Determine o valor da primeira derivada da função f (x )=2x/3x^2-2x+4 no ponto de abscissa x=1/3?

\[\eqalign{ & f\left( x \right) = \dfrac{{2x}}{{3{x^2} - 2x + 4}} \cr & f\left( x \right)' = \dfrac{{2\left( {3{x^2} - 2x + 4} \right) - 2x\left( {6x - 2} \right)}}{{{{\left( {3{x^2} - 2x + 4} \right)}^2}}} \cr & f\left( x \right)' = \dfrac{{\left( {6{x^2} - 4x + 8} \right) - \left( {12{x^2} - 4x} \right)}}{{{{\left( {3{x^2} - 2x + 4} \right)}^2}}} \cr & f\left( x \right)' = \dfrac{{\left( { - 6{x^2} + 8} \right)}}{{{{\left( {3{x^2} - 2x + 4} \right)}^2}}} }\]

Agora encontraremos o valor da derivada no ponto dado:

\[\eqalign{ & f\left( x \right)' = \dfrac{{\left( { - 6{x^2} + 8} \right)}}{{{{\left( {3{x^2} - 2x + 4} \right)}^2}}} \cr & f\left( {\dfrac{1}{3}} \right)' = \dfrac{{\left( { - 6{{\left( {\dfrac{1}{3}} \right)}^2} + 8} \right)}}{{{{\left( {3{{\left( {\dfrac{1}{3}} \right)}^2} - 2\left( {\dfrac{1}{3}} \right) + 4} \right)}^2}}} \cr & f\left( {\dfrac{1}{3}} \right)' = \dfrac{{\left( {\dfrac{{ - 6}}{9} + 8} \right)}}{{{{\left( {\dfrac{1}{3} - \dfrac{2}{3} + 4} \right)}^2}}} \cr & f\left( {\dfrac{1}{3}} \right)' = \dfrac{{\left( {\dfrac{{22}}{3}} \right)}}{{{{\left( {\dfrac{{11}}{3}} \right)}^2}}} \cr & f\left( {\dfrac{1}{3}} \right)' = \dfrac{{22}}{3} \cdot \dfrac{9}{{121}} \cr & f\left( {\dfrac{1}{3}} \right)' = \dfrac{{22 \cdot 3}}{{121}} \cr & f\left( {\dfrac{1}{3}} \right)' = \dfrac{{66}}{{121}} }\]

Portanto, obtemos que
\(\boxed{f\left( {\dfrac{1}{3}} \right)' = \dfrac{{66}}{{121}}}\)
.