点$$$x = c$$$における$$$e^{- x} \sin{\left(x \right)}$$$の微分係数

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 $$$f{\left(x \right)} = e^{- x}$$$ と $$$g{\left(x \right)} = \sin{\left(x \right)}$$$ に適用する:

$${\color{red}\left(\frac{d}{dx} \left(e^{- x} \sin{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(e^{- x}\right) \sin{\left(x \right)} + e^{- x} \frac{d}{dx} \left(\sin{\left(x \right)}\right)\right)}$$

関数$$$e^{- x}$$$は、2つの関数$$$f{\left(u \right)} = e^{u}$$$と$$$g{\left(x \right)} = - x$$$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:

$$\sin{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(e^{- x}\right)\right)} + e^{- x} \frac{d}{dx} \left(\sin{\left(x \right)}\right) = \sin{\left(x \right)} {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(- x\right)\right)} + e^{- x} \frac{d}{dx} \left(\sin{\left(x \right)}\right)$$

指数関数の微分は$$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$です:

$$\sin{\left(x \right)} {\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(- x\right) + e^{- x} \frac{d}{dx} \left(\sin{\left(x \right)}\right) = \sin{\left(x \right)} {\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(- x\right) + e^{- x} \frac{d}{dx} \left(\sin{\left(x \right)}\right)$$

元の変数に戻す:

$$e^{{\color{red}\left(u\right)}} \sin{\left(x \right)} \frac{d}{dx} \left(- x\right) + e^{- x} \frac{d}{dx} \left(\sin{\left(x \right)}\right) = e^{{\color{red}\left(- x\right)}} \sin{\left(x \right)} \frac{d}{dx} \left(- x\right) + e^{- x} \frac{d}{dx} \left(\sin{\left(x \right)}\right)$$

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 $$$c = -1$$$ と $$$f{\left(x \right)} = x$$$ に対して適用します:

$$e^{- x} \sin{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(- x\right)\right)} + e^{- x} \frac{d}{dx} \left(\sin{\left(x \right)}\right) = e^{- x} \sin{\left(x \right)} {\color{red}\left(- \frac{d}{dx} \left(x\right)\right)} + e^{- x} \frac{d}{dx} \left(\sin{\left(x \right)}\right)$$

正弦関数の導関数は$$$\frac{d}{dx} \left(\sin{\left(x \right)}\right) = \cos{\left(x \right)}$$$:

$$- e^{- x} \sin{\left(x \right)} \frac{d}{dx} \left(x\right) + e^{- x} {\color{red}\left(\frac{d}{dx} \left(\sin{\left(x \right)}\right)\right)} = - e^{- x} \sin{\left(x \right)} \frac{d}{dx} \left(x\right) + e^{- x} {\color{red}\left(\cos{\left(x \right)}\right)}$$

$$$n = 1$$$ を用いて冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を適用すると、すなわち $$$\frac{d}{dx} \left(x\right) = 1$$$:

$$- e^{- x} \sin{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + e^{- x} \cos{\left(x \right)} = - e^{- x} \sin{\left(x \right)} {\color{red}\left(1\right)} + e^{- x} \cos{\left(x \right)}$$

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、$$$\frac{d}{dx} \left(e^{- x} \sin{\left(x \right)}\right) = \sqrt{2} e^{- x} \cos{\left(x + \frac{\pi}{4} \right)}$$$。

最後に、$$$x = c$$$での導関数の値を求めます。

$$$\left(\frac{d}{dx} \left(e^{- x} \sin{\left(x \right)}\right)\right)|_{\left(x = c\right)} = \sqrt{2} e^{- c} \cos{\left(c + \frac{\pi}{4} \right)}$$$