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