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