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Câu hỏi: Hàm số $f\left( x \right)={{2}^{{{x}^{2}}-2x}}$ có đạo hàm
A. $f'\left( x \right)=2\left( x-1 \right){{.2}^{{{x}^{2}}-2x+1}}$
B. $f'\left( x \right)=\left( x-1 \right){{.2}^{{{x}^{2}}-2x+1}}.\ln 2$
C. $f'\left( x \right)=\left( x-1 \right){{.2}^{{{x}^{2}}-2x}}.\ln 2$
D. $f'\left( x \right)={{2}^{{{x}^{2}}-2x}}.\ln 2$
A. $f'\left( x \right)=2\left( x-1 \right){{.2}^{{{x}^{2}}-2x+1}}$
B. $f'\left( x \right)=\left( x-1 \right){{.2}^{{{x}^{2}}-2x+1}}.\ln 2$
C. $f'\left( x \right)=\left( x-1 \right){{.2}^{{{x}^{2}}-2x}}.\ln 2$
D. $f'\left( x \right)={{2}^{{{x}^{2}}-2x}}.\ln 2$
Áp dụng công thức: ${{\left( {{a}^{u}} \right)}^{ }}^{'}=u'{{a}^{u}}\ln a$, ta có:
$f'\left( x \right)={{\left( {{2}^{{{x}^{2}}-2x}} \right)}^{'}}=\left( 2x-2 \right){{.2}^{{{x}^{2}}-2x}}\ln 2=\left( x-1 \right){{.2}^{{{x}^{2}}-2x+1}}\ln 2$
$f'\left( x \right)={{\left( {{2}^{{{x}^{2}}-2x}} \right)}^{'}}=\left( 2x-2 \right){{.2}^{{{x}^{2}}-2x}}\ln 2=\left( x-1 \right){{.2}^{{{x}^{2}}-2x+1}}\ln 2$
Đáp án B.