Cho hàm số \(f\left( x \right) = \dfrac{x}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)...\left( {x - 2018} \right)}}\). Tính \(f'\left( 0 \right)\).

* Hướng dẫn giải

\(\begin{array}{l}f'\left( x \right) = \dfrac{\begin{array}{l}\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)...\left( {x - 2018} \right)\\\,\,\,\, - x\left[ {\left( {x - 2} \right)\left( {x - 3} \right)...\left( {x - 2018} \right) + \left( {x - 1} \right)\left( {x - 3} \right)...\left( {x - 2018} \right) + \left( {x - 2} \right)\left( {x - 3} \right)...\left( {x - 2017} \right)} \right]\end{array}}{{{{\left[ {\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)...\left( {x - 2018} \right)} \right]}^2}}}\\ \Rightarrow f'\left( 0 \right) = \dfrac{{\left( { - 1} \right)\left( { - 2} \right)...\left( { - 2018} \right)}}{{{{\left[ {\left( { - 1} \right)\left( { - 2} \right)...\left( { - 2018} \right)} \right]}^2}}} = \dfrac{1}{{\left( { - 1} \right)\left( { - 2} \right)...\left( { - 2018} \right)}} = \dfrac{1}{{2018!}}\end{array}\)

Chọn D.