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