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

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

Ta có

\(S = f'\left( 1 \right) + f'\left( 2 \right) + f'\left( 3 \right) + ... + f'\left( {2018} \right)\)

\( = \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + ... + \left( {\frac{1}{{2018}} - \frac{1}{{2019}}} \right)\)

\( = 1 - \frac{1}{{2019}} = \frac{{2018}}{{2019}}.\)

Đáp án D