$
\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}
$
