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