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Câu hỏi: Hàm số $f\left( x \right)={{2}^{3x-1}}$ có đạo hàm
A. ${f}'\left( x \right)={{3.2}^{3x-1}}$.
B. ${f}'\left( x \right)={{3.2}^{3x-1}}.\ln 2$.
C. ${f}'\left( x \right)=\left( 3x-1 \right){{.2}^{3x-1}}$.
D. ${f}'\left( x \right)=\left( 3x-1 \right){{.2}^{3x-1}}.\ln 2$.
A. ${f}'\left( x \right)={{3.2}^{3x-1}}$.
B. ${f}'\left( x \right)={{3.2}^{3x-1}}.\ln 2$.
C. ${f}'\left( x \right)=\left( 3x-1 \right){{.2}^{3x-1}}$.
D. ${f}'\left( x \right)=\left( 3x-1 \right){{.2}^{3x-1}}.\ln 2$.
${f}'\left( x \right)={{\left( 3x-1 \right)}^{\prime }}{{2}^{3x-1}}.\ln 2={{3.2}^{3x-1}}.\ln 2$
Vậy ${f}'\left( x \right)={{3.2}^{3x-1}}.\ln 2$.
Vậy ${f}'\left( x \right)={{3.2}^{3x-1}}.\ln 2$.
Đáp án B.