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Câu hỏi: Đạo hàm của hàm số $y=\dfrac{{{2}^{x+1}}}{{{3}^{x}}}$ bằng
A. $\dfrac{{{2}^{x+1}}}{{{3}^{x}}}\left( \ln 2-\ln 3 \right).$
B. $\dfrac{\left( x+1 \right){{.2}^{x}}}{x{{.3}^{x-1}}}.$
C. $\dfrac{{{2}^{x+1}}\ln 2}{{{3}^{x}}\ln 3}.$
D. $\dfrac{{{2}^{x}}}{{{3}^{x}}}\left( \ln 2-\ln 3 \right).$
A. $\dfrac{{{2}^{x+1}}}{{{3}^{x}}}\left( \ln 2-\ln 3 \right).$
B. $\dfrac{\left( x+1 \right){{.2}^{x}}}{x{{.3}^{x-1}}}.$
C. $\dfrac{{{2}^{x+1}}\ln 2}{{{3}^{x}}\ln 3}.$
D. $\dfrac{{{2}^{x}}}{{{3}^{x}}}\left( \ln 2-\ln 3 \right).$
${y}'={{\left( \dfrac{{{2}^{x+1}}}{{{3}^{x}}} \right)}^{\prime }}=2{{\left( {{\left( \dfrac{2}{3} \right)}^{x}} \right)}^{\prime }}=2.{{\left( \dfrac{2}{3} \right)}^{x}}.\ln \dfrac{2}{3}=\dfrac{{{2}^{x+1}}}{{{3}^{x}}}\left( \ln 2-\ln 3 \right).$
Đáp án A.