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Câu hỏi: Tính đạo hạm của hàm số $y={{\left( 2{{x}^{2}}-x+1 \right)}^{\dfrac{3}{2}}}$
A. ${y}'=\dfrac{3}{2}.{{\left( 2{{x}^{2}}-x+1 \right)}^{\dfrac{5}{2}}}$.
B. ${y}'=\dfrac{3}{2}.\left( 4x-1 \right)\sqrt{2{{x}^{2}}-x+1}.$.
C. ${y}'=\dfrac{2}{5}.{{\left( 2{{x}^{2}}-x+1 \right)}^{\dfrac{5}{2}}}$.
D. ${y}'=\dfrac{2}{3}.\left( 4x-1 \right){{\left( 2{{x}^{2}}-x+1 \right)}^{\dfrac{1}{2}}}$.
Ta có: $y={{\left( 2{{x}^{2}}-x+1 \right)}^{\dfrac{3}{2}}}\Rightarrow {y}'=\dfrac{3}{2}.{{\left( 2{{x}^{2}}-x+1 \right)}^{\dfrac{1}{2}}}.{{\left( 2{{x}^{2}}-x+1 \right)}^{\prime }}=\dfrac{3}{2}.\left( 4x-1 \right){{\left( 2{{x}^{2}}-x+1 \right)}^{\dfrac{1}{2}}}.$
A. ${y}'=\dfrac{3}{2}.{{\left( 2{{x}^{2}}-x+1 \right)}^{\dfrac{5}{2}}}$.
B. ${y}'=\dfrac{3}{2}.\left( 4x-1 \right)\sqrt{2{{x}^{2}}-x+1}.$.
C. ${y}'=\dfrac{2}{5}.{{\left( 2{{x}^{2}}-x+1 \right)}^{\dfrac{5}{2}}}$.
D. ${y}'=\dfrac{2}{3}.\left( 4x-1 \right){{\left( 2{{x}^{2}}-x+1 \right)}^{\dfrac{1}{2}}}$.
Ta có: $y={{\left( 2{{x}^{2}}-x+1 \right)}^{\dfrac{3}{2}}}\Rightarrow {y}'=\dfrac{3}{2}.{{\left( 2{{x}^{2}}-x+1 \right)}^{\dfrac{1}{2}}}.{{\left( 2{{x}^{2}}-x+1 \right)}^{\prime }}=\dfrac{3}{2}.\left( 4x-1 \right){{\left( 2{{x}^{2}}-x+1 \right)}^{\dfrac{1}{2}}}.$
Đáp án B.