次の関数を微分する問題です。 (1) $y = \log(2x + 1)$ (2) $y = \log|3x^2 - 1|$ (3) $y = x\log(5x + 3)$ (4) $y = \sqrt{x}\log|2x - 1|$ (5) $y = \frac{\log x}{x^2}$ (6) $y = \frac{1}{\log x + 1}$ (7) $y = (1 + \log x)^2$ (8) $y = \log \sqrt{x^2 + 1}$

1. 問題の内容

次の関数を微分する問題です。

(1) $y = \log(2x + 1)$

(2) $y = \log|3x^2 - 1|$

(3) $y = x\log(5x + 3)$

(4) $y = \sqrt{x}\log|2x - 1|$

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

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

(7) $y = (1 + \log x)^2$

(8) $y = \log \sqrt{x^2 + 1}$

2. 解き方の手順

(1) $y = \log(2x + 1)$

合成関数の微分法を使う。

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

(2) $y = \log|3x^2 - 1|$

合成関数の微分法を使う。

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

(3) $y = x\log(5x + 3)$

積の微分法と合成関数の微分法を使う。

$y' = (x)'\log(5x+3) + x(\log(5x+3))' = \log(5x+3) + x \cdot \frac{1}{5x+3} \cdot (5x+3)'$

$= \log(5x+3) + x \cdot \frac{1}{5x+3} \cdot 5 = \log(5x+3) + \frac{5x}{5x+3}$

(4) $y = \sqrt{x}\log|2x - 1|$

積の微分法と合成関数の微分法を使う。

$y' = (\sqrt{x})'\log|2x-1| + \sqrt{x}(\log|2x-1|)' = \frac{1}{2\sqrt{x}} \log|2x-1| + \sqrt{x} \cdot \frac{1}{2x-1} \cdot (2x-1)'$

$= \frac{\log|2x-1|}{2\sqrt{x}} + \sqrt{x} \cdot \frac{1}{2x-1} \cdot 2 = \frac{\log|2x-1|}{2\sqrt{x}} + \frac{2\sqrt{x}}{2x-1}$

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

商の微分法を使う。

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

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

合成関数の微分法または商の微分法を使う。

$y = (\log x + 1)^{-1}$ と考えると

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

または、商の微分法を使う。

$y' = \frac{(1)'(\log x + 1) - 1(\log x + 1)'}{(\log x + 1)^2} = \frac{0 - 1 \cdot \frac{1}{x}}{(\log x + 1)^2} = -\frac{1}{x(\log x + 1)^2}$

(7) $y = (1 + \log x)^2$

合成関数の微分法を使う。

$y' = 2(1 + \log x) \cdot (1 + \log x)' = 2(1 + \log x) \cdot \frac{1}{x} = \frac{2(1 + \log x)}{x}$

(8) $y = \log \sqrt{x^2 + 1}$

合成関数の微分法を使う。

$y = \log (x^2 + 1)^{\frac{1}{2}} = \frac{1}{2} \log (x^2 + 1)$ と変形してから微分すると、

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

3. 最終的な答え

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

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

(3) $y' = \log(5x+3) + \frac{5x}{5x+3}$

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

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

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

(7) $y' = \frac{2(1 + \log x)}{x}$

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