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}
