以下の6つの関数を微分します。 (1) $y = 4x^3 - 2$ (2) $y = x^2(x^3 + 5)$ (3) $y = \frac{x^2}{x-1}$ (4) $y = \tan x$ (5) $y = (x^2 + 5)^3$ (6) $y = (x + \frac{1}{x})^3$

はい、承知いたしました。与えられた関数の微分を計算します。

1. 問題の内容

以下の6つの関数を微分します。

(1) $y = 4x^3 - 2$

(2) $y = x^2(x^3 + 5)$

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

(4) $y = \tan x$

(5) $y = (x^2 + 5)^3$

(6) $y = (x + \frac{1}{x})^3$

2. 解き方の手順

(1) $y = 4x^3 - 2$ の微分

$y' = \frac{d}{dx}(4x^3 - 2) = 4 \cdot 3x^2 - 0 = 12x^2$

(2) $y = x^2(x^3 + 5) = x^5 + 5x^2$ の微分

$y' = \frac{d}{dx}(x^5 + 5x^2) = 5x^4 + 5 \cdot 2x = 5x^4 + 10x$

(3) $y = \frac{x^2}{x-1}$ の微分（商の微分法）

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

(4) $y = \tan x$ の微分

$y' = \frac{d}{dx}(\tan x) = \frac{1}{\cos^2 x} = \sec^2 x$

(5) $y = (x^2 + 5)^3$ の微分（合成関数の微分法）

$y' = 3(x^2 + 5)^2 \cdot (x^2 + 5)' = 3(x^2 + 5)^2 \cdot 2x = 6x(x^2 + 5)^2$

(6) $y = (x + \frac{1}{x})^3$ の微分（合成関数の微分法）

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

3. 最終的な答え

(1) $y' = 12x^2$

(2) $y' = 5x^4 + 10x$

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

(4) $y' = \sec^2 x$

(5) $y' = 6x(x^2 + 5)^2$

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