函數 $$$e^{- x} \sin{\left(x \right)}$$$ 在 $$$x = c$$$ 處的導數
您的輸入
求出 $$$\frac{d}{dx} \left(e^{- x} \sin{\left(x \right)}\right)$$$,並在 $$$x = c$$$ 處求值。
解答
將乘積法則 $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ 應用於 $$$f{\left(x \right)} = e^{- x}$$$ 和 $$$g{\left(x \right)} = \sin{\left(x \right)}$$$:
$${\color{red}\left(\frac{d}{dx} \left(e^{- x} \sin{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(e^{- x}\right) \sin{\left(x \right)} + e^{- x} \frac{d}{dx} \left(\sin{\left(x \right)}\right)\right)}$$函數 $$$e^{- x}$$$ 是兩個函數 $$$f{\left(u \right)} = e^{u}$$$ 與 $$$g{\left(x \right)} = - x$$$ 之複合 $$$f{\left(g{\left(x \right)} \right)}$$$。
應用鏈式法則 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$$\sin{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(e^{- x}\right)\right)} + e^{- x} \frac{d}{dx} \left(\sin{\left(x \right)}\right) = \sin{\left(x \right)} {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(- x\right)\right)} + e^{- x} \frac{d}{dx} \left(\sin{\left(x \right)}\right)$$指數函數的導數為 $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:
$$\sin{\left(x \right)} {\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(- x\right) + e^{- x} \frac{d}{dx} \left(\sin{\left(x \right)}\right) = \sin{\left(x \right)} {\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(- x\right) + e^{- x} \frac{d}{dx} \left(\sin{\left(x \right)}\right)$$返回原變數:
$$e^{{\color{red}\left(u\right)}} \sin{\left(x \right)} \frac{d}{dx} \left(- x\right) + e^{- x} \frac{d}{dx} \left(\sin{\left(x \right)}\right) = e^{{\color{red}\left(- x\right)}} \sin{\left(x \right)} \frac{d}{dx} \left(- x\right) + e^{- x} \frac{d}{dx} \left(\sin{\left(x \right)}\right)$$套用常數倍法則 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$,使用 $$$c = -1$$$ 與 $$$f{\left(x \right)} = x$$$:
$$e^{- x} \sin{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(- x\right)\right)} + e^{- x} \frac{d}{dx} \left(\sin{\left(x \right)}\right) = e^{- x} \sin{\left(x \right)} {\color{red}\left(- \frac{d}{dx} \left(x\right)\right)} + e^{- x} \frac{d}{dx} \left(\sin{\left(x \right)}\right)$$正弦函數的導數為$$$\frac{d}{dx} \left(\sin{\left(x \right)}\right) = \cos{\left(x \right)}$$$:
$$- e^{- x} \sin{\left(x \right)} \frac{d}{dx} \left(x\right) + e^{- x} {\color{red}\left(\frac{d}{dx} \left(\sin{\left(x \right)}\right)\right)} = - e^{- x} \sin{\left(x \right)} \frac{d}{dx} \left(x\right) + e^{- x} {\color{red}\left(\cos{\left(x \right)}\right)}$$套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$,取 $$$n = 1$$$,也就是 $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$- e^{- x} \sin{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + e^{- x} \cos{\left(x \right)} = - e^{- x} \sin{\left(x \right)} {\color{red}\left(1\right)} + e^{- x} \cos{\left(x \right)}$$化簡:
$$- e^{- x} \sin{\left(x \right)} + e^{- x} \cos{\left(x \right)} = \sqrt{2} e^{- x} \cos{\left(x + \frac{\pi}{4} \right)}$$因此,$$$\frac{d}{dx} \left(e^{- x} \sin{\left(x \right)}\right) = \sqrt{2} e^{- x} \cos{\left(x + \frac{\pi}{4} \right)}$$$。
最後,在$$$x = c$$$處計算導數。
$$$\left(\frac{d}{dx} \left(e^{- x} \sin{\left(x \right)}\right)\right)|_{\left(x = c\right)} = \sqrt{2} e^{- c} \cos{\left(c + \frac{\pi}{4} \right)}$$$
答案
$$$\frac{d}{dx} \left(e^{- x} \sin{\left(x \right)}\right) = \sqrt{2} e^{- x} \cos{\left(x + \frac{\pi}{4} \right)}$$$A
$$$\left(\frac{d}{dx} \left(e^{- x} \sin{\left(x \right)}\right)\right)|_{\left(x = c\right)} = \sqrt{2} e^{- c} \cos{\left(c + \frac{\pi}{4} \right)}\approx 1.414213562373095 e^{- c} \cos{\left(c + \frac{\pi}{4} \right)}$$$A