$$$3 e^{- 4 r} \sin{\left(3 \theta \right)}$$$ 关于 $$$r$$$ 的导数
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您的输入
求$$$\frac{d}{dr} \left(3 e^{- 4 r} \sin{\left(3 \theta \right)}\right)$$$。
解答
对 $$$c = 3 \sin{\left(3 \theta \right)}$$$ 和 $$$f{\left(r \right)} = e^{- 4 r}$$$ 应用常数倍法则 $$$\frac{d}{dr} \left(c f{\left(r \right)}\right) = c \frac{d}{dr} \left(f{\left(r \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dr} \left(3 e^{- 4 r} \sin{\left(3 \theta \right)}\right)\right)} = {\color{red}\left(3 \sin{\left(3 \theta \right)} \frac{d}{dr} \left(e^{- 4 r}\right)\right)}$$函数$$$e^{- 4 r}$$$是两个函数$$$f{\left(u \right)} = e^{u}$$$和$$$g{\left(r \right)} = - 4 r$$$的复合$$$f{\left(g{\left(r \right)} \right)}$$$。
应用链式法则 $$$\frac{d}{dr} \left(f{\left(g{\left(r \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dr} \left(g{\left(r \right)}\right)$$$:
$$3 \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{dr} \left(e^{- 4 r}\right)\right)} = 3 \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dr} \left(- 4 r\right)\right)}$$指数函数的导数为 $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:
$$3 \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dr} \left(- 4 r\right) = 3 \sin{\left(3 \theta \right)} {\color{red}\left(e^{u}\right)} \frac{d}{dr} \left(- 4 r\right)$$返回到原变量:
$$3 e^{{\color{red}\left(u\right)}} \sin{\left(3 \theta \right)} \frac{d}{dr} \left(- 4 r\right) = 3 e^{{\color{red}\left(- 4 r\right)}} \sin{\left(3 \theta \right)} \frac{d}{dr} \left(- 4 r\right)$$对 $$$c = -4$$$ 和 $$$f{\left(r \right)} = r$$$ 应用常数倍法则 $$$\frac{d}{dr} \left(c f{\left(r \right)}\right) = c \frac{d}{dr} \left(f{\left(r \right)}\right)$$$:
$$3 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{dr} \left(- 4 r\right)\right)} = 3 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(- 4 \frac{d}{dr} \left(r\right)\right)}$$应用幂法则 $$$\frac{d}{dr} \left(r^{n}\right) = n r^{n - 1}$$$,取 $$$n = 1$$$,也就是说,$$$\frac{d}{dr} \left(r\right) = 1$$$:
$$- 12 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(\frac{d}{dr} \left(r\right)\right)} = - 12 e^{- 4 r} \sin{\left(3 \theta \right)} {\color{red}\left(1\right)}$$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