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Câu hỏi: Đạo hàm của hàm số $y=\log _{2}^{2}\left( 2x+1 \right)$ là
A. $\dfrac{2{{\log }_{2}}\left( 2x+1 \right)}{\left( 2x+1 \right)\ln 2}$.
B. $\dfrac{4{{\log }_{2}}\left( 2x+1 \right)}{\left( 2x+1 \right)\ln 2}$.
C. $\dfrac{{{\log }_{2}}\left( 2x+1 \right)}{\left( 2x+1 \right)\ln 2}$.
D. $\dfrac{2}{\left( 2x+1 \right)\ln 2}$.
A. $\dfrac{2{{\log }_{2}}\left( 2x+1 \right)}{\left( 2x+1 \right)\ln 2}$.
B. $\dfrac{4{{\log }_{2}}\left( 2x+1 \right)}{\left( 2x+1 \right)\ln 2}$.
C. $\dfrac{{{\log }_{2}}\left( 2x+1 \right)}{\left( 2x+1 \right)\ln 2}$.
D. $\dfrac{2}{\left( 2x+1 \right)\ln 2}$.
Ta có: ${y}'={{\left[ \log _{2}^{2}\left( 2x+1 \right) \right]}^{\prime }}=2{{\log }_{2}}\left( 2x+1 \right).{{\left[ {{\log }_{2}}\left( 2x+1 \right) \right]}^{\prime }}=2{{\log }_{2}}\left( 2x+1 \right).\dfrac{2}{\left( 2x+1 \right).\ln 2}=\dfrac{4{{\log }_{2}}\left( 2x+1 \right)}{\left( 2x+1 \right).\ln 2}$.
Đáp án B.