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