(1 điểm) Tìm đạo hàm các hàm số

a) \[y = - {x^4} + \frac{3}{2}{x^2} + 2020x\];                                     

b) \[y = \frac{{\sqrt x + 2}}{{x + 1}}\].

a) \[y' = {\left( { - {x^4}} \right)^\prime } + {\left( {\frac{3}{2}{x^2}} \right)^\prime } + {\left( {2020x} \right)^\prime } = - 4{x^3} + 3x + 2020\].

b) \[y' = \frac{{{{\left( {\sqrt x + 2} \right)}^\prime } \cdot \left( {x + 1} \right) - \left( {\sqrt x + 2} \right){{\left( {x + 1} \right)}^\prime }}}{{{{\left( {x + 1} \right)}^2}}}\]

\[ = \frac{{\frac{1}{{2\sqrt x }}.\left( {x + 1} \right) - \left( {\sqrt x + 2} \right)}}{{{{\left( {x + 1} \right)}^2}}}\]

\[ = \frac{{x + 1 - 2x - 4\sqrt x }}{{2\sqrt x {{\left( {x + 1} \right)}^2}}}\]

\[ = \frac{{1 - x - 4\sqrt x }}{{2\sqrt x {{\left( {x + 1} \right)}^2}}}\].