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Câu hỏi: Hàm số $f(x)={{\log }_{2}}\left( {{x}^{2}}-2x \right)$ có đạo hàm bằng
A. $f'(x)=\dfrac{\ln 2}{{{x}^{2}}-2x}.$
B. $f'(x)=\dfrac{1}{\left( {{x}^{2}}-2x \right)\ln 2}.$
C. $f'(x)=\dfrac{\left( 2x-2 \right)\ln 2}{{{x}^{2}}-2x}.$
D. $f'(x)=\dfrac{\left( 2x-2 \right)}{\left( {{x}^{2}}-2x \right)\ln 2}.$
A. $f'(x)=\dfrac{\ln 2}{{{x}^{2}}-2x}.$
B. $f'(x)=\dfrac{1}{\left( {{x}^{2}}-2x \right)\ln 2}.$
C. $f'(x)=\dfrac{\left( 2x-2 \right)\ln 2}{{{x}^{2}}-2x}.$
D. $f'(x)=\dfrac{\left( 2x-2 \right)}{\left( {{x}^{2}}-2x \right)\ln 2}.$
Hướng Dẫn. Ta có $f'(x)=\left( {{\log }_{2}}\left( {{x}^{2}}-2x \right) \right)'=\dfrac{\left( {{x}^{2}}-2x \right)'}{\left( {{x}^{2}}-2x \right)\ln 2}=\dfrac{2x-2}{({{x}^{2}}-2x)\ln 2}.$
Đáp án D.