$$$e^{x} \cos{\left(x \right)}$$$の導関数
入力内容
$$$\frac{d}{dx} \left(e^{x} \cos{\left(x \right)}\right)$$$ 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 $$$f{\left(x \right)} = \cos{\left(x \right)}$$$ と $$$g{\left(x \right)} = e^{x}$$$ に適用する:
$${\color{red}\left(\frac{d}{dx} \left(e^{x} \cos{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(\cos{\left(x \right)}\right) e^{x} + \cos{\left(x \right)} \frac{d}{dx} \left(e^{x}\right)\right)}$$余弦関数の導関数は$$$\frac{d}{dx} \left(\cos{\left(x \right)}\right) = - \sin{\left(x \right)}$$$:
$$e^{x} {\color{red}\left(\frac{d}{dx} \left(\cos{\left(x \right)}\right)\right)} + \cos{\left(x \right)} \frac{d}{dx} \left(e^{x}\right) = e^{x} {\color{red}\left(- \sin{\left(x \right)}\right)} + \cos{\left(x \right)} \frac{d}{dx} \left(e^{x}\right)$$指数関数の微分は$$$\frac{d}{dx} \left(e^{x}\right) = e^{x}$$$です:
$$- e^{x} \sin{\left(x \right)} + \cos{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(e^{x}\right)\right)} = - e^{x} \sin{\left(x \right)} + \cos{\left(x \right)} {\color{red}\left(e^{x}\right)}$$簡単化せよ:
$$- e^{x} \sin{\left(x \right)} + e^{x} \cos{\left(x \right)} = \sqrt{2} e^{x} \cos{\left(x + \frac{\pi}{4} \right)}$$したがって、$$$\frac{d}{dx} \left(e^{x} \cos{\left(x \right)}\right) = \sqrt{2} e^{x} \cos{\left(x + \frac{\pi}{4} \right)}$$$。
解答
$$$\frac{d}{dx} \left(e^{x} \cos{\left(x \right)}\right) = \sqrt{2} e^{x} \cos{\left(x + \frac{\pi}{4} \right)}$$$A