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sigmoid函数求导(sigmoid函数求导表达式)
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sigmoid函数： f(x)=11+e−xf(x)= \frac{1}{1+e^{-x}}f(x)=1+e−x1 sigmoid函数的导数： f′(x)=f(x)(1−f(x))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)=11+e−x=11+1ex=(1+1ex)−1=(exex+1ex)−1=(ex+1ex)−1=exex+1=(ex+1)−1ex+1=ex+1ex+1−1ex+1=1−1ex+1=1−(ex+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
f′(x)=(1−(ex+1)−1)′=(−1)(−1)(ex+1)−2ex=(ex+1)−2ex=(ex+1)−1(ex+1)−1ex\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f(x)f(x)f(x)的变形可知： f(x)=11+e−x=(1+e−x)−1=exex+1=ex(ex+1)−1\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 f′(x)=(ex+1)−1⋅(ex+1)−1ex=(ex+1)−1⋅ex(ex+1)−1=(ex+1)−1⋅(1+e−x)−1=1ex+1⋅11+e−x=(ex+1)−exex+1⋅11+e−x=(ex+1ex+1−exex+1)⋅11+e−x=(1−exex+1)⋅11+e−x=(1−11+e−x)⋅11+e−x=(1−f(x))⋅f(x)=f(x)(1−f(x))\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))
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