Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
b,\\
y = {x^2}\left( {{x^2} - 1} \right)\left( {{x^2} - 4} \right) = {x^2}.\left( {{x^4} - 5{x^2} + 4} \right) = {x^6} - 5{x^4} + 4{x^2}\\
\Rightarrow y' = 6{x^5} - 5.4{x^3} + 4.2x = 6{x^5} - 20{x^3} + 8x\\
c,\\
y = \frac{{1 + x - {x^2}}}{{1 - x + {x^2}}}\\
\Rightarrow y' = \frac{{\left( {1 + x - {x^2}} \right)'.\left( {1 - x + {x^2}} \right) - \left( {1 - x + {x^2}} \right)'.\left( {1 + x - {x^2}} \right)}}{{{{\left( {1 - x + {x^2}} \right)}^2}}}\\
= \frac{{\left( {1 - 2x} \right)\left( {1 - x + {x^2}} \right) - \left( { - 1 + 2x} \right)\left( {1 + x - {x^2}} \right)}}{{{{\left( {1 - x + {x^2}} \right)}^2}}}\\
= \frac{{\left( {1 - x + {x^2} - 2x + 2{x^2} - 2{x^3}} \right) - \left( { - 1 - x + {x^2} + 2x + 2{x^2} - 2{x^3}} \right)}}{{{{\left( {1 - x + {x^2}} \right)}^2}}}\\
= \frac{{\left( { - 2{x^3} + 3{x^2} - 3x + 1} \right) - \left( { - 2{x^3} + 3{x^2} + x - 1} \right)}}{{{{\left( {1 - x + {x^2}} \right)}^2}}}\\
= \frac{{ - 4x + 2}}{{{{\left( {1 - x + {x^2}} \right)}^2}}}
\end{array}\)
