Câu hỏi 33.4

\(y = \frac{{4x + 1}}{{\sqrt {{x^2} - x + 2} }}\)

\(\begin{array}{l}y' = \frac{{4\sqrt {{x^2} - x + 2}  - \left( {4x + 1} \right).\frac{{2x - 1}}{{2\sqrt {{x^2} - x + 2} }}}}{{{x^2} - x + 2}}\\\,\,\,\,\, = \frac{{8\left( {{x^2} - x + 2} \right) - \left( {4x + 1} \right).\left( {2x - 1} \right)}}{{2\sqrt {{x^2} - x + 2} \left( {{x^2} - x + 2} \right)}}\\\,\,\,\,\, = \frac{{8{x^2} - 8x + 16 - 8{x^2} + 4x - 2x + 1}}{{2\sqrt {{x^2} - x + 2} \left( {{x^2} - x + 2} \right)}}\\\,\,\,\,\, = \frac{{ - 6x + 17}}{{2\sqrt {{x^2} - x + 2} \left( {{x^2} - x + 2} \right)}}\end{array}\)

Chọn A.