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Câu hỏi: Hàm số $f\left( x \right)={{2}^{3x+4}}$ có đạo hàm là
A. ${f}'\left( x \right)=\dfrac{{{3.2}^{3x+4}}}{\ln 2}$.
B. ${f}'\left( x \right)={{3.2}^{3x+4}}\ln 2$.
C. ${f}'\left( x \right)={{2}^{3x+4}}\ln 2$.
D. ${f}'\left( x \right)=\dfrac{{{2}^{3x+4}}}{\ln 2}$.
Ta có: ${f}'\left( x \right)={{\left( {{2}^{3x+4}} \right)}^{\prime }}={{\left( 3x+4 \right)}^{\prime }}{{.2}^{3x+4}}\ln 2={{3.2}^{3x+4}}\ln 2$.
A. ${f}'\left( x \right)=\dfrac{{{3.2}^{3x+4}}}{\ln 2}$.
B. ${f}'\left( x \right)={{3.2}^{3x+4}}\ln 2$.
C. ${f}'\left( x \right)={{2}^{3x+4}}\ln 2$.
D. ${f}'\left( x \right)=\dfrac{{{2}^{3x+4}}}{\ln 2}$.
Ta có: ${f}'\left( x \right)={{\left( {{2}^{3x+4}} \right)}^{\prime }}={{\left( 3x+4 \right)}^{\prime }}{{.2}^{3x+4}}\ln 2={{3.2}^{3x+4}}\ln 2$.
Đáp án B.