与えられた4つの関数を微分する問題です。 (1) $y = x^2(2x^3 - 1)$ (2) $y = (-x + 1)(x^2 - 3x + 5)$ (3) $y = (3x^4 + 2)(4x^2 - 1)$ (4) $y = \frac{(x^3 - 5x^2)(2x + 5)}{6}$

1. 問題の内容

与えられた4つの関数を微分する問題です。

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

(2) $y = (-x + 1)(x^2 - 3x + 5)$

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

(4) $y = \frac{(x^3 - 5x^2)(2x + 5)}{6}$

2. 解き方の手順

(1) $y = x^2(2x^3 - 1) = 2x^5 - x^2$

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

(2) $y = (-x + 1)(x^2 - 3x + 5) = -x^3 + 3x^2 - 5x + x^2 - 3x + 5 = -x^3 + 4x^2 - 8x + 5$

$y' = \frac{d}{dx}(-x^3 + 4x^2 - 8x + 5) = -3x^2 + 8x - 8$

(3) $y = (3x^4 + 2)(4x^2 - 1) = 12x^6 - 3x^4 + 8x^2 - 2$

$y' = \frac{d}{dx}(12x^6 - 3x^4 + 8x^2 - 2) = 72x^5 - 12x^3 + 16x$

(4) $y = \frac{(x^3 - 5x^2)(2x + 5)}{6} = \frac{2x^4 + 5x^3 - 10x^3 - 25x^2}{6} = \frac{2x^4 - 5x^3 - 25x^2}{6}$

$y' = \frac{d}{dx}(\frac{2x^4 - 5x^3 - 25x^2}{6}) = \frac{8x^3 - 15x^2 - 50x}{6}$

3. 最終的な答え

(1) $y' = 10x^4 - 2x$

(2) $y' = -3x^2 + 8x - 8$

(3) $y' = 72x^5 - 12x^3 + 16x$

(4) $y' = \frac{8x^3 - 15x^2 - 50x}{6}$