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