Uitwerkingen
a.
$
\eqalign{
& f(x) = \frac{{1 - x}}
{{\sqrt x }} \cr
& f'(x) = \frac{{ - 1 \cdot \sqrt x - \left( {1 - x} \right) \cdot \frac{1}
{{2\sqrt x }}}}
{{\left( {\sqrt x } \right)^2 }} \cr
& f'(x) = \frac{{ - \sqrt x - \frac{1}
{{2\sqrt x }} + \frac{x}
{{2\sqrt x }}}}
{x} \cr
& f'(x) = \frac{{ - x - \frac{1}
{2} + \frac{x}
{2}}}
{{x\sqrt x }} = - \frac{{x + 1}}
{{2x\sqrt x }} \cr}
$
Soms kan het handig zijn om eerst het functievoorschrift anders te schrijven:
$
\eqalign{
& f(x) = \frac{{1 - x}}
{{\sqrt x }} = \frac{1}
{{\sqrt x }} - \sqrt x = x^{ - \frac{1}
{2}} - x^{\frac{1}
{2}} \cr
& f'(x) = - \frac{1}
{2}x^{ - 1\frac{1}
{2}} - \frac{1}
{2}x^{ - \frac{1}
{2}} \cr
& f'(x) = - \frac{1}
{{2x\sqrt x }} - \frac{1}
{{2\sqrt x }} = - \frac{{x + 1}}
{{2x\sqrt x }} \cr}
$
Maar of dat hier nu handig is...
b.
$
\eqalign{
& f(x) = \frac{{2x - 3}}
{{\sqrt {5x + 1} }} \cr
& f'(x) = \frac{{2\sqrt {5x + 1} - \left( {2x - 3} \right) \cdot \frac{1}
{{2\sqrt {5x + 1} }} \cdot 5}}
{{\left( {\sqrt {5x + 1} } \right)^2 }} \cr
& f'(x) = \frac{{2\sqrt {5x + 1} - \frac{{5\left( {2x - 3} \right)}}
{{2\sqrt {5x + 1} }}}}
{{5x + 1}} \cr
& f'(x) = \frac{{\left( {2\sqrt {5x + 1} } \right)^2 - 5\left( {2x - 3} \right)}}
{{2\left( {5x + 1} \right)\sqrt {5x + 1} }} \cr
& f'(x) = \frac{{4\left( {5x + 1} \right) - 5\left( {2x - 3} \right)}}
{{2\left( {5x + 1} \right)\sqrt {5x + 1} }} \cr
& f'(x) = \frac{{10x + 19}}
{{2\left( {5x + 1} \right)\sqrt {5x + 1} }} \cr}
$
c.
$\eqalign{
& f(x) = \frac{{\sqrt {2x + 5} }}{{3x - 1}} \cr
& f'(x) = \frac{{\frac{1}{{2\sqrt {2x + 5} }} \cdot 2 \cdot (3x - 1) - \sqrt {2x + 5} \cdot 3}}{{{{(3x - 1)}^2}}} \cr
& f'(x) = \frac{{\frac{{3x - 1}}{{\sqrt {2x + 5} }} - 3 \cdot \sqrt {2x + 5} }}{{{{(3x - 1)}^2}}} \cr
& f'(x) = \frac{{3x - 1 - 3(2x + 5)}}{{\sqrt {2x + 5} \cdot {{(3x - 1)}^2}}} \cr
& f'(x) = \frac{{3x - 1 - 6x - 15}}{{\sqrt {2x + 5} \cdot {{(3x - 1)}^2}}} \cr
& f'(x) = \frac{{ - 3x - 16}}{{\sqrt {2x + 5} \cdot {{(3x - 1)}^2}}} \cr
& f'(x) = - \frac{{3x + 16}}{{\sqrt {2x + 5} \cdot {{(3x - 1)}^2}}} \cr} $
d.
$\eqalign{
& f(x) = \frac{{2x + 3}}{{\sqrt {x - 1} }} \cr
& f'(x) = \frac{{2\sqrt {x - 1} - \left( {2x + 3} \right) \cdot \frac{1}{{2\sqrt {x - 1} }}}}{{x - 1}} \cr
& f'(x) = \frac{{2\sqrt {x - 1} - \frac{{2x + 3}}{{2\sqrt {x - 1} }}}}{{x - 1}} \cr
& f'(x) = \frac{{2\left( {x - 1} \right) - \frac{{2x + 3}}{2}}}{{\left( {x - 1} \right)\sqrt {x - 1} }} \cr
& f'(x) = \frac{{4\left( {x - 1} \right) - \left( {2x + 3} \right)}}{{2\left( {x - 1} \right)\sqrt {x - 1} }} \cr
& f'(x) = \frac{{4x - 4 - 2x - 3}}{{2\left( {x - 1} \right)\sqrt {x - 1} }} \cr
& f'(x) = \frac{{2x - 7}}{{2\left( {x - 1} \right)\sqrt {x - 1} }} \cr} $
e.
