Cho \[y = \ln \left( {{e^{f\left( {2x - 1} \right)}} - 1} \right)\], tính y’:
A.
\[\frac{{2f'\left( {2x - 1} \right)}}{{{e^{f\left( {2x - 1} \right)}} - 1}}\]
B.
\[\frac{{{e^{f\left( {2x - 1} \right)}}f'\left( {2x - 1} \right)}}{{{e^{f\left( {2x - 1} \right)}} - 1}}\]
C.
\[\frac{{2.{e^{f\left( {2x - 1} \right)}}f'\left( {2x - 1} \right)}}{{{e^{f\left( {2x - 1} \right)}} - 1}}\]
D.
\[ - \frac{{2.{e^{f\left( {2x - 1} \right)}}f'\left( {2x - 1} \right)}}{{{e^{f\left( {2x - 1} \right)}} - 1}}\]
Trả lời:
Đáp án đúng: B
Ta có:
\[y' = \frac{{\left( {{e^{f\left( {2x - 1} \right)}} - 1} \right)'}}{{{e^{f\left( {2x - 1} \right)}} - 1}} = \frac{{{e^{f\left( {2x - 1} \right)}}.f'\left( {2x - 1} \right).\left( {2x - 1} \right)'}}{{{e^{f\left( {2x - 1} \right)}} - 1}} = \frac{{2{e^{f\left( {2x - 1} \right)}}.f'\left( {2x - 1} \right)}}}{{{e^{f\left( {2x - 1} \right)}} - 1}}.\]
Vậy đáp án đúng là C.