$$$\tan{\left(x \right)} \sec{\left(x \right)}$$$の導関数
関連する計算機: 対数微分計算機, 陰関数微分計算機(手順付き)
入力内容
$$$\frac{d}{dx} \left(\tan{\left(x \right)} \sec{\left(x \right)}\right)$$$ を求めよ。
解答
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 $$$f{\left(x \right)} = \sec{\left(x \right)}$$$ と $$$g{\left(x \right)} = \tan{\left(x \right)}$$$ に適用する:
$${\color{red}\left(\frac{d}{dx} \left(\tan{\left(x \right)} \sec{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(\sec{\left(x \right)}\right) \tan{\left(x \right)} + \sec{\left(x \right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)}$$正割関数の導関数は$$$\frac{d}{dx} \left(\sec{\left(x \right)}\right) = \tan{\left(x \right)} \sec{\left(x \right)}$$$:
$$\tan{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\sec{\left(x \right)}\right)\right)} + \sec{\left(x \right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right) = \tan{\left(x \right)} {\color{red}\left(\tan{\left(x \right)} \sec{\left(x \right)}\right)} + \sec{\left(x \right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right)$$正接関数の導関数は$$$\frac{d}{dx} \left(\tan{\left(x \right)}\right) = \sec^{2}{\left(x \right)}$$$:
$$\tan^{2}{\left(x \right)} \sec{\left(x \right)} + \sec{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)} = \tan^{2}{\left(x \right)} \sec{\left(x \right)} + \sec{\left(x \right)} {\color{red}\left(\sec^{2}{\left(x \right)}\right)}$$簡単化せよ:
$$\tan^{2}{\left(x \right)} \sec{\left(x \right)} + \sec^{3}{\left(x \right)} = \left(-1 + \frac{2}{\cos^{2}{\left(x \right)}}\right) \sec{\left(x \right)}$$したがって、$$$\frac{d}{dx} \left(\tan{\left(x \right)} \sec{\left(x \right)}\right) = \left(-1 + \frac{2}{\cos^{2}{\left(x \right)}}\right) \sec{\left(x \right)}$$$。
解答
$$$\frac{d}{dx} \left(\tan{\left(x \right)} \sec{\left(x \right)}\right) = \left(-1 + \frac{2}{\cos^{2}{\left(x \right)}}\right) \sec{\left(x \right)}$$$A