$$$\operatorname{asec}{\left(x \right)}$$$ の二階導関数

一階導関数 $$$\frac{d}{dx} \left(\operatorname{asec}{\left(x \right)}\right)$$$ を求めよ

逆正割関数の導関数は $$$\frac{d}{dx} \left(\operatorname{asec}{\left(x \right)}\right) = \frac{1}{x^{2} \sqrt{1 - \frac{1}{x^{2}}}}$$$:

$${\color{red}\left(\frac{d}{dx} \left(\operatorname{asec}{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{1}{x^{2} \sqrt{1 - \frac{1}{x^{2}}}}\right)}$$

簡単化せよ:

$$\frac{1}{x^{2} \sqrt{1 - \frac{1}{x^{2}}}} = \frac{\left|{x}\right|}{x^{2} \sqrt{x^{2} - 1}}$$

したがって、$$$\frac{d}{dx} \left(\operatorname{asec}{\left(x \right)}\right) = \frac{\left|{x}\right|}{x^{2} \sqrt{x^{2} - 1}}$$$。

次に、$$$\frac{d^{2}}{dx^{2}} \left(\operatorname{asec}{\left(x \right)}\right) = \frac{d}{dx} \left(\frac{\left|{x}\right|}{x^{2} \sqrt{x^{2} - 1}}\right)$$$

$$$f{\left(x \right)} = \left|{x}\right|$$$ と $$$g{\left(x \right)} = x^{2} \sqrt{x^{2} - 1}$$$ に対して商の微分法則 $$$\frac{d}{dx} \left(\frac{f{\left(x \right)}}{g{\left(x \right)}}\right) = \frac{\frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} - f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)}{g^{2}{\left(x \right)}}$$$ を適用する：

$${\color{red}\left(\frac{d}{dx} \left(\frac{\left|{x}\right|}{x^{2} \sqrt{x^{2} - 1}}\right)\right)} = {\color{red}\left(\frac{\frac{d}{dx} \left(\left|{x}\right|\right) x^{2} \sqrt{x^{2} - 1} - \left|{x}\right| \frac{d}{dx} \left(x^{2} \sqrt{x^{2} - 1}\right)}{\left(x^{2} \sqrt{x^{2} - 1}\right)^{2}}\right)}$$

積の微分法 $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ を $$$f{\left(x \right)} = x^{2}$$$ と $$$g{\left(x \right)} = \sqrt{x^{2} - 1}$$$ に適用する:

$$\frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left|{x}\right|\right) - \left|{x}\right| {\color{red}\left(\frac{d}{dx} \left(x^{2} \sqrt{x^{2} - 1}\right)\right)}}{x^{4} \left(x^{2} - 1\right)} = \frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left|{x}\right|\right) - \left|{x}\right| {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) \sqrt{x^{2} - 1} + x^{2} \frac{d}{dx} \left(\sqrt{x^{2} - 1}\right)\right)}}{x^{4} \left(x^{2} - 1\right)}$$

冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を $$$n = 2$$$ に対して適用する:

$$\frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left|{x}\right|\right) - \left(x^{2} \frac{d}{dx} \left(\sqrt{x^{2} - 1}\right) + \sqrt{x^{2} - 1} {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)}\right) \left|{x}\right|}{x^{4} \left(x^{2} - 1\right)} = \frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left|{x}\right|\right) - \left(x^{2} \frac{d}{dx} \left(\sqrt{x^{2} - 1}\right) + \sqrt{x^{2} - 1} {\color{red}\left(2 x\right)}\right) \left|{x}\right|}{x^{4} \left(x^{2} - 1\right)}$$

関数$$$\sqrt{x^{2} - 1}$$$は、2つの関数$$$f{\left(u \right)} = \sqrt{u}$$$と$$$g{\left(x \right)} = x^{2} - 1$$$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:

$$\frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left|{x}\right|\right) - \left(x^{2} {\color{red}\left(\frac{d}{dx} \left(\sqrt{x^{2} - 1}\right)\right)} + 2 x \sqrt{x^{2} - 1}\right) \left|{x}\right|}{x^{4} \left(x^{2} - 1\right)} = \frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left|{x}\right|\right) - \left(x^{2} {\color{red}\left(\frac{d}{du} \left(\sqrt{u}\right) \frac{d}{dx} \left(x^{2} - 1\right)\right)} + 2 x \sqrt{x^{2} - 1}\right) \left|{x}\right|}{x^{4} \left(x^{2} - 1\right)}$$

冪法則 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ を $$$n = \frac{1}{2}$$$ に対して適用する:

$$\frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left|{x}\right|\right) - \left(x^{2} {\color{red}\left(\frac{d}{du} \left(\sqrt{u}\right)\right)} \frac{d}{dx} \left(x^{2} - 1\right) + 2 x \sqrt{x^{2} - 1}\right) \left|{x}\right|}{x^{4} \left(x^{2} - 1\right)} = \frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left|{x}\right|\right) - \left(x^{2} {\color{red}\left(\frac{1}{2 \sqrt{u}}\right)} \frac{d}{dx} \left(x^{2} - 1\right) + 2 x \sqrt{x^{2} - 1}\right) \left|{x}\right|}{x^{4} \left(x^{2} - 1\right)}$$

