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