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Câu hỏi: Tính đạo hàm ${f}'\left( x \right)$ của hàm số $f\left( x \right)={{\log }_{2}}\left( 3x-1 \right)$ với $x>\dfrac{1}{3}.$
A. ${f}'\left( x \right)=\dfrac{1}{\left( 3x-1 \right)\ln 2}$.
B. ${f}'\left( x \right)=\dfrac{3}{\left( 3x-1 \right)\ln 2}$.
C. ${f}'\left( x \right)=\dfrac{3}{\left( 3x-1 \right)}$.
D. ${f}'\left( x \right)=\dfrac{3\ln 2}{\left( 3x-1 \right)}$.
A. ${f}'\left( x \right)=\dfrac{1}{\left( 3x-1 \right)\ln 2}$.
B. ${f}'\left( x \right)=\dfrac{3}{\left( 3x-1 \right)\ln 2}$.
C. ${f}'\left( x \right)=\dfrac{3}{\left( 3x-1 \right)}$.
D. ${f}'\left( x \right)=\dfrac{3\ln 2}{\left( 3x-1 \right)}$.
Ta có: ${f}'\left( x \right)={{\left( {{\log }_{2}}\left( 3x-1 \right) \right)}^{\prime }}=\dfrac{{{\left( 3x-1 \right)}^{\prime }}}{\left( 3x-1 \right)\ln 2}=\dfrac{3}{\left( 3x-1 \right)\ln 2}$.
Đáp án B.