元の変数に戻す:

$$\frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left|{x}\right|\right) - \left(\frac{x^{2} \frac{d}{dx} \left(x^{2} - 1\right)}{2 \sqrt{{\color{red}\left(u\right)}}} + 2 x \sqrt{x^{2} - 1}\right) \left|{x}\right|}{x^{4} \left(x^{2} - 1\right)} = \frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left|{x}\right|\right) - \left(\frac{x^{2} \frac{d}{dx} \left(x^{2} - 1\right)}{2 \sqrt{{\color{red}\left(x^{2} - 1\right)}}} + 2 x \sqrt{x^{2} - 1}\right) \left|{x}\right|}{x^{4} \left(x^{2} - 1\right)}$$

和/差の導関数は、導関数の和/差である:

$$\frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left|{x}\right|\right) - \left(\frac{x^{2} {\color{red}\left(\frac{d}{dx} \left(x^{2} - 1\right)\right)}}{2 \sqrt{x^{2} - 1}} + 2 x \sqrt{x^{2} - 1}\right) \left|{x}\right|}{x^{4} \left(x^{2} - 1\right)} = \frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left|{x}\right|\right) - \left(\frac{x^{2} {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(1\right)\right)}}{2 \sqrt{x^{2} - 1}} + 2 x \sqrt{x^{2} - 1}\right) \left|{x}\right|}{x^{4} \left(x^{2} - 1\right)}$$

冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を $$$n = 2$$$ に対して適用する:

$$\frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left|{x}\right|\right) - \left(\frac{x^{2} \left({\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} - \frac{d}{dx} \left(1\right)\right)}{2 \sqrt{x^{2} - 1}} + 2 x \sqrt{x^{2} - 1}\right) \left|{x}\right|}{x^{4} \left(x^{2} - 1\right)} = \frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left|{x}\right|\right) - \left(\frac{x^{2} \left({\color{red}\left(2 x\right)} - \frac{d}{dx} \left(1\right)\right)}{2 \sqrt{x^{2} - 1}} + 2 x \sqrt{x^{2} - 1}\right) \left|{x}\right|}{x^{4} \left(x^{2} - 1\right)}$$

定数の導数は$$$0$$$です:

$$\frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left|{x}\right|\right) - \left(\frac{x^{2} \left(2 x - {\color{red}\left(\frac{d}{dx} \left(1\right)\right)}\right)}{2 \sqrt{x^{2} - 1}} + 2 x \sqrt{x^{2} - 1}\right) \left|{x}\right|}{x^{4} \left(x^{2} - 1\right)} = \frac{x^{2} \sqrt{x^{2} - 1} \frac{d}{dx} \left(\left|{x}\right|\right) - \left(\frac{x^{2} \left(2 x - {\color{red}\left(0\right)}\right)}{2 \sqrt{x^{2} - 1}} + 2 x \sqrt{x^{2} - 1}\right) \left|{x}\right|}{x^{4} \left(x^{2} - 1\right)}$$

絶対値の導関数は$$$\frac{d}{dx} \left(\left|{x}\right|\right) = \frac{x}{\left|{x}\right|}$$$:

$$\frac{x^{2} \sqrt{x^{2} - 1} {\color{red}\left(\frac{d}{dx} \left(\left|{x}\right|\right)\right)} - \left(\frac{x^{3}}{\sqrt{x^{2} - 1}} + 2 x \sqrt{x^{2} - 1}\right) \left|{x}\right|}{x^{4} \left(x^{2} - 1\right)} = \frac{x^{2} \sqrt{x^{2} - 1} {\color{red}\left(\frac{x}{\left|{x}\right|}\right)} - \left(\frac{x^{3}}{\sqrt{x^{2} - 1}} + 2 x \sqrt{x^{2} - 1}\right) \left|{x}\right|}{x^{4} \left(x^{2} - 1\right)}$$

簡単化せよ:

$$\frac{\frac{x^{3} \sqrt{x^{2} - 1}}{\left|{x}\right|} - \left(\frac{x^{3}}{\sqrt{x^{2} - 1}} + 2 x \sqrt{x^{2} - 1}\right) \left|{x}\right|}{x^{4} \left(x^{2} - 1\right)} = \frac{1 - 2 x^{2}}{x \left(x^{2} - 1\right)^{\frac{3}{2}} \left|{x}\right|}$$

したがって、$$$\frac{d}{dx} \left(\frac{\left|{x}\right|}{x^{2} \sqrt{x^{2} - 1}}\right) = \frac{1 - 2 x^{2}}{x \left(x^{2} - 1\right)^{\frac{3}{2}} \left|{x}\right|}$$$。

したがって、$$$\frac{d^{2}}{dx^{2}} \left(\operatorname{asec}{\left(x \right)}\right) = \frac{1 - 2 x^{2}}{x \left(x^{2} - 1\right)^{\frac{3}{2}} \left|{x}\right|}$$$。