Voorbeeld 1
$
\eqalign{
& f(x) = \frac{1}
{{\root 4 \of {x^3 } }} = \frac{1}
{{x^{\frac{3}
{4}} }} = x^{ - \frac{3}
{4}} \cr
& f'(x) = - \frac{3}
{4}x^{ - 1\frac{3}
{4}} = - \frac{3}
{4}x^{ - \frac{7}
{4}} = - \frac{3}
{{4x^{\frac{7}
{4}} }} = - \frac{3}
{{4\root 4 \of {x^7 } }} \cr}
$
Voorbeeld 2
$
\eqalign{
& f(x) = \frac{{x^2 + 1}}
{{\sqrt x }} = x^{1\frac{1}
{2}} + x^{ - \frac{1}
{2}} \cr
& f'(x) = 1\frac{1}
{2}x^{\frac{1}
{2}} - \frac{1}
{2}x^{ - 1\frac{1}
{2}} = \frac{3}
{2}\sqrt x - \frac{1}
{{2x\sqrt x }} \cr
& f'(x) = \frac{3}
{2}\sqrt x \frac{{x\sqrt x }}
{{x\sqrt x }} - \frac{1}
{{2x\sqrt x }} = \frac{{3x^2 }}
{{2x\sqrt x }} - \frac{1}
{{2x\sqrt x }} = \frac{{3x^2 - 1}}
{{2x\sqrt x }} \cr}
$
Voorbeeld 3
$
\eqalign{
& f(x) = \sqrt {2x} = \sqrt 2 \cdot \sqrt x = \sqrt 2 \cdot x^{\frac{1}
{2}} \cr
& f'(x) = \frac{1}
{2}\sqrt 2 \cdot x^{ - \frac{1}
{2}} = \frac{{\sqrt 2 }}
{{2\sqrt x }} = \frac{1}
{{\sqrt 2 \cdot \sqrt x }} = \frac{1}
{{\sqrt {2x} }} \cr
& of... \cr
& f(x) = \sqrt {2x} = \left( {2x} \right)^{\frac{1}
{2}} \cr
& f'(x) = \frac{1}
{2}\left( {2x} \right)^{ - \frac{1}
{2}} \cdot 2 = \left( {2x} \right)^{ - \frac{1}
{2}} = \frac{1}
{{\sqrt {2x} }} \cr}
$
