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$$$e^{- \frac{1}{u^{2}}}$$$的导数
您的输入
求$$$\frac{d}{du} \left(e^{- \frac{1}{u^{2}}}\right)$$$。
解答
函数$$$e^{- \frac{1}{u^{2}}}$$$是两个函数$$$f{\left(v \right)} = e^{v}$$$和$$$g{\left(u \right)} = - \frac{1}{u^{2}}$$$的复合$$$f{\left(g{\left(u \right)} \right)}$$$。
应用链式法则 $$$\frac{d}{du} \left(f{\left(g{\left(u \right)} \right)}\right) = \frac{d}{dv} \left(f{\left(v \right)}\right) \frac{d}{du} \left(g{\left(u \right)}\right)$$$：
$${\color{red}\left(\frac{d}{du} \left(e^{- \frac{1}{u^{2}}}\right)\right)} = {\color{red}\left(\frac{d}{dv} \left(e^{v}\right) \frac{d}{du} \left(- \frac{1}{u^{2}}\right)\right)}$$指数函数的导数为 $$$\frac{d}{dv} \left(e^{v}\right) = e^{v}$$$：
$${\color{red}\left(\frac{d}{dv} \left(e^{v}\right)\right)} \frac{d}{du} \left(- \frac{1}{u^{2}}\right) = {\color{red}\left(e^{v}\right)} \frac{d}{du} \left(- \frac{1}{u^{2}}\right)$$返回到原变量：
$$e^{{\color{red}\left(v\right)}} \frac{d}{du} \left(- \frac{1}{u^{2}}\right) = e^{{\color{red}\left(- \frac{1}{u^{2}}\right)}} \frac{d}{du} \left(- \frac{1}{u^{2}}\right)$$对 $$$c = -1$$$ 和 $$$f{\left(u \right)} = \frac{1}{u^{2}}$$$ 应用常数倍法则 $$$\frac{d}{du} \left(c f{\left(u \right)}\right) = c \frac{d}{du} \left(f{\left(u \right)}\right)$$$：
$$e^{- \frac{1}{u^{2}}} {\color{red}\left(\frac{d}{du} \left(- \frac{1}{u^{2}}\right)\right)} = e^{- \frac{1}{u^{2}}} {\color{red}\left(- \frac{d}{du} \left(\frac{1}{u^{2}}\right)\right)}$$应用幂次法则 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$，其中 $$$n = -2$$$:
$$- e^{- \frac{1}{u^{2}}} {\color{red}\left(\frac{d}{du} \left(\frac{1}{u^{2}}\right)\right)} = - e^{- \frac{1}{u^{2}}} {\color{red}\left(- \frac{2}{u^{3}}\right)}$$因此，$$$\frac{d}{du} \left(e^{- \frac{1}{u^{2}}}\right) = \frac{2 e^{- \frac{1}{u^{2}}}}{u^{3}}$$$。
答案
$$$\frac{d}{du} \left(e^{- \frac{1}{u^{2}}}\right) = \frac{2 e^{- \frac{1}{u^{2}}}}{u^{3}}$$$A