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Câu hỏi: Tìm đạo hàm của hàm số $y=\left(1+x+x^2\right)^{\dfrac{2}{5}}$.
A. $y^{\prime}=\dfrac{2}{5} \cdot(1+2 x) \cdot\left(1+x+x^2\right)^{\dfrac{3}{5}}$.
B. $y^{\prime}=\dfrac{5}{2} \cdot(1+2 x)\left(1+x+x^2\right)^{-\dfrac{3}{5}}$.
C. $y^{\prime}=\dfrac{2}{5} \cdot(1+2 x) \cdot\left(1+x+x^2\right)^{-\dfrac{3}{5}}$.
D. $y^{\prime}=\dfrac{2}{5} \cdot\left(1+x+x^2\right)^{-\dfrac{3}{5}}$.
A. $y^{\prime}=\dfrac{2}{5} \cdot(1+2 x) \cdot\left(1+x+x^2\right)^{\dfrac{3}{5}}$.
B. $y^{\prime}=\dfrac{5}{2} \cdot(1+2 x)\left(1+x+x^2\right)^{-\dfrac{3}{5}}$.
C. $y^{\prime}=\dfrac{2}{5} \cdot(1+2 x) \cdot\left(1+x+x^2\right)^{-\dfrac{3}{5}}$.
D. $y^{\prime}=\dfrac{2}{5} \cdot\left(1+x+x^2\right)^{-\dfrac{3}{5}}$.
$
y^{\prime}=\dfrac{2}{5} \cdot\left(1+x+x^2\right)^{\dfrac{2}{5}-1} \cdot\left(1+x+x^2\right)^{\prime}=\dfrac{2}{5} \cdot(1+2 x) \cdot\left(1+x+x^2\right)^{-\dfrac{3}{5}} .
$
y^{\prime}=\dfrac{2}{5} \cdot\left(1+x+x^2\right)^{\dfrac{2}{5}-1} \cdot\left(1+x+x^2\right)^{\prime}=\dfrac{2}{5} \cdot(1+2 x) \cdot\left(1+x+x^2\right)^{-\dfrac{3}{5}} .
$
Đáp án C.