Cho hàm số $y = \dfrac{{{x^2}}}{{1 - x}}$. Đạo h...

B

- Hướng dẫn giải

Phương pháp giải:

$\left( {\dfrac{1}{u}} \right)' = \dfrac{{u'}}{u}$.

Giải chi tiết:

Ta có:

$\begin{array}{l}f\left( x \right) = \dfrac{{{x^2}}}{{1 - x}} = \dfrac{{{x^2} - 1 + 1}}{{1 - x}} =  - x - 1 + \dfrac{1}{{1 - x}}\\ \Rightarrow f'\left( x \right) =  - 1 + \dfrac{1}{{{{\left( {x - 1} \right)}^2}}}\\\,\,\,\,\,\,f''\left( x \right) = \dfrac{{ - 2\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^4}}} = \dfrac{{ - 2}}{{{{\left( {x - 1} \right)}^3}}}\\\,\,\,\,\,f'''\left( x \right) = \dfrac{{2.3{{\left( {x - 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^6}}} = \dfrac{{2.3}}{{{{\left( {x - 1} \right)}^4}}}\\.......\\ \Rightarrow {f^{\left( {2018} \right)}}\left( x \right) = \dfrac{{ - 2.3...2018}}{{{{\left( {x - 1} \right)}^{2019}}}} =  - \dfrac{{2018!}}{{{{\left( {x - 1} \right)}^{2019}}}} = \dfrac{{2018!}}{{{{\left( {1 - x} \right)}^{2019}}}}\end{array}$

Chọn B.