1
$f(x) = x^4 \cdot 2\sqrt x $
$f'(x) = 4x^3 \cdot 2\sqrt x + x^4 \cdot 2 \cdot \frac{1}{{2\sqrt x }}$
$f'(x) = 8x^3 \sqrt x + x^3 \sqrt x $
$f'(x) = 9x^3 \sqrt x $
Alternatieve oplossing:
$f(x) = x^4 \cdot 2\sqrt x = 2x^{4\frac{1}{2}} $
$f'(x) = 2 \cdot 4\frac{1}{2}x^{3\frac{1}{2}} = 9x^3 \sqrt x $
2
$
f(x) = 5x\sqrt {x - 4}
$
$
f'(x) = 5\sqrt {x - 4} + 5x \cdot \frac{1}{{2\sqrt {x - 4} }}
$
$
f'(x) = 5\sqrt {x - 4} + \frac{{5x}}{{2\sqrt {x - 4} }}
$
$
f'(x) = 5\sqrt {x - 4} \cdot \frac{{2\sqrt {x - 4} }}{{2\sqrt {x - 4} }} + \frac{{5x}}{{2\sqrt {x - 4} }}
$
$
f'(x) = \frac{{10\left( {x - 4} \right)}}{{2\sqrt {x - 4} }} + \frac{{5x}}{{2\sqrt {x - 4} }}
$
$
f'(x) = \frac{{10x - 40 + 5x}}{{2\sqrt {x - 4} }}
$
$
f'(x) = \frac{{15x - 40}}{{2\sqrt {x - 4} }}
$
Alternatieve oplossing:
$
f(x) = 5x\sqrt {x - 4} = 5\sqrt {x^3 - 4x^2 }
$
$
f'(x) = \frac{5}{{2\sqrt {x^3 - 4x^2 } }} \cdot \left( {3x^2 - 8x} \right)
$
$
f'(x) = \frac{{15x^2 - 40x}}{{2\sqrt {x^3 - 4x^2 } }}
$
$
f'(x) = \frac{{15x^2 - 40x}}{{2x\sqrt {x - 4} }}
$
$
f'(x) = \frac{{15x - 40}}{{2\sqrt {x - 4} }}
$
Die andere twee moet je dan maar 's zelf doen...
Zie eventueel ook extra oefeningen wortelvormen
WvR
12-6-2012