$
\eqalign{f(x) = \frac{{\root 4 \of x }}
{{x\sqrt x - 2}}}
$
Je kunt nu de afgeleide bepalen met de quotientregel. Je kunt de term in de teller niet samen nemen met die in de noemer, helaas...
Er zit niets anders op dan hard werken...
$
\eqalign{
& f(x) = \frac{{x^{\frac{1}
{4}} }}
{{x^{\frac{3}
{2}} - 2}} \cr
& f'(x) = \frac{{\frac{1}
{4}x^{ - \frac{3}
{4}} \cdot \left( {x^{\frac{3}
{2}} - 2} \right) - x^{\frac{1}
{4}} \cdot \frac{3}
{2}x^{\frac{1}
{2}} }}
{{\left( {x^{\frac{3}
{2}} - 2} \right)^2 }} \cr
& f'(x) = \frac{{\frac{1}
{4}x^{\frac{3}
{4}} - \frac{1}
{2}x^{ - \frac{3}
{4}} - \frac{3}
{2}x^{\frac{3}
{4}} }}
{{\left( {x^{\frac{3}
{2}} - 2} \right)^2 }} \cr
& f'(x) = \frac{{ - \frac{5}
{4}x^{\frac{3}
{4}} - \frac{1}
{2}x^{ - \frac{3}
{4}} }}
{{\left( {x^{\frac{3}
{2}} - 2} \right)^2 }} \cr
& f'(x) = \frac{{ - 5x^{\frac{3}
{4}} - 2x^{ - \frac{3}
{4}} }}
{{4\left( {x^{\frac{3}
{2}} - 2} \right)^2 }} \cr
& f'(x) = \frac{{ - 5x^{\frac{3}
{4}} - \frac{2}
{{x^{\frac{3}
{4}} }}}}
{{4\left( {x^{\frac{3}
{2}} - 2} \right)^2 }} \cr
& f'(x) = \frac{{ - 5x^{\frac{3}
{2}} - 2}}
{{4x^{\frac{3}
{4}} \left( {x^{\frac{3}
{2}} - 2} \right)^2 }} \cr
& f'(x) = \frac{{ - 5\sqrt {x^3 } - 2}}
{{4\root 4 \of {x^3 } \left( {\sqrt {x^3 } - 2} \right)^2 }} \cr
& f'(x) = - \frac{{5\sqrt {x^3 } + 2}}
{{4\root 4 \of {x^3 } \left( {\sqrt {x^3 } - 2} \right)^2 }} \cr}
$
Was dat het idee?
WvR
23-1-2018