\require{AMSmath}

Voorbeeld 2

\eqalign{   & f(x) = \frac{{x^2  + 2x - 12}} {{3x(x + 2)}}  \cr   & f'(x) = \frac{{\left( {2x + 2} \right)\left( {3x(x + 2)} \right) - \left( {x^2  + 2x - 12} \right)\left( {6x + 6} \right)}} {{\left( {3x\left( {x + 2} \right)} \right)^2 }}  \cr   & f'(x) = \frac{{6\left( {x + 1} \right)\left( {x(x + 2)} \right) - 6\left( {x^2  + 2x - 12} \right)\left( {x + 1} \right)}} {{9\left( {x\left( {x + 2} \right)} \right)^2 }}  \cr   & f'(x) = \frac{{6\left( {x + 1} \right)\left( {\left( {x(x + 2)} \right) - \left( {x^2  + 2x - 12} \right)} \right)}} {{9\left( {x\left( {x + 2} \right)} \right)^2 }}  \cr   & f'(x) = \frac{{6\left( {x + 1} \right)\left( {\left( {x^2  + 2x} \right) - \left( {x^2  + 2x - 12} \right)} \right)}} {{9\left( {x\left( {x + 2} \right)} \right)^2 }}  \cr   & f'(x) = \frac{{72\left( {x + 1} \right)}} {{9\left( {x\left( {x + 2} \right)} \right)^2 }}  \cr   & f'(x) = \frac{{8\left( {x + 1} \right)}} {{x^2 \left( {x + 2} \right)^2 }} \cr}

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