Cho hàm số f(x)=ln2018x/x+1. Tính tổng S=f'(1) +f'(2) +...+f'(2018)

* Hướng dẫn giải

Chọn D.

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

Ta có

$$\begin{gathered} S=f\text{'}\left(1\right)+f\text{'}\left(2\right)+f\text{'}\left(3\right)+...+f\text{'}\left(2018\right) \\ =\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{2018}-\frac{1}{2019}\right) \\ =1-\frac{1}{2019}=\frac{2018}{2019}. \end{gathered}$$