与えられた6つの関数をそれぞれ微分せよ。 (1) $y = \sqrt[3]{x^2}$ (2) $y = 6x\sqrt[3]{x^2}$ (3) $y = (2x-3)\sqrt{x}$ (4) $y = \frac{3x+2}{\sqrt{x}}$ (5) $y = \frac{1}{2\sqrt[3]{x^2}+1}$ (6) $y = \frac{2x}{\sqrt{x}+1}$

1. 問題の内容

与えられた6つの関数をそれぞれ微分せよ。

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

(2) $y = 6x\sqrt[3]{x^2}$

(3) $y = (2x-3)\sqrt{x}$

(4) $y = \frac{3x+2}{\sqrt{x}}$

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

(6) $y = \frac{2x}{\sqrt{x}+1}$

2. 解き方の手順

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

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

(2) $y = 6x\sqrt[3]{x^2} = 6x \cdot x^{\frac{2}{3}} = 6x^{\frac{5}{3}}$

$y' = 6 \cdot \frac{5}{3}x^{\frac{5}{3}-1} = 10x^{\frac{2}{3}} = 10\sqrt[3]{x^2}$

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

$y' = 2 \cdot \frac{3}{2}x^{\frac{3}{2}-1} - 3 \cdot \frac{1}{2}x^{\frac{1}{2}-1} = 3x^{\frac{1}{2}} - \frac{3}{2}x^{-\frac{1}{2}} = 3\sqrt{x} - \frac{3}{2\sqrt{x}} = \frac{6x-3}{2\sqrt{x}} = \frac{3(2x-1)}{2\sqrt{x}}$

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

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

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

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

(6) $y = \frac{2x}{\sqrt{x}+1}$

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

3. 最終的な答え

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

(2) $y' = 10\sqrt[3]{x^2}$

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

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

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

(6) $y' = \frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}$