Bewijs van de quotiëntregel
Stelling
\eqalign{ & Als{\text{ }}g(a) \ne 0{\text{ }}dan{\text{ }}is{\text{ }}\frac{f} {g}{\text{ }}differentieerbaar{\text{ }}in{\text{ }}a{\text{ }}met{\text{ }}afgeleide: \cr & \left( {\frac{f} {g}} \right)'\left( a \right) = \frac{{f'\left( a \right)g\left( a \right) - f\left( a \right)g'\left( a \right)}} {{\left( {g\left( a \right)} \right)^2 }} \cr}
Bewijs
\eqalign{ & Omdat{\text{ }}g{\text{ }}continu{\text{ }}is{\text{ }}in{\text{ }}a{\text{ }}is{\text{ }}\mathop {\lim {\text{ }}}\limits_{x \to a} g(x) = g(a) \ne 0,{\text{ }}dus{\text{ }}is \cr & er{\text{ }}een{\text{ }}r - omgeving{\text{ }}U_r (a){\text{ }}van{\text{ }}a{\text{ }}zo{\text{ }}dat{\text{ }}g(x) \ne 0{\text{ }}voor \cr & alle{\text{ }}x \in U_r (a)\,en\,\,U_r (a) \subset D. \cr & Dus\,\,\frac{1} {g}\,\,is\,\,gedefinieerd\,\,op\,\,U_r (a)\,\,en: \cr & \mathop {\lim }\limits_{x \to a} \frac{{\frac{1} {{g(x)}} - \frac{1} {{g(a)}}}} {{x - a}} = \mathop {\lim }\limits_{x \to a} \frac{{g(a) - g(x)}} {{g(a)g(x)\left( {x - a} \right)}} = \cr & \frac{{ - 1}} {{g(a)}}\mathop {\lim }\limits_{x \to a} \frac{1} {{g(x)}}\mathop {\lim }\limits_{x \to a} \frac{{g(x) - g(a)}} {{x - a}} = - \frac{{g'(a)}} {{\left( {g(a)} \right)^2 }} \cr}
Met de productregel:
\eqalign{ & \left( {\frac{f} {g}} \right)'\left( a \right) = f'\left( a \right)\left( {\frac{1} {g}} \right)\left( a \right) + f\left( a \right)\left( {\frac{1} {g}} \right)'\left( a \right) = \cr & \frac{{f'\left( a \right)}} {{g\left( a \right)}} + f\left( a \right) \cdot \frac{{ - g'\left( a \right)}} {{\left( {g\left( a \right)} \right)^2 }} = \frac{{f'\left( a \right)g\left( a \right) - f\left( a \right)g'\left( a \right)}} {{\left( {g\left( a \right)} \right)^2 }} \cr}
bron: Almering e.a. pag.141

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