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Câu hỏi: Đạo hàm của hàm số $y=\sqrt[3]{{{\left( 1-3x \right)}^{5}}}$ là
A. ${y}'=-5{{\left( 1-3x \right)}^{\dfrac{4}{3}}}$.
B. ${y}'=\dfrac{5}{3}{{\left( 1-3x \right)}^{\dfrac{2}{3}}}$.
C. ${y}'=\dfrac{5}{3}{{\left( 1-3x \right)}^{\dfrac{4}{3}}}$.
D. ${y}'=-5{{\left( 1-3x \right)}^{\dfrac{2}{3}}}$.
A. ${y}'=-5{{\left( 1-3x \right)}^{\dfrac{4}{3}}}$.
B. ${y}'=\dfrac{5}{3}{{\left( 1-3x \right)}^{\dfrac{2}{3}}}$.
C. ${y}'=\dfrac{5}{3}{{\left( 1-3x \right)}^{\dfrac{4}{3}}}$.
D. ${y}'=-5{{\left( 1-3x \right)}^{\dfrac{2}{3}}}$.
Ta có $y=\sqrt[3]{{{\left( 1-3x \right)}^{5}}}={{\left( 1-3x \right)}^{\dfrac{5}{3}}}$. Ta suy ra ${y}'=\dfrac{5}{3}{{\left( 1-3x \right)}^{\dfrac{2}{3}}}.{{\left( 1-3x \right)}^{\prime }}=-5{{\left( 1-3x \right)}^{\dfrac{2}{3}}}$.
Đáp án D.