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Câu hỏi: Tính đạo hàm của hàm số $y=\dfrac{{{4}^{x+1}}}{{{9}^{x}}}.$
A. ${y}'=\dfrac{{{2}^{2x+4}}}{{{3}^{2x}}}\ln \dfrac{2}{3}.$
B. ${y}'=\dfrac{{{2}^{2x+3}}}{{{3}^{2x}}}\ln \dfrac{2}{3}.$
C. ${y}'=\dfrac{{{2}^{{{x}^{2}}+3}}}{{{3}^{{{x}^{2}}}}}\ln \dfrac{2}{3}.$
D. ${y}'=\dfrac{{{2}^{{{x}^{2}}+4}}}{{{3}^{{{x}^{2}}}}}\ln \dfrac{2}{3}.$
A. ${y}'=\dfrac{{{2}^{2x+4}}}{{{3}^{2x}}}\ln \dfrac{2}{3}.$
B. ${y}'=\dfrac{{{2}^{2x+3}}}{{{3}^{2x}}}\ln \dfrac{2}{3}.$
C. ${y}'=\dfrac{{{2}^{{{x}^{2}}+3}}}{{{3}^{{{x}^{2}}}}}\ln \dfrac{2}{3}.$
D. ${y}'=\dfrac{{{2}^{{{x}^{2}}+4}}}{{{3}^{{{x}^{2}}}}}\ln \dfrac{2}{3}.$
Ta có $y=\dfrac{{{4}^{x+1}}}{{{9}^{x}}}=4.{{\left( \dfrac{4}{9} \right)}^{x}}\Rightarrow {y}'=4.{{\left( \dfrac{4}{9} \right)}^{x}}.\ln \dfrac{4}{9}=4.{{\left[ {{\left( \dfrac{2}{3} \right)}^{2}} \right]}^{x}}.$
$\Rightarrow {y}'=4.{{\left( \dfrac{2}{3} \right)}^{2x}}.2\ln \dfrac{2}{3}=8.\dfrac{{{2}^{2x}}}{{{3}^{2x}}}\ln \dfrac{2}{3}=\dfrac{{{2}^{3}}{{.2}^{2x}}}{{{3}^{2x}}}\ln \dfrac{2}{3}=\dfrac{{{2}^{2x+3}}}{{{3}^{2x}}}\ln \dfrac{2}{3}.$
$\Rightarrow {y}'=4.{{\left( \dfrac{2}{3} \right)}^{2x}}.2\ln \dfrac{2}{3}=8.\dfrac{{{2}^{2x}}}{{{3}^{2x}}}\ln \dfrac{2}{3}=\dfrac{{{2}^{3}}{{.2}^{2x}}}{{{3}^{2x}}}\ln \dfrac{2}{3}=\dfrac{{{2}^{2x+3}}}{{{3}^{2x}}}\ln \dfrac{2}{3}.$
Đáp án B.