f(x)=Ln 2018x/x+1 tính S=f'(1) +f'(2)+...+f'(2019)

Đáp án:$\frac{{2019}}{{2020}}$

Giải thích các bước giải: $\eqalign{
  & \ln \left( {\frac{{2018x}}{{x + 1}}} \right) = \ln (2018x) - \ln (x + 1)  \cr 
  & \ln '\left( {\frac{{2018x}}{{x + 1}}} \right) = \ln '(2018x) - \ln '(x + 1) = \frac{1}{x} - \frac{1}{{x + 1}}  \cr 
  & f'(1) = 1 - \frac{1}{2}  \cr 
  & f'(2) = \frac{1}{2} - \frac{1}{3}  \cr 
  & f'(3) = \frac{1}{3} - \frac{1}{4}  \cr 
  & ...  \cr 
  & f'(2019) = \frac{1}{{2019}} - \frac{1}{{2020}}  \cr 
  & S = f'(1) + f'(2) + f'(3) + ... + f'(2019) = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{2019}} - \frac{1}{{2020}} = 1 - \frac{1}{{2020}} = \frac{{2019}}{{2020}} \cr} $