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