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