Als ik verder geen rekenfouten heb gemaakt krijg je zoiets als dit:
$
\eqalign{
& f(x) = \frac{{\root 4 \of x }}
{{x\sqrt {x - 2} }} \cr
& f(x) = \frac{1}
{{x^{\frac{3}
{4}} \cdot \left( {x - 2} \right)^{\frac{1}
{2}} }} = \frac{1}
{{\left( {x^{\frac{5}
{2}} - 2x^{\frac{3}
{2}} } \right)^{\frac{1}
{2}} }} = \left( {x^{\frac{5}
{2}} - 2x^{\frac{3}
{2}} } \right)^{ - \frac{1}
{2}} \cr
& f'(x) = - \frac{1}
{2}\left( {x^{\frac{5}
{2}} - 2x^{\frac{3}
{2}} } \right)^{ - \frac{3}
{2}} \left( {\frac{5}
{2}x^{\frac{3}
{2}} - 3x^{\frac{1}
{2}} } \right) \cr
& f'(x) = \frac{{ - \frac{1}
{2}\left( {\frac{5}
{2}x^{\frac{3}
{2}} - 3x^{\frac{1}
{2}} } \right)}}
{{\left( {x^{\frac{5}
{2}} - 2x^{\frac{3}
{2}} } \right)^{\frac{3}
{2}} }} \cr
& f'(x) = \frac{{ - \left( {5x^{\frac{3}
{2}} - 6x^{\frac{1}
{2}} } \right)}}
{{4\left( {x^{\frac{5}
{2}} - 2x^{\frac{3}
{2}} } \right)^{\frac{3}
{2}} }} \cr
& f'(x) = \frac{{ - 5x + 6}}
{{4x^{\frac{7}
{4}} \left( {x - 2} \right)^{\frac{3}
{2}} }} \cr
& f'(x) = \frac{{ - 5x + 6}}
{{4\root 4 \of {x^7 } \sqrt {\left( {x - 2} \right)^3 } }} \cr}
$
Helpt dat? Anders nog maar verder vragen.
WvR
23-1-2018