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\begin{aligned} \sigma^{\prime}(x) &=\frac{\partial}{\partial x} \frac{1}{1+e^{-x}} \\ &=\frac{e^{-x}}{\left(1+e^{-x}\right)^{2}} \\ &=\frac{1}{1+e^{-x}} \cdot \frac{e^{-x}}{1+e^{-x}} \\ &=\sigma(x)(1-\sigma(x)) \end{aligned}

I do not understand what is the last step in this derivative. How can you factor out the last term?

\begin{aligned} \sigma(x)(1-\sigma(x)) \end{aligned}

David C. Ullrich
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Mark P
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1 Answers1

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$$\frac{e^{-x}}{1+e^{-x}} = \frac{(1+e^{-x})-1}{1+e^{-x}} = 1 - \frac{1}{1+e^{-x}} = 1- \sigma(x)$$

19aksh
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