Hàm số nào sau đây có $y' = 2x + \dfrac{1}{{{x^2}}}$?

$y' = 4{x^3} - 6x + 3$ 

$y' = 4{x^4} - 6x + 2$ 

$y' = 4{x^3} - 3x + 2$            

$y' = 4{x^3} - 6x + 2$

$- \dfrac{3}{{{{\left( {x + 2} \right)}^2}}}$             

$\dfrac{3}{{x + 2}}$ 

$\dfrac{3}{{{{\left( {x + 2} \right)}^2}}}$ 

$\dfrac{2}{{{{\left( {x + 2} \right)}^2}}}$

$\dfrac{1}{6}$ 

$\dfrac{1}{{12}}$ 

$- \dfrac{1}{6}$ 

$- \dfrac{1}{{12}}$

$\{1\}$

$\mathbb{R}\backslash \left\{ 1 \right\}$

$\emptyset$ 

$R$

$y'=- \dfrac{3}{{{x^4}}} + \dfrac{1}{{{x^3}}}$

$y'=\dfrac{{ - 3}}{{{x^4}}} + \dfrac{2}{{{x^3}}}$ 

$y'=\dfrac{{ - 3}}{{{x^4}}} - \dfrac{2}{{{x^3}}}$         

$y'=\dfrac{3}{{{x^4}}} - \dfrac{1}{{{x^3}}}$

$\dfrac{a}{c}$ 

$\dfrac{{ad - bc}}{{{{\left( {cx + d} \right)}^2}}}$ 

$\dfrac{{ad + bc}}{{{{\left( {cx + d} \right)}^2}}}$                         

$\dfrac{{ad - bc}}{{cx + d}}$