Cho hàm số \(f\left( x \right) = \frac{1}.\) Tính \(f''\left( { - 1} \right).\)

Ta có \[f\left( x \right) = \frac \Rightarrow f'\left( x \right) = \frac{{{{\left( {2x - 1} \right)}^2}}} = - \frac{2}{{{{\left( {2x - 1} \right)}^2}}}\]

Suy ra \(f''\left( x \right) = {\left[ { - \frac{2}{{{{\left( {2x - 1} \right)}^2}}}} \right]^\prime } = \frac \right]}^\prime }}}{{{{\left( {2x - 1} \right)}^4}}} = \frac{{8\left( {2x - 1} \right)}}{{{{\left( {2x - 1} \right)}^4}}} = \frac{8}{{{{\left( {2x - 1} \right)}^3}}}.\)

Khi đó \[f''\left( { - 1} \right) = \frac{8}{{{{\left[ {2 \cdot \left( { - 1} \right) - 1} \right]}^3}}} = \frac{8}{{{{\left( { - 3} \right)}^3}}} = - \frac{8}.\] Chọn A.