Cho ham số f(x)=ln 2018x/x+1. Tính tổng S=f'(1)+f'(2)+...+f'(2018)?

Ta có \(f'(x) = {\left( {\ln \frac{{2018x}}{{x + 1}}} \right)^\prime } = \frac{1}{{\frac{{2018x}}{{x + 1}}}} \cdot {\left( {\frac{{2018x}}{{x + 1}}} \right)^\prime } = \frac{{x + 1}}{{2018x}} \cdot \frac{{2018}}{{{{\left( {x + 1} \right)}^2}}} = \frac{1}{{x\left( {x + 1} \right)}}\).

Vậy \[S = f'\left( 1 \right) + f'\left( 2 \right) +  \ldots  + f'\left( {2018} \right)\]\( = \frac{1}{{1.2}} + \frac{1}{{2.3}} + .. + \frac{1}{{2018.2019}}\)

\( = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + .. + \frac{1}{{2018}} - \frac{1}{{2019}}\)\( = 1 - \frac{1}{{2019}} = \frac{{2018}}{{2019}}.\) Chọn D.