Met de productregel

$
\eqalign{
  & f(x) = \sqrt {2x^2  + x}   \cr
  & f(x) = \sqrt {x(2x + 1)}   \cr
  & f(x) = \sqrt x  \cdot \sqrt {2x + 1}   \cr
  & f'(x) = \frac{1}
{{2\sqrt x }} \cdot \sqrt {2x + 1}  + \sqrt x  \cdot \frac{1}
{{2\sqrt {2x + 1} }} \cdot 2  \cr
  & f'(x) = \frac{{\sqrt {2x + 1} }}
{{2\sqrt x }} + \frac{{\sqrt x }}
{{\sqrt {2x + 1} }}  \cr
  & f'(x) = \frac{{\sqrt {2x + 1} }}
{{2\sqrt x }} \cdot \frac{{\sqrt {2x + 1} }}
{{\sqrt {2x + 1} }} + \frac{{\sqrt x }}
{{\sqrt {2x + 1} }} \cdot \frac{{2\sqrt x }}
{{2\sqrt x }}  \cr
  & f'(x) = \frac{{2x + 1}}
{{2\sqrt x  \cdot \sqrt {2x + 1} }} + \frac{{2x}}
{{2\sqrt x \sqrt {2x + 1} }}  \cr
  & f'(x) = \frac{{4x + 1}}
{{2\sqrt x  \cdot \sqrt {2x + 1} }}  \cr
  & f'(x) = \frac{{4x + 1}}
{{2\sqrt {2x^2  + x} }} \cr}
$

Echt handig is het niet maar 't kan...:-)

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