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$$$\tanh{\left(x \right)}$$$ 的二階導數
您的輸入
求$$$\frac{d^{2}}{dx^{2}} \left(\tanh{\left(x \right)}\right)$$$。
解答
求第一階導數 $$$\frac{d}{dx} \left(\tanh{\left(x \right)}\right)$$$
雙曲正切的導數為 $$$\frac{d}{dx} \left(\tanh{\left(x \right)}\right) = \operatorname{sech}^{2}{\left(x \right)}$$$：
$${\color{red}\left(\frac{d}{dx} \left(\tanh{\left(x \right)}\right)\right)} = {\color{red}\left(\operatorname{sech}^{2}{\left(x \right)}\right)}$$因此，$$$\frac{d}{dx} \left(\tanh{\left(x \right)}\right) = \operatorname{sech}^{2}{\left(x \right)}$$$。
接下來，$$$\frac{d^{2}}{dx^{2}} \left(\tanh{\left(x \right)}\right) = \frac{d}{dx} \left(\operatorname{sech}^{2}{\left(x \right)}\right)$$$
函數 $$$\operatorname{sech}^{2}{\left(x \right)}$$$ 是兩個函數 $$$f{\left(u \right)} = u^{2}$$$ 與 $$$g{\left(x \right)} = \operatorname{sech}{\left(x \right)}$$$ 之複合 $$$f{\left(g{\left(x \right)} \right)}$$$。
應用鏈式法則 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dx} \left(\operatorname{sech}^{2}{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(\operatorname{sech}{\left(x \right)}\right)\right)}$$套用冪次法則 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$，取 $$$n = 2$$$：
$${\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(\operatorname{sech}{\left(x \right)}\right) = {\color{red}\left(2 u\right)} \frac{d}{dx} \left(\operatorname{sech}{\left(x \right)}\right)$$返回原變數：
$$2 {\color{red}\left(u\right)} \frac{d}{dx} \left(\operatorname{sech}{\left(x \right)}\right) = 2 {\color{red}\left(\operatorname{sech}{\left(x \right)}\right)} \frac{d}{dx} \left(\operatorname{sech}{\left(x \right)}\right)$$雙曲正割的導數為 $$$\frac{d}{dx} \left(\operatorname{sech}{\left(x \right)}\right) = - \tanh{\left(x \right)} \operatorname{sech}{\left(x \right)}$$$：
$$2 \operatorname{sech}{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\operatorname{sech}{\left(x \right)}\right)\right)} = 2 \operatorname{sech}{\left(x \right)} {\color{red}\left(- \tanh{\left(x \right)} \operatorname{sech}{\left(x \right)}\right)}$$因此，$$$\frac{d}{dx} \left(\operatorname{sech}^{2}{\left(x \right)}\right) = - 2 \tanh{\left(x \right)} \operatorname{sech}^{2}{\left(x \right)}$$$。
因此，$$$\frac{d^{2}}{dx^{2}} \left(\tanh{\left(x \right)}\right) = - 2 \tanh{\left(x \right)} \operatorname{sech}^{2}{\left(x \right)}$$$。
答案
$$$\frac{d^{2}}{dx^{2}} \left(\tanh{\left(x \right)}\right) = - 2 \tanh{\left(x \right)} \operatorname{sech}^{2}{\left(x \right)}$$$A