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Câu hỏi: Đạo hàm của hàm số $y={{\left( 2x-1 \right)}^{\dfrac{1}{3}}}$ là:
A. ${y}'=\dfrac{1}{3}{{\left( 2x-1 \right)}^{-\dfrac{2}{3}}}$.
B. ${y}'={{\left( 2x-1 \right)}^{\dfrac{1}{3}}}\cdot \ln \left| 2x-1 \right|$.
C. ${y}'=\dfrac{2}{3}{{\left( 2x-1 \right)}^{\dfrac{4}{3}}}$.
D. ${y}'=\dfrac{2}{3}{{\left( 2x-1 \right)}^{-\dfrac{2}{3}}}$.
A. ${y}'=\dfrac{1}{3}{{\left( 2x-1 \right)}^{-\dfrac{2}{3}}}$.
B. ${y}'={{\left( 2x-1 \right)}^{\dfrac{1}{3}}}\cdot \ln \left| 2x-1 \right|$.
C. ${y}'=\dfrac{2}{3}{{\left( 2x-1 \right)}^{\dfrac{4}{3}}}$.
D. ${y}'=\dfrac{2}{3}{{\left( 2x-1 \right)}^{-\dfrac{2}{3}}}$.
Ta có: ${y}'=\dfrac{1}{3}{{\left( 2x-1 \right)}^{-\dfrac{2}{3}}}\cdot {{\left( 2x-1 \right)}^{\prime }}=\dfrac{2}{3}{{\left( 2x-1 \right)}^{-\dfrac{2}{3}}}$.Đáp án D.