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$$$\frac{1}{\sin{\left(x \right)}}$$$の導関数
入力内容
$$$\frac{d}{dx} \left(\frac{1}{\sin{\left(x \right)}}\right)$$$ を求めよ。
解答
関数$$$\frac{1}{\sin{\left(x \right)}}$$$は、2つの関数$$$f{\left(u \right)} = \frac{1}{u}$$$と$$$g{\left(x \right)} = \sin{\left(x \right)}$$$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:
$${\color{red}\left(\frac{d}{dx} \left(\frac{1}{\sin{\left(x \right)}}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\frac{1}{u}\right) \frac{d}{dx} \left(\sin{\left(x \right)}\right)\right)}$$冪法則 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ を $$$n = -1$$$ に対して適用する:
$${\color{red}\left(\frac{d}{du} \left(\frac{1}{u}\right)\right)} \frac{d}{dx} \left(\sin{\left(x \right)}\right) = {\color{red}\left(- \frac{1}{u^{2}}\right)} \frac{d}{dx} \left(\sin{\left(x \right)}\right)$$元の変数に戻す:
$$- \frac{\frac{d}{dx} \left(\sin{\left(x \right)}\right)}{{\color{red}\left(u\right)}^{2}} = - \frac{\frac{d}{dx} \left(\sin{\left(x \right)}\right)}{{\color{red}\left(\sin{\left(x \right)}\right)}^{2}}$$正弦関数の導関数は$$$\frac{d}{dx} \left(\sin{\left(x \right)}\right) = \cos{\left(x \right)}$$$:
$$- \frac{{\color{red}\left(\frac{d}{dx} \left(\sin{\left(x \right)}\right)\right)}}{\sin^{2}{\left(x \right)}} = - \frac{{\color{red}\left(\cos{\left(x \right)}\right)}}{\sin^{2}{\left(x \right)}}$$したがって、$$$\frac{d}{dx} \left(\frac{1}{\sin{\left(x \right)}}\right) = - \frac{\cos{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$$。
解答
$$$\frac{d}{dx} \left(\frac{1}{\sin{\left(x \right)}}\right) = - \frac{\cos{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$$A