Đạo hàm của hàm số \(f \left( x \right) = { \left( { \dfrac{{2x + 1}}{{x + 1}}} \right)^{2018}} \) là:
A.\(f'\left( x \right) = 2018{\left( {\dfrac{{2x + 1}}{{x + 1}}} \right)^{2017}}\left( {\dfrac{{ - 1}}{{x + 1}}} \right)\)
B.\(f'\left( x \right) = 2018\dfrac{{{{\left( {2x + 1} \right)}^{2017}}}}{{{{\left( {x + 1} \right)}^{2019}}}}\)
C.\(f'\left( x \right) = 2018{\left( {\dfrac{{2x + 1}}{{x + 1}}} \right)^{2017}}\)
D.\(f'\left( x \right) = {\left( {\dfrac{{2x + 1}}{{x + 1}}} \right)^{2017}}{\left( {\dfrac{1}{{x + 1}}} \right)^2}\)
A.\(f'\left( x \right) = 2018{\left( {\dfrac{{2x + 1}}{{x + 1}}} \right)^{2017}}\left( {\dfrac{{ - 1}}{{x + 1}}} \right)\)
B.\(f'\left( x \right) = 2018\dfrac{{{{\left( {2x + 1} \right)}^{2017}}}}{{{{\left( {x + 1} \right)}^{2019}}}}\)
C.\(f'\left( x \right) = 2018{\left( {\dfrac{{2x + 1}}{{x + 1}}} \right)^{2017}}\)
D.\(f'\left( x \right) = {\left( {\dfrac{{2x + 1}}{{x + 1}}} \right)^{2017}}{\left( {\dfrac{1}{{x + 1}}} \right)^2}\)
Loga
Hóa Học lớp 12
