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
