Đề bài
Tìm \(f'\left( 1 \right),f'\left( 2 \right),f'\left( 3 \right)\) nếu \(f\left( x \right) = \left( {x - 1} \right){\left( {x - 2} \right)^2}{\left( {x - 3} \right)^3}.\)
Lời giải chi tiết
\(\begin{array}{l}
f'\left( x \right) = \left( {x - 1} \right)'{\left( {x - 2} \right)^2}{\left( {x - 3} \right)^3}\\
+ \left( {x - 1} \right)\left[ {{{\left( {x - 2} \right)}^2}} \right]'{\left( {x - 3} \right)^3}\\
+ \left( {x - 1} \right){\left( {x - 2} \right)^2}\left[ {{{\left( {x - 3} \right)}^3}} \right]'\\
= {\left( {x - 2} \right)^2}{\left( {x - 3} \right)^3}\\
+ \left( {x - 1} \right).2\left( {x - 2} \right){\left( {x - 3} \right)^3}\\
+ \left( {x - 1} \right){\left( {x - 2} \right)^2}.3{\left( {x - 3} \right)^2}\\
\Rightarrow f'\left( 1 \right) = {\left( {1 - 2} \right)^2}{\left( {1 - 3} \right)^3} + 0 = - 8\\
f'\left( 2 \right) = 0 + 0 + 0 = 0\\
f'\left( 3 \right) = 0 + 0 + 0 = 0
\end{array}\)
