\[\begin{array}{l}y' = \frac{{\left[ {2{x^2} + x + 1} \right]'\left[ {{x^2} - x + 1} \right] - \left[ {2{x^2} + x + 1} \right]\left[ {{x^2} - x + 1} \right]'}}{{{{\left[ {{x^2} - x + 1} \right]}^2}}}\\ = \frac{{\left[ {4x + 1} \right]\left[ {{x^2} - x + 1} \right] - \left[ {2{x^2} + x + 1} \right]\left[ {2x - 1} \right]}}{{{{\left[ {{x^2} - x + 1} \right]}^2}}}\\ = \frac{{4{x^3} + {x^2} - 4{x^2} - x + 4x + 1 - \left[ {4{x^3} + 2{x^2} + 2x - 2{x^2} - x - 1} \right]}}{{{{\left[ {{x^2} - x + 1} \right]}^2}}}\\ = \frac{{ - 3{x^2} + 2x + 2}}{{{{\left[ {{x^2} - x + 1} \right]}^2}}}\end{array}\]
Đề bài
Tìm đạo hàm của hàm số sau:
\[y = {{2{x^2} + x + 1} \over {{x^2} - x + 1}}.\]
Lời giải chi tiết
\[\begin{array}{l}
y' = \frac{{\left[ {2{x^2} + x + 1} \right]'\left[ {{x^2} - x + 1} \right] - \left[ {2{x^2} + x + 1} \right]\left[ {{x^2} - x + 1} \right]'}}{{{{\left[ {{x^2} - x + 1} \right]}^2}}}\\
= \frac{{\left[ {4x + 1} \right]\left[ {{x^2} - x + 1} \right] - \left[ {2{x^2} + x + 1} \right]\left[ {2x - 1} \right]}}{{{{\left[ {{x^2} - x + 1} \right]}^2}}}\\
= \frac{{4{x^3} + {x^2} - 4{x^2} - x + 4x + 1 - \left[ {4{x^3} + 2{x^2} + 2x - 2{x^2} - x - 1} \right]}}{{{{\left[ {{x^2} - x + 1} \right]}^2}}}\\
= \frac{{ - 3{x^2} + 2x + 2}}{{{{\left[ {{x^2} - x + 1} \right]}^2}}}
\end{array}\]