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