The Chain Rule, using Rudin’s notation:
Let \(y=f(x)\). From the definition of the derivative we have:
\[\begin{align}
f(t)-f(x)=&(t-x)[f'(x)+u(t)] \\
g(s)-g(y) =& (s-y)[g'(y)+v(y)]
\end{align}
\]
where \(u(t) \to 0\) as \(t \to x\) and \(v(s) \to 0\) as \(s\to y = f(x)\)
Let \(h(x) = g(f(x))\)
\[\begin{align}
h(t) – h(x) =& g(f(t)) – g(f(x)) \\
=& (s-y)[g'(y) + v(s)] \\
=& (f(t) – f(x))[g'(y) + v(s)] \\
=& (t-x)[f'(x)+u(t)][g'(y)+v(s)]
\end{align}
\]
Therfore
\[
\displaystyle \frac{h(t)-h(x)}{t-x} \, = \, \frac{(t-x)[f'(x)+u(t)][g'(y)+v(s)]}{(t-x)}
\]
as \(t \to x\) then \(s \to y\), \(u(t) \to 0\), and \(v(s) \to 0\) so
\[
h'(x) = f'(x) \, g'(f(x)) =g'(f(x)) \, f'(x)
\]