Tính đạo hàm của hàm số sau: (y = dfrac{{{x^2} + 2x – 1}}{{x + 1}})

(y = dfrac{{{x^2} + 2x – 1}}{{x + 1}})

(begin{array}{l}y’ = dfrac{{left( {2{x^2} + 2x – 1} right)’left( {x + 1} right) – left( {x + 1} right)’left( {2{x^2} + 2x – 1} right)}}{{{{left( {x + 1} right)}^2}}}\ = dfrac{{left( {4{rm{x}} + 2} right)left( {x + 1} right) – left( {2{{rm{x}}^2} + 2x – 1} right)}}{{{{left( {x + 1} right)}^2}}}\ = dfrac{{2{{rm{x}}^2} + 4{rm{x}} + 3}}{{{{left( {x + 1} right)}^2}}}end{array})