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$$$\ln\left(\frac{2}{x}\right)$$$の導関数
入力内容
$$$\frac{d}{dx} \left(\ln\left(\frac{2}{x}\right)\right)$$$ を求めよ。
解答
関数$$$\ln\left(\frac{2}{x}\right)$$$は、2つの関数$$$f{\left(u \right)} = \ln\left(u\right)$$$と$$$g{\left(x \right)} = \frac{2}{x}$$$の合成$$$f{\left(g{\left(x \right)} \right)}$$$である。
連鎖律 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ を適用する:
$${\color{red}\left(\frac{d}{dx} \left(\ln\left(\frac{2}{x}\right)\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(\frac{2}{x}\right)\right)}$$自然対数の導関数は $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:
$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(\frac{2}{x}\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(\frac{2}{x}\right)$$元の変数に戻す:
$$\frac{\frac{d}{dx} \left(\frac{2}{x}\right)}{{\color{red}\left(u\right)}} = \frac{\frac{d}{dx} \left(\frac{2}{x}\right)}{{\color{red}\left(\frac{2}{x}\right)}}$$定数倍の法則 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ を $$$c = 2$$$ と $$$f{\left(x \right)} = \frac{1}{x}$$$ に対して適用します:
$$\frac{x {\color{red}\left(\frac{d}{dx} \left(\frac{2}{x}\right)\right)}}{2} = \frac{x {\color{red}\left(2 \frac{d}{dx} \left(\frac{1}{x}\right)\right)}}{2}$$冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を $$$n = -1$$$ に対して適用する:
$$x {\color{red}\left(\frac{d}{dx} \left(\frac{1}{x}\right)\right)} = x {\color{red}\left(- \frac{1}{x^{2}}\right)}$$したがって、$$$\frac{d}{dx} \left(\ln\left(\frac{2}{x}\right)\right) = - \frac{1}{x}$$$。
解答
$$$\frac{d}{dx} \left(\ln\left(\frac{2}{x}\right)\right) = - \frac{1}{x}$$$A