sigmoid函数求导
sigmoid函数：
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f(x)= \frac{1}{1+e^{-x}}
f(x)=1+e−x1
sigmoid函数的导数：
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f'(x)=f(x)(1-f(x))
f′(x)=f(x)(1−f(x))
推导过程
- 首先，对 f ( x ) f(x) f(x)进行变形： f ( x ) = 1 1 + e − x = 1 1 + 1 e x = ( 1 + 1 e x ) − 1 = ( e x e x + 1 e x ) − 1 = ( e x + 1 e x ) − 1 = e x e x + 1 = ( e x + 1 ) − 1 e x + 1 = e x + 1 e x + 1 − 1 e x + 1 = 1 − 1 e x + 1 = 1 − ( e x + 1 ) − 1 \begin{aligned} f(x)&= \frac{1}{1+e^{-x}} \ &= \frac{1}{1+\frac{1}{e^x}} \ &=(1+\frac{1}{e^x})^{-1} \ &=(\frac{e^x}{e^x}+\frac{1}{e^x})^{-1} \ &=(\frac{e^{x}+1}{e^x})^{-1} \ &=\frac{e^x}{e^{x}+1} \ &=\frac{(e^{x}+1)-1}{e^{x}+1} \ &=\frac{e^{x}+1}{e^{x}+1}-\frac{1}{e^{x}+1} \ &=1-\frac{1}{e^{x}+1} \ &=1-(e^{x}+1)^{-1} \end{aligned} f(x)=1+e−x1=1+ex11=(1+ex1)−1=(exex+ex1)−1=(exex+1)−1=ex+1ex=ex+1(ex+1)−1=ex+1ex+1−ex+11=1−ex+11=1−(ex+1)−1
- 求导：
注意使用链式法则求导
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\begin{aligned} f'(x)&=(1-(e^{x}+1)^{-1})' \\ &=(-1)(-1)(e^{x}+1)^{-2} e^{x}\\ &=(e^{x}+1)^{-2} e^{x}\\ &=(e^{x}+1)^{-1}(e^{x}+1)^{-1} e^{x} \end{aligned}
f′(x)=(1−(ex+1)−1)′=(−1)(−1)(ex+1)−2ex=(ex+1)−2ex=(ex+1)−1(ex+1)−1ex
由前面提到的
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f(x)的变形可知：
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\begin{aligned} f(x)&=\frac{1}{1+e^{-x}} =(1+e^{-x})^{-1}=\frac{e^{x}}{e^{x}+1}=e^{x}(e^{x}+1)^{-1} \end{aligned}
f(x)=1+e−x1=(1+e−x)−1=ex+1ex=ex(ex+1)−1
所以
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\begin{aligned} f'(x)&=(e^{x}+1)^{-1} \cdot (e^{x}+1)^{-1} e^{x} \\ &= (e^{x}+1)^{-1} \cdot e^{x}(e^{x}+1)^{-1} \\ &=(e^{x}+1)^{-1} \cdot (1+e^{-x})^{-1} \\ &=\frac{1}{e^{x}+1} \cdot \frac{1}{1+e^{-x}} \\ &=\frac{(e^{x}+1)-e^{x}}{e^{x}+1} \cdot \frac{1}{1+e^{-x}} \\ &=(\frac{e^{x}+1}{e^{x}+1}-\frac{e^{x}}{e^{x}+1}) \cdot \frac{1}{1+e^{-x}} \\ &=(1-\frac{e^{x}}{e^{x}+1}) \cdot \frac{1}{1+e^{-x}} \\ &=(1-\frac{1}{1+e^{-x}}) \cdot \frac{1}{1+e^{-x}} \\ &=(1-f(x)) \cdot f(x) \\ &=f(x)(1-f(x)) \end{aligned}
f′(x)=(ex+1)−1⋅(ex+1)−1ex=(ex+1)−1⋅ex(ex+1)−1=(ex+1)−1⋅(1+e−x)−1=ex+11⋅1+e−x1=ex+1(ex+1)−ex⋅1+e−x1=(ex+1ex+1−ex+1ex)⋅1+e−x1=(1−ex+1ex)⋅1+e−x1=(1−1+e−x1)⋅1+e−x1=(1−f(x))⋅f(x)=f(x)(1−f(x))
本文转载自: https://blog.csdn.net/qq_35229591/article/details/128752369
版权归原作者 卡洛驰 所有， 如有侵权，请联系我们删除。
版权归原作者 卡洛驰 所有， 如有侵权，请联系我们删除。