関数 $y = \frac{\log(x^2 + x)}{\log(x+1)}$ を微分せよ。

1. 問題の内容

関数 $y = \frac{\log(x^2 + x)}{\log(x+1)}$ を微分せよ。

2. 解き方の手順

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

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

$\frac{d}{dx} (\log x) = \frac{1}{x}$

$\frac{d}{dx} (\log(x+1)) = \frac{1}{x+1}$

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

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

3. 最終的な答え

$\frac{(x+1)\log(x+1) - x\log x}{x(x+1)\{\log(x+1)\}^2}$

選択肢の3が正しい。