与えられた関数を微分する問題です。具体的には、以下の6つの関数をそれぞれ微分します。 (1) $y = \sqrt[3]{x} + 1$ (2) $y = \sqrt[3]{\frac{x+1}{x+4}}$ (3) $y = \sqrt[4]{2x-3}$ (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

与えられた関数を微分する問題です。具体的には、以下の6つの関数をそれぞれ微分します。

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

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

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

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

微分すると、

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

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

$u = \frac{x+1}{x+4}$ とすると、$y = u^{\frac{1}{3}}$.

$y' = \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} = \frac{1}{3}u^{-\frac{2}{3}} \cdot \frac{(x+4) - (x+1)}{(x+4)^2} = \frac{1}{3}(\frac{x+1}{x+4})^{-\frac{2}{3}} \cdot \frac{3}{(x+4)^2} = (\frac{x+4}{x+1})^{\frac{2}{3}} \cdot \frac{1}{(x+4)^2} = \frac{1}{(x+4)^{\frac{4}{3}}(x+1)^{\frac{2}{3}}}$

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

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

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

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

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

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

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

$y' = 0$

3. 最終的な答え

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

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

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

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

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

(6) $y' = 0$