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