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

Trả lời: \( - \frac{{2018}}{{2019}}\)

Lời giải

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

Khi đó \(S = \frac{1}{2} - 1 + \frac{1}{3} - \frac{1}{2} +  \ldots  + \frac{1}{{2018}} - \frac{1}{{2017}} + \frac{1}{{2019}} - \frac{1}{{2018}} = \frac{1}{{2019}} - 1 =  - \frac{{2018}}{{2019}}\).