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