$$$\tan^{2}{\left(x \right)}$$$ の二階導関数
入力内容
$$$\frac{d^{2}}{dx^{2}} \left(\tan^{2}{\left(x \right)}\right)$$$ を求めよ。
解答
一階導関数 $$$\frac{d}{dx} \left(\tan^{2}{\left(x \right)}\right)$$$ を求めよ
関数$$$\tan^{2}{\left(x \right)}$$$は、2つの関数$$$f{\left(u \right)} = u^{2}$$$と$$$g{\left(x \right)} = \tan{\left(x \right)}$$$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:
$${\color{red}\left(\frac{d}{dx} \left(\tan^{2}{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)}$$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:
$${\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right) = {\color{red}\left(2 u\right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right)$$元の変数に戻す:
$$2 {\color{red}\left(u\right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right) = 2 {\color{red}\left(\tan{\left(x \right)}\right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right)$$正接関数の導関数は$$$\frac{d}{dx} \left(\tan{\left(x \right)}\right) = \sec^{2}{\left(x \right)}$$$:
$$2 \tan{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)} = 2 \tan{\left(x \right)} {\color{red}\left(\sec^{2}{\left(x \right)}\right)}$$したがって、$$$\frac{d}{dx} \left(\tan^{2}{\left(x \right)}\right) = 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}$$$。
次に、$$$\frac{d^{2}}{dx^{2}} \left(\tan^{2}{\left(x \right)}\right) = \frac{d}{dx} \left(2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)$$$
定数倍の法則 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ を $$$c = 2$$$ と $$$f{\left(x \right)} = \tan{\left(x \right)} \sec^{2}{\left(x \right)}$$$ に対して適用します:
$${\color{red}\left(\frac{d}{dx} \left(2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)\right)} = {\color{red}\left(2 \frac{d}{dx} \left(\tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)\right)}$$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 $$$f{\left(x \right)} = \sec^{2}{\left(x \right)}$$$ と $$$g{\left(x \right)} = \tan{\left(x \right)}$$$ に適用する:
$$2 {\color{red}\left(\frac{d}{dx} \left(\tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)\right)} = 2 {\color{red}\left(\frac{d}{dx} \left(\sec^{2}{\left(x \right)}\right) \tan{\left(x \right)} + \sec^{2}{\left(x \right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)}$$正接関数の導関数は$$$\frac{d}{dx} \left(\tan{\left(x \right)}\right) = \sec^{2}{\left(x \right)}$$$:
$$2 \tan{\left(x \right)} \frac{d}{dx} \left(\sec^{2}{\left(x \right)}\right) + 2 \sec^{2}{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)} = 2 \tan{\left(x \right)} \frac{d}{dx} \left(\sec^{2}{\left(x \right)}\right) + 2 \sec^{2}{\left(x \right)} {\color{red}\left(\sec^{2}{\left(x \right)}\right)}$$関数$$$\sec^{2}{\left(x \right)}$$$は、2つの関数$$$f{\left(u \right)} = u^{2}$$$と$$$g{\left(x \right)} = \sec{\left(x \right)}$$$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:
$$2 \tan{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\sec^{2}{\left(x \right)}\right)\right)} + 2 \sec^{4}{\left(x \right)} = 2 \tan{\left(x \right)} {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(\sec{\left(x \right)}\right)\right)} + 2 \sec^{4}{\left(x \right)}$$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:
$$2 \tan{\left(x \right)} {\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(\sec{\left(x \right)}\right) + 2 \sec^{4}{\left(x \right)} = 2 \tan{\left(x \right)} {\color{red}\left(2 u\right)} \frac{d}{dx} \left(\sec{\left(x \right)}\right) + 2 \sec^{4}{\left(x \right)}$$元の変数に戻す:
$$4 \tan{\left(x \right)} {\color{red}\left(u\right)} \frac{d}{dx} \left(\sec{\left(x \right)}\right) + 2 \sec^{4}{\left(x \right)} = 4 \tan{\left(x \right)} {\color{red}\left(\sec{\left(x \right)}\right)} \frac{d}{dx} \left(\sec{\left(x \right)}\right) + 2 \sec^{4}{\left(x \right)}$$正割関数の導関数は$$$\frac{d}{dx} \left(\sec{\left(x \right)}\right) = \tan{\left(x \right)} \sec{\left(x \right)}$$$:
$$4 \tan{\left(x \right)} \sec{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\sec{\left(x \right)}\right)\right)} + 2 \sec^{4}{\left(x \right)} = 4 \tan{\left(x \right)} \sec{\left(x \right)} {\color{red}\left(\tan{\left(x \right)} \sec{\left(x \right)}\right)} + 2 \sec^{4}{\left(x \right)}$$簡単化せよ:
$$4 \tan^{2}{\left(x \right)} \sec^{2}{\left(x \right)} + 2 \sec^{4}{\left(x \right)} = \left(-4 + \frac{6}{\cos^{2}{\left(x \right)}}\right) \sec^{2}{\left(x \right)}$$したがって、$$$\frac{d}{dx} \left(2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}\right) = \left(-4 + \frac{6}{\cos^{2}{\left(x \right)}}\right) \sec^{2}{\left(x \right)}$$$。
したがって、$$$\frac{d^{2}}{dx^{2}} \left(\tan^{2}{\left(x \right)}\right) = \left(-4 + \frac{6}{\cos^{2}{\left(x \right)}}\right) \sec^{2}{\left(x \right)}$$$。
解答
$$$\frac{d^{2}}{dx^{2}} \left(\tan^{2}{\left(x \right)}\right) = \left(-4 + \frac{6}{\cos^{2}{\left(x \right)}}\right) \sec^{2}{\left(x \right)}$$$A