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$$$\tanh{\left(x \right)}$$$ の二階導関数
入力内容
$$$\frac{d^{2}}{dx^{2}} \left(\tanh{\left(x \right)}\right)$$$ を求めよ。
解答
一階導関数 $$$\frac{d}{dx} \left(\tanh{\left(x \right)}\right)$$$ を求めよ
双曲線正接関数の導関数は$$$\frac{d}{dx} \left(\tanh{\left(x \right)}\right) = \operatorname{sech}^{2}{\left(x \right)}$$$です:
$${\color{red}\left(\frac{d}{dx} \left(\tanh{\left(x \right)}\right)\right)} = {\color{red}\left(\operatorname{sech}^{2}{\left(x \right)}\right)}$$したがって、$$$\frac{d}{dx} \left(\tanh{\left(x \right)}\right) = \operatorname{sech}^{2}{\left(x \right)}$$$。
次に、$$$\frac{d^{2}}{dx^{2}} \left(\tanh{\left(x \right)}\right) = \frac{d}{dx} \left(\operatorname{sech}^{2}{\left(x \right)}\right)$$$
関数$$$\operatorname{sech}^{2}{\left(x \right)}$$$は、2つの関数$$$f{\left(u \right)} = u^{2}$$$と$$$g{\left(x \right)} = \operatorname{sech}{\left(x \right)}$$$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:
$${\color{red}\left(\frac{d}{dx} \left(\operatorname{sech}^{2}{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(\operatorname{sech}{\left(x \right)}\right)\right)}$$冪法則 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ を $$$n = 2$$$ に対して適用する:
$${\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(\operatorname{sech}{\left(x \right)}\right) = {\color{red}\left(2 u\right)} \frac{d}{dx} \left(\operatorname{sech}{\left(x \right)}\right)$$元の変数に戻す:
$$2 {\color{red}\left(u\right)} \frac{d}{dx} \left(\operatorname{sech}{\left(x \right)}\right) = 2 {\color{red}\left(\operatorname{sech}{\left(x \right)}\right)} \frac{d}{dx} \left(\operatorname{sech}{\left(x \right)}\right)$$双曲線正割関数の導関数は$$$\frac{d}{dx} \left(\operatorname{sech}{\left(x \right)}\right) = - \tanh{\left(x \right)} \operatorname{sech}{\left(x \right)}$$$です:
$$2 \operatorname{sech}{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\operatorname{sech}{\left(x \right)}\right)\right)} = 2 \operatorname{sech}{\left(x \right)} {\color{red}\left(- \tanh{\left(x \right)} \operatorname{sech}{\left(x \right)}\right)}$$したがって、$$$\frac{d}{dx} \left(\operatorname{sech}^{2}{\left(x \right)}\right) = - 2 \tanh{\left(x \right)} \operatorname{sech}^{2}{\left(x \right)}$$$。
したがって、$$$\frac{d^{2}}{dx^{2}} \left(\tanh{\left(x \right)}\right) = - 2 \tanh{\left(x \right)} \operatorname{sech}^{2}{\left(x \right)}$$$。
解答
$$$\frac{d^{2}}{dx^{2}} \left(\tanh{\left(x \right)}\right) = - 2 \tanh{\left(x \right)} \operatorname{sech}^{2}{\left(x \right)}$$$A