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