Google
Câu hỏi: Đạo hàm của hàm số $y={{\left( {{x}^{2}}+x+1 \right)}^{\dfrac{1}{3}}}$ là
A. ${y}'=\dfrac{1}{3}{{\left( {{x}^{2}}+x+1 \right)}^{\dfrac{8}{3}}}$.
B. ${y}'=\dfrac{2x+1}{2\sqrt[3]{{{x}^{2}}+x+1}}$.
C. ${y}'=\dfrac{2x+1}{3\sqrt[3]{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}}$.
D. ${y}'=\dfrac{1}{3}{{\left( {{x}^{2}}+x+1 \right)}^{\dfrac{2}{3}}}$.
A. ${y}'=\dfrac{1}{3}{{\left( {{x}^{2}}+x+1 \right)}^{\dfrac{8}{3}}}$.
B. ${y}'=\dfrac{2x+1}{2\sqrt[3]{{{x}^{2}}+x+1}}$.
C. ${y}'=\dfrac{2x+1}{3\sqrt[3]{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}}$.
D. ${y}'=\dfrac{1}{3}{{\left( {{x}^{2}}+x+1 \right)}^{\dfrac{2}{3}}}$.
Ta có ${y}'=\dfrac{1}{3}{{\left( {{x}^{2}}+x+1 \right)}^{\dfrac{1}{3}-1}}{{\left( {{x}^{2}}+x+1 \right)}^{\prime }}=\dfrac{2x+1}{3\sqrt[3]{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}}$.
Đáp án C.