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Câu hỏi: Tìm \(f'\left( 2 \right)\) nếu
\(f\left( x \right) = {x^2}\sin \left({x - 2} \right).\)
\(f\left( x \right) = {x^2}\sin \left({x - 2} \right).\)
Lời giải chi tiết
\(\begin{array}{l}
f'\left(x \right) = \left({{x^2}} \right)'\sin \left({x - 2} \right)\\
+ {x^2}\left[ {\sin \left({x - 2} \right)} \right]'\\
= 2x\sin \left({x - 2} \right) + {x^2}.\cos \left({x - 2} \right)\\
\Rightarrow f'\left(2 \right) = 2.2\sin 0 + {2^2}\cos 0\\
= 0 + 4.1 = 4
\end{array}\)
\(\begin{array}{l}
f'\left(x \right) = \left({{x^2}} \right)'\sin \left({x - 2} \right)\\
+ {x^2}\left[ {\sin \left({x - 2} \right)} \right]'\\
= 2x\sin \left({x - 2} \right) + {x^2}.\cos \left({x - 2} \right)\\
\Rightarrow f'\left(2 \right) = 2.2\sin 0 + {2^2}\cos 0\\
= 0 + 4.1 = 4
\end{array}\)