与えられた9つの関数を $x$ で微分する問題です。

1. 問題の内容

与えられた9つの関数を $x$ で微分する問題です。

2. 解き方の手順

関数の微分は、次の公式を利用します。

$\frac{d}{dx}x^n = nx^{n-1}$

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

$\frac{dy}{dx} = \frac{2}{5} x^{\frac{2}{5}-1} = \frac{2}{5} x^{-\frac{3}{5}}$

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

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

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

$\frac{dy}{dx} = \frac{1}{4} x^{\frac{1}{4}-1} = \frac{1}{4} x^{-\frac{3}{4}} = \frac{1}{4\sqrt[4]{x^3}}$

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

$\frac{dy}{dx} = \frac{5}{3} x^{\frac{5}{3}-1} = \frac{5}{3} x^{\frac{2}{3}} = \frac{5}{3}\sqrt[3]{x^2}$

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

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

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

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

(7) $y = -\frac{3}{x^3} = -3x^{-3}$

$\frac{dy}{dx} = -3 \cdot (-3) x^{-3-1} = 9 x^{-4} = \frac{9}{x^4}$

(8) $y = x^2 \sqrt{x} = x^2 \cdot x^{\frac{1}{2}} = x^{\frac{5}{2}}$

$\frac{dy}{dx} = \frac{5}{2} x^{\frac{5}{2}-1} = \frac{5}{2} x^{\frac{3}{2}} = \frac{5}{2} x\sqrt{x}$

(9) $y = \frac{\sqrt{x}}{\sqrt[3]{x}} = \frac{x^{\frac{1}{2}}}{x^{\frac{1}{3}}} = x^{\frac{1}{2}-\frac{1}{3}} = x^{\frac{1}{6}}$

$\frac{dy}{dx} = \frac{1}{6} x^{\frac{1}{6}-1} = \frac{1}{6} x^{-\frac{5}{6}} = \frac{1}{6\sqrt[6]{x^5}}$

3. 最終的な答え

(1) $\frac{dy}{dx} = \frac{2}{5} x^{-\frac{3}{5}}$

(2) $\frac{dy}{dx} = -\frac{1}{x^2}$

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

(4) $\frac{dy}{dx} = \frac{5}{3} x^{\frac{2}{3}}$

(5) $\frac{dy}{dx} = \frac{1}{2} x^{-\frac{1}{2}}$

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

(7) $\frac{dy}{dx} = \frac{9}{x^4}$

(8) $\frac{dy}{dx} = \frac{5}{2} x^{\frac{3}{2}}$

(9) $\frac{dy}{dx} = \frac{1}{6} x^{-\frac{5}{6}}$