Hoe moeilijk kan 't zijn?

\( \eqalign{ & f(x) = x\root 3 \of {(x^3 + 3x)^2 } \cr & f(x) = x\left( {x{}^3 + 3x} \right)^{\frac{2} {3}} \cr & f'(x) = 1 \cdot \left( {x{}^3 + 3x} \right)^{\frac{2} {3}} + x \cdot \frac{2} {3}\left( {x{}^3 + 3x} \right)^{ - \frac{1} {3}} \cdot \left( {3x^2 + 3} \right) \cr & f'(x) = \left( {x{}^3 + 3x} \right)^{\frac{2} {3}} + \frac{{2x\left( {3x^2 + 3} \right)}} {{3\left( {x{}^3 + 3x} \right)^{\frac{1} {3}} }} \cr & f'(x) = \left( {x{}^3 + 3x} \right)^{\frac{2} {3}} \cdot \frac{{3\left( {x{}^3 + 3x} \right)^{\frac{1} {3}} }} {{3\left( {x{}^3 + 3x} \right)^{\frac{1} {3}} }} + \frac{{2x\left( {3x^2 + 3} \right)}} {{3\left( {x{}^3 + 3x} \right)^{\frac{1} {3}} }} \cr & f'(x) = \frac{{3(x{}^3 + 3x)}} {{3\left( {x{}^3 + 3x} \right)^{\frac{1} {3}} }} + \frac{{2x\left( {3x^2 + 3} \right)}} {{3\left( {x{}^3 + 3x} \right)^{\frac{1} {3}} }} \cr & f'(x) = \frac{{3x^3 + 9x + 6x^3 + 6x}} {{3\left( {x{}^3 + 3x} \right)^{\frac{1} {3}} }} \cr & f'(x) = \frac{{9x^3 + 15x}} {{3\left( {x{}^3 + 3x} \right)^{\frac{1} {3}} }} \cr & f'(x) = \frac{{3x^3 + 5x}} {{\root 3 \of {x{}^3 + 3x} }} \cr} \)