与えられた6つの関数について、それぞれ微分を求めます。 (1) $y = \sqrt[3]{x+1}$ (2) $y = \sqrt[3]{\frac{x+1}{x+4}}$ (3) $y = 4\sqrt{2x-3}$ (4) $y = \sqrt[3]{x^4+\sqrt{x}}$ (5) $y = \frac{x^2+x+\sqrt{x}}{x\sqrt{x}}$ (6) $y = \frac{\sqrt{x}+1}{\sqrt{x}+1}$

1. 問題の内容

与えられた6つの関数について、それぞれ微分を求めます。

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

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

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

(4) $y = \sqrt[3]{x^4+\sqrt{x}}$

(5) $y = \frac{x^2+x+\sqrt{x}}{x\sqrt{x}}$

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

2. 解き方の手順

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

$\frac{dy}{dx} = \frac{1}{3}(x+1)^{\frac{1}{3}-1} \cdot 1 = \frac{1}{3}(x+1)^{-\frac{2}{3}} = \frac{1}{3\sqrt[3]{(x+1)^2}}$

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

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

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

$\frac{dy}{dx} = 4 \cdot \frac{1}{2}(2x-3)^{-\frac{1}{2}} \cdot 2 = 4(2x-3)^{-\frac{1}{2}} = \frac{4}{\sqrt{2x-3}}$

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

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

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

$\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}x^{-\frac{3}{2}} - x^{-2} = \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}} - \frac{1}{x^2} = \frac{x^2 - x - 2}{2x^2\sqrt{x}}$

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

$\frac{dy}{dx} = 0$

3. 最終的な答え

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

(2) $\frac{dy}{dx} = \frac{1}{\sqrt[3]{(x+1)^2(x+4)^4}}$

(3) $\frac{dy}{dx} = \frac{4}{\sqrt{2x-3}}$

(4) $\frac{dy}{dx} = \frac{4x^3+\frac{1}{2\sqrt{x}}}{3\sqrt[3]{(x^4+\sqrt{x})^2}}$

(5) $\frac{dy}{dx} = \frac{x^2 - x - 2}{2x^2\sqrt{x}}$

(6) $\frac{dy}{dx} = 0$