Demonstração da regra da cadeia.

Sejam $f$ e $g$ funções diferenciáreis,

$f'(u) = \displaystyle\lim_{\Delta u \rightarrow 0} \dfrac{f(u + \Delta u) - f(u)}{\Delta u} =  \displaystyle\lim_{\Delta u \rightarrow 0} \dfrac{\Delta f(u)}{\Delta u}$;

$g'(x) = \displaystyle\lim_{\Delta x \rightarrow 0} \dfrac{g(x + \Delta x) - g(x)}{\Delta x} = \displaystyle\lim_{\Delta x \rightarrow 0} \dfrac{\Delta g(x)}{\Delta x}$.

Observemos que $\Delta g(x) \rightarrow 0\ \Leftrightarrow\ \Delta x \rightarrow 0$.

 

$(f \circ g)'(x) = \displaystyle\lim_{\Delta g(x) \rightarrow 0} \dfrac{\Delta (f \circ g)(x)}{\Delta g(x)} = \displaystyle\lim_{\Delta g(x) \rightarrow 0} \left[\dfrac{\Delta (f \circ g)(x)}{\Delta x} \cdot \dfrac{\Delta g(x)}{\Delta g(x)}\right] =$

$= \displaystyle\lim_{\Delta g(x) \rightarrow 0} \dfrac{\Delta (f \circ g)(x)}{\Delta g(x)} \cdot \displaystyle\lim_{\Delta g(x) \rightarrow 0} \dfrac{\Delta g(x)}{\Delta x} = f'[g(x)] \cdot g'(x)$

Quod Erat Demonstrandum.