Beweis: Kettenregel

\begin{align*} f(x) \coloneqq& g(h(x)) \\ f'(x) =& \lim_{\delta \to 0}\left(\frac{f(x + \delta) - f(x)}{\delta}\right) \\ =& \lim_{\delta \to 0}\left(\frac{g(h(x + \delta)) - g(h(x))}{\delta} \cdot \frac{h(x + \delta) - h(x)}{h(x + \delta) - h(x)}\right) \\ =& \lim_{\delta \to 0}\left( \frac{g(h(x + \delta)) - g(h(x))}{h(x + \delta) - h(x)} \cdot \frac{h(x + \delta) - h(x)}{\delta}\right) \\ =& \lim_{\delta \to 0}\left( \frac{g(h(x + \delta)) - g(h(x))}{h(x + \delta) - h(x)}\right) \cdot \lim_{\delta \to 0}\left(\frac{h(x + \delta) - h(x)}{\delta}\right) \\ =& g'(h(x)) \cdot h'(x) \end{align*}