$
\eqalign{
& f(x) = (3x + 4)\root 4 \of {\left( {2x + 1} \right)^3 } \cr
& f(x) = (3x + 4)\left( {2x + 1} \right)^{\frac{3}
{4}} \cr
& f'(x) = 3\left( {2x + 1} \right)^{\frac{3}
{4}} + (3x + 4) \cdot \frac{3}
{4}(2x + 1)^{ - \frac{1}
{4}} \cdot 2 \cr
& f'(x) = 3\left( {2x + 1} \right)^{\frac{3}
{4}} + (3x + 4) \cdot \frac{3}
{{2\left( {2x + 1} \right)^{\frac{1}
{4}} }} \cr
& f'(x) = 3\left( {2x + 1} \right)^{\frac{3}
{4}} + \frac{{9x + 12}}
{{2\left( {2x + 1} \right)^{\frac{1}
{4}} }} \cr
& f'(x) = 3\left( {2x + 1} \right)^{\frac{3}
{4}} \cdot \frac{{2\left( {2x + 1} \right)^{\frac{1}
{4}} }}
{{2\left( {2x + 1} \right)^{\frac{1}
{4}} }} + \frac{{9x + 12}}
{{2\left( {2x + 1} \right)^{\frac{1}
{4}} }} \cr
& f'(x) = \frac{{6\left( {2x + 1} \right)}}
{{2\left( {2x + 1} \right)^{\frac{1}
{4}} }} + \frac{{9x + 12}}
{{2\left( {2x + 1} \right)^{\frac{1}
{4}} }} \cr
& f'(x) = \frac{{12x + 6 + 9x + 12}}
{{2\left( {2x + 1} \right)^{\frac{1}
{4}} }} \cr
& f'(x) = \frac{{21x + 18}}
{{2\root 4 \of {2x + 1} }} \cr}
$
Bij wiskunde komt altijd alles wel ergens van pas.
WvR
22-10-2020