Cho hàm số \(f\left( x \right) = \left( {2018 + x} \right)\left( {2017 + 2x} \right)\left( {2016 + 3x} \right)....\left( {1 + 2018x} \right)\). Tính \(f'\left( 1 \right)\).
Trả lời bởi giáo viên
\(f'\left( x \right) = \left( {2017 + 2x} \right)\left( {2016 + 3x} \right)....\left( {1 + 2018x} \right)\) \( + \left( {2018 + x} \right).2.\left( {2016 + 3x} \right)....\left( {1 + 2018x} \right)\) \( + ... +\left( {2018 + x} \right)\left( {2017 + 2x} \right)\left( {2016 + 3x} \right)....2018\)
Suy ra
\(f'\left( 1 \right) = {2019^{2017}} + {2.2019^{2017}} + {3.2019^{2017}} + ... + {2018.2019^{2017}}\)
\( = {2019^{2017}}\left( {1 + 2 + 3 + ... + 2018} \right)\)
\( = {2019^{2017}}.\dfrac{{2018.2019}}{2}\)\( = {1009.2019^{2018}}\).
Hướng dẫn giải:
- Tính \(f'\left( x \right)\)
- Cho \(x = 1\) tìm \(f'\left( 1 \right)\)