与えられた6つの関数を微分する問題です。関数はすべて対数関数で、以下の通りです。 (1) $y = \log(2x-5)$ (2) $y = \log(e^x + 1)$ (3) $y = \log(\sin x)$ (4) $y = \log(\frac{1}{x})$ (5) $y = \log((x+3)(x+1))$ (6) $y = \log(\frac{x+2}{x+1})$

1. 問題の内容

与えられた6つの関数を微分する問題です。関数はすべて対数関数で、以下の通りです。

(1) $y = \log(2x-5)$

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

(3) $y = \log(\sin x)$

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

(5) $y = \log((x+3)(x+1))$

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

2. 解き方の手順

対数関数の微分公式 $(\log x)' = \frac{1}{x}$ と合成関数の微分法を利用して解きます。

(1) $y = \log(2x-5)$

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

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

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

(3) $y = \log(\sin x)$

$y' = \frac{1}{\sin x} \cdot (\sin x)' = \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x} = \cot x$

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

$y' = -\frac{1}{x}$

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

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

または、$y = \log(x+3) + \log(x+1)$ より $y' = \frac{1}{x+3} + \frac{1}{x+1} = \frac{x+1 + x+3}{(x+3)(x+1)} = \frac{2x+4}{(x+3)(x+1)}$

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

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

3. 最終的な答え

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

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

(3) $y' = \cot x$

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

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

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