a. Ta có: \(y = {x^{14}} + 2{x^8} + {x^2} \Rightarrow y' = 14{x^{13}} + 16{x^7} + 2x\)
b.
\(\eqalign{ & y' = \left( {{x^2} + 1} \right)'\left( {5 - 3{x^2}} \right) + \left( {{x^2} + 1} \right)\left( {5 - 3{x^2}} \right)' \cr & = 2x\left( {5 - 3{x^2}} \right) - 6x\left( {{x^2} + 1} \right) = 4x - 12{x^3} \cr} \)
c. \(y' = {{2\left( {{x^2} - 1} \right) - 2x\left( {2x} \right)} \over {{{\left( {{x^2} - 1} \right)}^2}}} = {{ - 2\left( {{x^2} + 1} \right)} \over {{{\left( {{x^2} - 1} \right)}^2}}}\)
d. \(y' = {{5\left( {{x^2} + x + 1} \right) - \left( {5x - 3} \right)\left( {2x + 1} \right)} \over {{{\left( {{x^2} + x + 1} \right)}^2}}} = {{ - 5{x^2} + 6x + 8} \over {{{\left( {{x^2} + x + 1} \right)}^2}}}\)
e. \(y' = {{\left( {2x + 2} \right)\left( {x + 1} \right) - \left( {{x^2} + 2x + 2} \right)} \over {{{\left( {x + 1} \right)}^2}}} = {{{x^2} + 2x} \over {{{\left( {x + 1} \right)}^2}}}\)
f.
\(\eqalign{ & y = \left( {2{x^2} - x} \right)\left( {3x + 2} \right) \cr & \Rightarrow y' = \left( {4x - 1} \right)\left( {3x + 2} \right) + \left( {2{x^2} - x} \right)3 \cr & = 18{x^2} + 2x - 2 \cr} \)
