\eqalign{ & De{\text{ }}afgeleide{\text{ }}f'(x){\text{ }}wordt{\text{ }}gedefinieerd{\text{ }}als: \cr & f'(x) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f(x + \Delta x) - f(x)}} {{\Delta x}} \cr & Voorbeeld:f(x) = \sqrt x \cr & f'(x) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f(x + \Delta x) - f(x)}} {{\Delta x}} \cr & De\,\,worteltruuk! \cr & \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sqrt {x + \Delta x} - \sqrt x }} {{\Delta x}} = \cr & \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sqrt {x + \Delta x} - \sqrt x }} {{\Delta x}} \cdot \frac{{\sqrt {x + \Delta x} + \sqrt x }} {{\sqrt {x + \Delta x} + \sqrt x }} \cr & \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta x}} {{\Delta x \cdot \left( {\sqrt {x + \Delta x} + \sqrt x } \right)}} \cr & Omdat\,\,\Delta x \to 0\,\,kan\,\,je\,\,\Delta x\,\,wegdelen\,: \cr & \mathop {\lim }\limits_{\Delta x \to 0} \frac{1} {{\sqrt {x + \Delta x} + \sqrt x }} = \frac{1} {{2\sqrt x }} \cr}