$\eqalign{
& f(x) = \frac{{\sqrt x - 1}}{{3{x^2} - 4}} \cr
& f'(x) = \frac{{\frac{1}{{2\sqrt x }}\left( {3{x^2} - 4} \right) - \left( {\sqrt x - 1} \right) \cdot 6x}}{{{{\left( {3{x^2} - 4} \right)}^2}}} \cr
& f'(x) = \frac{{\frac{{3{x^2} - 4}}{{2\sqrt x }} - \left( {\sqrt x - 1} \right) \cdot 6x}}{{{{\left( {3{x^2} - 4} \right)}^2}}} \cr
& f'(x) = \frac{{3{x^2} - 4 - \left( {\sqrt x - 1} \right) \cdot 6x \cdot 2\sqrt x }}{{2\sqrt x \cdot {{\left( {3{x^2} - 4} \right)}^2}}} \cr
& f'(x) = \frac{{3{x^2} - 4 - 12{x^2} + 12x\sqrt x }}{{2\sqrt x \cdot {{\left( {3{x^2} - 4} \right)}^2}}} \cr
& f'(x) = \frac{{ - 9{x^2} + 12x\sqrt x - 4}}{{2\sqrt x \cdot {{\left( {3{x^2} - 4} \right)}^2}}} \cr
& f'(x) = - \frac{{9{x^2} - 12x\sqrt x + 4}}{{2\sqrt x \cdot {{\left( {3{x^2} - 4} \right)}^2}}} \cr} $
f.
$\eqalign{
& f(x) = (2{x^3} - 2) \cdot \sqrt {x + 1} \cr
& f'(x) = 6{x^2} \cdot \sqrt {x + 1} + (2{x^3} - 2) \cdot \frac{1}{{2\sqrt {x + 1} }} \cr
& f'(x) = 6{x^2} \cdot \sqrt {x + 1} + \frac{{2{x^3} - 2}}{{2\sqrt {x + 1} }} \cr
& f'(x) = 6{x^2} \cdot \sqrt {x + 1} + \frac{{{x^3} - 1}}{{\sqrt {x + 1} }} \cr
& f'(x) = 6{x^2} \cdot \sqrt {x + 1} \cdot \frac{{\sqrt {x + 1} }}{{\sqrt {x + 1} }} + \frac{{{x^3} - 1}}{{\sqrt {x + 1} }} \cr
& f'(x) = \frac{{6{x^2} \cdot (x + 1)}}{{\sqrt {x + 1} }} + \frac{{{x^3} - 1}}{{\sqrt {x + 1} }} \cr
& f'(x) = \frac{{6{x^3} + 6{x^2} + {x^3} - 1}}{{\sqrt {x + 1} }} \cr
& f'(x) = \frac{{7{x^3} + 6{x^2} - 1}}{{\sqrt {x + 1} }} \cr} $
g.
$ \eqalign{ & f(x) = \left( {\frac{{x(x - 1)}} {{5 - x}}} \right)^5 \cr & f(x) = \left( {\frac{{x^2 - x}} {{5 - x}}} \right)^5 \cr & f'(x) = 5\left( {\frac{{x^2 - x}} {{5 - x}}} \right)^4 \cdot ... \cr & f'(x) = 5\left( {\frac{{x^2 - x}} {{5 - x}}} \right)^4 \cdot \frac{{\left( {2x - 1} \right)\left( {5 - x} \right) - \left( {x^2 - x} \right) \cdot - 1}} {{\left( {5 - x} \right)^2 }} \cr & f'(x) = 5\left( {\frac{{x^2 - x}} {{5 - x}}} \right)^4 \cdot \frac{{ - 2x^2 + 11x - 5 + x^2 - x}} {{\left( {5 - x} \right)^2 }} \cr & f'(x) = 5\left( {\frac{{x(x - 1)}} {{5 - x}}} \right)^4 \cdot \frac{{ - x^2 + 10x - 5}} {{\left( {5 - x} \right)^2 }} \cr & f'(x) = \frac{{5x^4 (x - 1)^4 }} {{\left( {5 - x} \right)^4 }} \cdot \frac{{ - x^2 + 10x - 5}} {{\left( {5 - x} \right)^2 }} \cr & f'(x) = \frac{{5x^4 (x - 1)^4 \left( { - x^2 + 10x - 5} \right)}} {{\left( {5 - x} \right)^6 }} \cr & f'(x) = - \frac{{5x^4 (x - 1)^4 \left( {x^2 - 10x + 5} \right)}} {{\left( {5 - x} \right)^6 }} \cr} $