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