以下の関数を微分し、$y'$ を求める問題です。 (15) $y = -\frac{5}{2x^2}$ (16) $y = -\frac{1}{x^4}$ (17) $y = x^{\frac{3}{4}}$ (18) $y = \sqrt[5]{x}$ (19) $y = \sqrt[5]{x^2}$ (20) $y = x\sqrt[4]{x}$

はい、承知しました。次の関数の微分を求めます。

1. 問題の内容

以下の関数を微分し、$y'$ を求める問題です。

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

(16) $y = -\frac{1}{x^4}$

(17) $y = x^{\frac{3}{4}}$

(18) $y = \sqrt[5]{x}$

(19) $y = \sqrt[5]{x^2}$

(20) $y = x\sqrt[4]{x}$

2. 解き方の手順

微分公式 $y = x^n$ のとき $y' = nx^{n-1}$ を利用します。

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

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

(16) $y = -\frac{1}{x^4} = -x^{-4}$

$y' = -(-4)x^{-4-1} = 4x^{-5} = \frac{4}{x^5}$

(17) $y = x^{\frac{3}{4}}$

$y' = \frac{3}{4}x^{\frac{3}{4}-1} = \frac{3}{4}x^{-\frac{1}{4}} = \frac{3}{4x^{\frac{1}{4}}}$

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

$y' = \frac{1}{5}x^{\frac{1}{5}-1} = \frac{1}{5}x^{-\frac{4}{5}} = \frac{1}{5x^{\frac{4}{5}}}$

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

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

(20) $y = x\sqrt[4]{x} = x \cdot x^{\frac{1}{4}} = x^{1+\frac{1}{4}} = x^{\frac{5}{4}}$

$y' = \frac{5}{4}x^{\frac{5}{4}-1} = \frac{5}{4}x^{\frac{1}{4}}$

3. 最終的な答え

(15) $y' = \frac{5}{x^3}$

(16) $y' = \frac{4}{x^5}$

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

(18) $y' = \frac{1}{5}x^{-\frac{4}{5}}$

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

(20) $y' = \frac{5}{4}x^{\frac{1}{4}}$