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Câu hỏi: Đạo hàm của hàm số $y={{\log }_{3}}(2+{{e}^{2x}})$ là
A. ${y}'=\dfrac{2{{\text{e}}^{2\text{x}}}\ln 3}{2+{{e}^{2\text{x}}}}$
B. ${y}'=\dfrac{{{e}^{2\text{x}}}}{2+{{e}^{2\text{x}}}}$
C. ${y}'=\dfrac{2{{\text{e}}^{2\text{x}}}}{(2+{{e}^{2\text{x}}})\ln 3}$
D. ${y}'=\dfrac{{{e}^{2\text{x}}}}{(2+{{e}^{2\text{x}}})\ln 3}$
A. ${y}'=\dfrac{2{{\text{e}}^{2\text{x}}}\ln 3}{2+{{e}^{2\text{x}}}}$
B. ${y}'=\dfrac{{{e}^{2\text{x}}}}{2+{{e}^{2\text{x}}}}$
C. ${y}'=\dfrac{2{{\text{e}}^{2\text{x}}}}{(2+{{e}^{2\text{x}}})\ln 3}$
D. ${y}'=\dfrac{{{e}^{2\text{x}}}}{(2+{{e}^{2\text{x}}})\ln 3}$
Sử dụng công thức ${{\left( {{\log }_{a}}u \right)}^{\prime }}=\dfrac{{{u}'}}{u\ln a}$, suy ra ${y}'{{\left( {{\log }_{3}}(2+{{e}^{2\text{x}}}) \right)}^{\prime }}=\dfrac{2{{\text{e}}^{2x}}}{(2+{{e}^{2\text{x}}})\ln 3}$.
Đáp án C.