Met machten

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