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Câu hỏi: Hàm số $f\left( x \right)={{\log }_{3}}\left( {{x}^{3}}-7{{x}^{2}}+1 \right)$ có đạo hàm
A. ${f}'\left( x \right)=\frac{3{{x}^{2}}-14x}{\left( {{x}^{3}}-7{{x}^{2}}+1 \right)\ln 3}$.
B. ${f}'\left( x \right)=\frac{\ln 3}{{{x}^{3}}-7{{x}^{2}}+1}$.
C. ${f}'\left( x \right)=\frac{\left( 3{{x}^{2}}-14x \right)\ln 3}{{{x}^{3}}-7{{x}^{2}}+1}$.
D. ${f}'\left( x \right)=\frac{1}{\left( {{x}^{3}}-7{{x}^{2}}+1 \right)\ln 3}$.
A. ${f}'\left( x \right)=\frac{3{{x}^{2}}-14x}{\left( {{x}^{3}}-7{{x}^{2}}+1 \right)\ln 3}$.
B. ${f}'\left( x \right)=\frac{\ln 3}{{{x}^{3}}-7{{x}^{2}}+1}$.
C. ${f}'\left( x \right)=\frac{\left( 3{{x}^{2}}-14x \right)\ln 3}{{{x}^{3}}-7{{x}^{2}}+1}$.
D. ${f}'\left( x \right)=\frac{1}{\left( {{x}^{3}}-7{{x}^{2}}+1 \right)\ln 3}$.
Ta có ${f}'\left( x \right)=\frac{{{\left( {{x}^{3}}-7{{x}^{2}}+1 \right)}^{\prime }}}{\left( {{x}^{3}}-7{{x}^{2}}+1 \right)\ln 3}$ $=$ $\frac{3{{x}^{2}}-14x}{\left( {{x}^{3}}-7{{x}^{2}}+1 \right)\ln 3}$.
Vậy ${f}'\left( x \right)=\frac{3{{x}^{2}}-14x}{\left( {{x}^{3}}-7{{x}^{2}}+1 \right)\ln 3}$.
Vậy ${f}'\left( x \right)=\frac{3{{x}^{2}}-14x}{\left( {{x}^{3}}-7{{x}^{2}}+1 \right)\ln 3}$.
Đáp án A.