² ³ $\frac{(2-x²)(3-x³)}{1-x+x²}$ tính đạo hàm của hàm số

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
y = \frac{{\left( {2 – {x^2}} \right)\left( {3 – {x^3}} \right)}}{{1 – x + {x^2}}} = \frac{{6 – 2{x^3} – 3{x^2} + {x^5}}}{{1 – x + {x^2}}} = \frac{{{x^5} – 2{x^3} – 3{x^2} + 6}}{{{x^2} – x + 1}}\\
 \Rightarrow y’ = \frac{{\left( {{x^5} – 2{x^3} – 3{x^2} + 6} \right)’.\left( {{x^2} – x + 1} \right) – \left( {{x^2} – x + 1} \right)’.\left( {{x^5} – 2{x^3} – 3{x^2} + 6} \right)}}{{{{\left( {{x^2} – x + 1} \right)}^2}}}\\
 = \frac{{\left( {5{x^4} – 6{x^2} – 6x + 6} \right)\left( {{x^2} – x + 1} \right) – \left( {2x – 1} \right)\left( {{x^5} – 2{x^3} – 3{x^2} + 6} \right)}}{{{{\left( {{x^2} – x + 1} \right)}^2}}}\\
 = \frac{{\left( {5{x^6} – 5{x^5} – {x^4} + 6{x^2} – 12x + 6} \right) – \left( {2{x^6} – {x^5} – 2{x^4} – 4{x^3} + 3{x^2} + 12x – 6} \right)}}{{{{\left( {{x^2} – x + 1} \right)}^2}}}\\
 = \frac{{2{x^6} – 4{x^5} + {x^4} + 4{x^3} + 3{x^2} – 24x + 12}}{{{{\left( {{x^2} – x + 1} \right)}^2}}}
\end{array}\)