\require{AMSmath} Uitwerkingen 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} $ ©2004-2024 WisFaq
\require{AMSmath}
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} $
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} $
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