some ''k''. In the above equation, the correct ''k'' varies with ''h''. Set {{math|1=''k''<sub>''h''</sub> = ''g''′(''a'') ''h'' + ''ε''(''h'') ''h''}} and the right hand side becomes {{math|''f''(''g''(''a'') + ''k''<sub>''h''</sub>) − ''f''(''g''(''a''))}}. Applying the definition of the derivative gives:
<math display="block">f(g(a) + k_h) - f(g(a)) = f'(g(a)) k_h + \eta(k_h) k_h.</math>
To study the behavior of this expression as ''h'' tends to zero, expand ''k''<sub>''h''</sub>. After regrouping the terms, the right-hand side becomes:
<math display="block">f'(g(a)) g'(a)h + [f'(g(a)) \varepsilon(h) + \eta(k_h) g'(a) + \eta(k_h) \varepsilon(h)] h.</math>
Because ''ε''(''h'') and ''η''(''k''<sub>''h''</sub>) tend to zero as ''h'' tends to zero, the first two bracketed terms tend to zero as ''h'' tends to zero. Applying the same theorem on products of limits as in the first proof, the third bracketed term also tends zero. Because the above expression is equal to the difference {{math|''f''(''g''(''a'' + ''h'')) − ''f''(''g''(''a''))}}, by the definition of the derivative {{math|''f'' ∘ ''g''}} is differentiable at ''a'' and its derivative is {{math|''f''′(''g''(''a'')) ''g''′(''a'').}}

The role of ''Q'' in the first proof is played by ''η'' in this proof. They are related by the equation:
<math display="block">Q(y) = f'(g(a)) + \eta(y - g(a)). </math>
The need to define ''Q'' at ''g''(''a'') is analogous to the need to define ''η'' at zero.