$$$3 e^{- 4 r} \sin{\left(3 \theta \right)}$$$ 對 $$$r$$$ 的導數
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求$$$\frac{d}{dr} \left(3 e^{- 4 r} \sin{\left(3 \theta \right)}\right)$$$。
解答
套用常數倍法則 $$$\frac{d}{dr} \left(c f{\left(r \right)}\right) = c \frac{d}{dr} \left(f{\left(r \right)}\right)$$$，使用 $$$c = 3 \sin{\left(3 \theta \right)}$$$ 與 $$$f{\left(r \right)} = e^{- 4 r}$$$：
$${\color{red}\left(\frac{d}{dr} \left(3 e^{- 4 r} \sin{\left(3 \theta \right)}\right)\right)} = {\color{red}\left(3 \sin{\left(3 \theta \right)} \frac{d}{dr} \left(e^{- 4 r}\right)\right)}$$函數 $$$e^{- 4 r}$$$ 是兩個函數 $$$f{\left(u \right)} = e^{u}$$$ 與 $$$g{\left(r \right)} = - 4 r$$$ 之複合 $$$f{\left(g{\left(r \right)} \right)}$$$。
應用鏈式法則 $$$\frac{d}{dr} \left(f{\left(g{\left(r \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dr} \left(g{\left(r \right)}\right)$$$：
$$3 \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{dr} \left(e^{- 4 r}\right)\right)} = 3 \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dr} \left(- 4 r\right)\right)}$$指數函數的導數為 $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$：
$$3 \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dr} \left(- 4 r\right) = 3 \sin{\left(3 \theta \right)} {\color{red}\left(e^{u}\right)} \frac{d}{dr} \left(- 4 r\right)$$返回原變數：
$$3 e^{{\color{red}\left(u\right)}} \sin{\left(3 \theta \right)} \frac{d}{dr} \left(- 4 r\right) = 3 e^{{\color{red}\left(- 4 r\right)}} \sin{\left(3 \theta \right)} \frac{d}{dr} \left(- 4 r\right)$$套用常數倍法則 $$$\frac{d}{dr} \left(c f{\left(r \right)}\right) = c \frac{d}{dr} \left(f{\left(r \right)}\right)$$$，使用 $$$c = -4$$$ 與 $$$f{\left(r \right)} = r$$$：
$$3 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{dr} \left(- 4 r\right)\right)} = 3 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(- 4 \frac{d}{dr} \left(r\right)\right)}$$套用冪次法則 $$$\frac{d}{dr} \left(r^{n}\right) = n r^{n - 1}$$$，取 $$$n = 1$$$，也就是 $$$\frac{d}{dr} \left(r\right) = 1$$$：
$$- 12 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{dr} \left(r\right)\right)} = - 12 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(1\right)}$$因此，$$$\frac{d}{dr} \left(3 e^{- 4 r} \sin{\left(3 \theta \right)}\right) = - 12 e^{- 4 r} \sin{\left(3 \theta \right)}$$$。
答案
$$$\frac{d}{dr} \left(3 e^{- 4 r} \sin{\left(3 \theta \right)}\right) = - 12 e^{- 4 r} \sin{\left(3 \theta \right)}$$$A