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