与えられた関数 $y$ を、積の微分公式を用いて微分する問題です。 (1) $y = (4x+3)(5x-2)$ (2) $y = (x^3+1)(x^2+x+1)$ (3) $y = (\frac{1}{x}-2)(x+\frac{3}{x})$ (4) $y = (\sqrt{x}^3 - 1)(\frac{1}{\sqrt{x}}+2)$

1. 問題の内容

与えられた関数 $y$ を、積の微分公式を用いて微分する問題です。

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

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

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

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

2. 解き方の手順

積の微分公式: $(uv)' = u'v + uv'$ を用います。

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

$u = 4x+3$, $v = 5x-2$ とすると、$u' = 4$, $v' = 5$

$y' = u'v + uv' = 4(5x-2) + (4x+3)5 = 20x - 8 + 20x + 15 = 40x + 7$

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

$u = x^3+1$, $v = x^2+x+1$ とすると、$u' = 3x^2$, $v' = 2x+1$

$y' = u'v + uv' = 3x^2(x^2+x+1) + (x^3+1)(2x+1) = 3x^4 + 3x^3 + 3x^2 + 2x^4 + x^3 + 2x + 1 = 5x^4 + 4x^3 + 3x^2 + 2x + 1$

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

$u = x^{-1}-2$, $v = x+3x^{-1}$ とすると、$u' = -x^{-2} = -\frac{1}{x^2}$, $v' = 1-3x^{-2} = 1-\frac{3}{x^2}$

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

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

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

$u = x^{3/2}-1$, $v = x^{-1/2}+2$ とすると、$u' = \frac{3}{2}x^{1/2}$, $v' = -\frac{1}{2}x^{-3/2}$

$y' = u'v + uv' = \frac{3}{2}x^{1/2}(x^{-1/2}+2) + (x^{3/2}-1)(-\frac{1}{2}x^{-3/2}) = \frac{3}{2} + 3x^{1/2} - \frac{1}{2} + \frac{1}{2}x^{-3/2} = 1 + 3\sqrt{x} + \frac{1}{2x\sqrt{x}}$

3. 最終的な答え

(1) $y' = 40x + 7$

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

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

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