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$f(x)$进行变形：
  
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  \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+ex1​1​=(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$f(x)$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))​