Cho hàm số \[f(x) = \frac{{{x^2} + x - 1}}\]. Xét hai câu sau: \[(I):f'(x) = 1 - \frac{1}{{{{(x - 1)}^2}}},\]\[\forall x \ne 1.\] \[(II):f'(x) = \frac{{{x^2} - 2x}}{{{{(x - 1)}^2}}},\]\[\forall x \ne 1.\] Hãy chọn câu đúng:

Hướng dẫn giải:

Chọn D

Áp dụng công thức \[{\left( {\frac{u}{v}} \right)^\prime } = \frac{{u'.v - v'.u}}{{{v^2}}}\] ta có:

\[\forall x \ne 1\], ta có: \[f(x) = \frac{{{x^2} + x - 1}}\]\[ \Rightarrow \]\[f'(x) = \frac{{({x^2} + x - 1)'.(x - 1) - (x - 1)'.({x^2} + x - 1)}}{{{{(x - 1)}^2}}}\]

\[ \Rightarrow \]\[f'(x)\]\[ = \]\[\frac{{(2x + 1).(x - 1) - 1.({x^2} + x - 1)}}{{{{(x - 1)}^2}}}\]\[ = \]\[\frac{{2{x^2} - 2x + x - 1 - {x^2} - x + 1}}{{{{(x - 1)}^2}}}\]\[ = \]\[\frac{{{x^2} - 2x}}{{{{(x - 1)}^2}}}\] \[ \Rightarrow \]\[(II)\]đúng.

Mặt khác:\[f'(x)\]\[ = \]\[\frac{{{x^2} - 2x}}{{{{(x - 1)}^2}}} = \frac{{{x^2} - 2x + 1 - 1}}{{{{(x - 1)}^2}}} = \frac{{{{(x - 1)}^2} - 1}}{{{{(x - 1)}^2}}} = 1 - \frac{1}{{{{(x - 1)}^2}}}\] \[ \Rightarrow \]\[(I)\]đúng.