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Câu hỏi: Trên khoảng $\left( \dfrac{1}{2};+\infty \right)$, đạo hàm của hàm số $y={{\left( 2x-1 \right)}^{\dfrac{3}{2}}}$ là
A. $\dfrac{5}{2}{{\left( 2x-1 \right)}^{\dfrac{2}{5}}}$.
B. $\dfrac{3}{2}{{\left( 2x-1 \right)}^{\dfrac{1}{2}}}$.
C. $3{{\left( 2x-1 \right)}^{\dfrac{1}{2}}}$.
D. $\dfrac{3}{2}{{\left( 2x-1 \right)}^{-\dfrac{1}{2}}}$.
A. $\dfrac{5}{2}{{\left( 2x-1 \right)}^{\dfrac{2}{5}}}$.
B. $\dfrac{3}{2}{{\left( 2x-1 \right)}^{\dfrac{1}{2}}}$.
C. $3{{\left( 2x-1 \right)}^{\dfrac{1}{2}}}$.
D. $\dfrac{3}{2}{{\left( 2x-1 \right)}^{-\dfrac{1}{2}}}$.
Ta có: ${{\left[ {{\left( 2x-1 \right)}^{\dfrac{3}{2}}} \right]}^{\prime }}=\dfrac{3}{2}{{\left( 2x-1 \right)}^{\dfrac{1}{2}}}{{\left( 2x-1 \right)}^{\prime }}=\dfrac{3}{2}.2.{{\left( 2x-1 \right)}^{\dfrac{1}{2}}}=3{{\left( 2x-1 \right)}^{\dfrac{1}{2}}}$.Đáp án C.