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