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Câu hỏi: Tính đạo hàm của hàm số $y={{\left( 2{{\text{x}}^{2}}+x-1 \right)}^{\dfrac{2}{3}}}$
A. ${y}'=\dfrac{2\left( 4\text{x}+1 \right)}{3\sqrt[3]{2{{\text{x}}^{2}}+x-1}}$
B. ${y}'=\dfrac{2\left( 4\text{x}+1 \right)}{3\sqrt[3]{{{\left( 2{{\text{x}}^{2}}+x-1 \right)}^{2}}}}$
C. ${y}'=\dfrac{3\left( 4\text{x}+1 \right)}{2\sqrt[3]{2{{\text{x}}^{2}}+x-1}}$
D. ${y}'=\dfrac{3\left( 4\text{x}+1 \right)}{2\sqrt[3]{{{\left( 2{{\text{x}}^{2}}+x-1 \right)}^{2}}}}$
A. ${y}'=\dfrac{2\left( 4\text{x}+1 \right)}{3\sqrt[3]{2{{\text{x}}^{2}}+x-1}}$
B. ${y}'=\dfrac{2\left( 4\text{x}+1 \right)}{3\sqrt[3]{{{\left( 2{{\text{x}}^{2}}+x-1 \right)}^{2}}}}$
C. ${y}'=\dfrac{3\left( 4\text{x}+1 \right)}{2\sqrt[3]{2{{\text{x}}^{2}}+x-1}}$
D. ${y}'=\dfrac{3\left( 4\text{x}+1 \right)}{2\sqrt[3]{{{\left( 2{{\text{x}}^{2}}+x-1 \right)}^{2}}}}$
Ta có ${y}'=\dfrac{2}{3}{{\left( 2{{\text{x}}^{2}}+x-1 \right)}^{-\dfrac{1}{3}}}{{\left( 2{{\text{x}}^{2}}+x-1 \right)}^{\prime }}=\dfrac{2}{3}.\dfrac{1}{\sqrt[3]{2{{\text{x}}^{2}}+x-1}}\left( 4\text{x}+1 \right)=\dfrac{2\left( 4\text{x}+1 \right)}{3\sqrt[3]{2{{\text{x}}^{2}}+x-1}}$.
| Đạo hàm ${{\left( {{u}^{\alpha }} \right)}^{\prime }}=\alpha .{{u}^{\alpha -1}}.{u}'$. |
Đáp án A.