第2章 · 第2-1-9题

exercise

📝 题目

9.求下列函数的导数: (1)$y=x^{4}$ ; (2)$y=\sqrt[3]{x^{2}}$ ; (3)$y=x^{1.6}$ ; (4)$y=\frac{1}{\sqrt{x}}$ ; (5)$y=\frac{1}{x^{2}}$ ; (6)$y=x^{3} \sqrt[5]{x}$ ; (7)$y=\frac{x^{2} \sqrt[3]{x^{2}}}{\sqrt{x^{5}}}$ .

💡 答案与解析

[AI解答]

[AI解答] 我们使用幂函数求导公式:若 $y = x^n$,则 $y' = n x^{n-1}$。 以下逐题求解。

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(1)$y = x^{4}$ $$ y' = 4x^{4-1} = 4x^{3} $$

(2)$y = \sqrt[3]{x^{2}} = x^{\frac{2}{3}}$ $$ y' = \frac{2}{3} x^{\frac{2}{3} - 1} = \frac{2}{3} x^{-\frac{1}{3}} = \frac{2}{3\sqrt[3]{x}} $$

(3)$y = x^{1.6}$ $$ y' = 1.6 x^{1.6-1} = 1.6 x^{0.6} $$

(4)$y = \frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}$ $$ y' = -\frac{1}{2} x^{-\frac{1}{2} - 1} = -\frac{1}{2} x^{-\frac{3}{2}} = -\frac{1}{2x^{\frac{3}{2}}} $$

(5)$y = \frac{1}{x^{2}} = x^{-2}$ $$ y' = -2 x^{-3} = -\frac{2}{x^{3}} $$

(6)$y = x^{3} \sqrt[5]{x} = x^{3} \cdot x^{\frac{1}{5}} = x^{\frac{16}{5}}$ $$ y' = \frac{16}{5} x^{\frac{16}{5} - 1} = \frac{16}{5} x^{\frac{11}{5}} = \frac{16}{5} x^{2.2} $$

(7)$y = \frac{x^{2} \sqrt[3]{x^{2}}}{\sqrt{x^{5}}}$ 先化简: 分子:$x^{2} \cdot x^{\frac{2}{3}} = x^{\frac{8}{3}}$ 分母:$\sqrt{x^{5}} = x^{\frac{5}{2}}$ 所以 $$ y = x^{\frac{8}{3} - \frac{5}{2}} = x^{\frac{16}{6} - \frac{15}{6}} = x^{\frac{1}{6}} $$ 求导: $$ y' = \frac{1}{6} x^{\frac{1}{6} - 1} = \frac{1}{6} x^{-\frac{5}{6}} = \frac{1}{6 \sqrt[6]{x^{5}}} $$

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难度评级:★☆☆☆☆ (全部为幂函数直接求导,仅需掌握基本公式与指数化简)