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
\[\begin{array}{l}
f'\left[ x \right] = \left[ {x - 1} \right]'{\left[ {x - 2} \right]^2}{\left[ {x - 3} \right]^3}\\
+ \left[ {x - 1} \right]\left[ {{{\left[ {x - 2} \right]}^2}} \right]'{\left[ {x - 3} \right]^3}\\
+ \left[ {x - 1} \right]{\left[ {x - 2} \right]^2}\left[ {{{\left[ {x - 3} \right]}^3}} \right]'\\
= {\left[ {x - 2} \right]^2}{\left[ {x - 3} \right]^3}\\
+ \left[ {x - 1} \right].2\left[ {x - 2} \right]{\left[ {x - 3} \right]^3}\\
+ \left[ {x - 1} \right]{\left[ {x - 2} \right]^2}.3{\left[ {x - 3} \right]^2}\\
\Rightarrow f'\left[ 1 \right] = {\left[ {1 - 2} \right]^2}{\left[ {1 - 3} \right]^3} + 0 = - 8\\
f'\left[ 2 \right] = 0 + 0 + 0 = 0\\
f'\left[ 3 \right] = 0 + 0 + 0 = 0
\end{array}\]