乙状结肠的导数 [英] Derivative of sigmoid

问题描述

我正在使用反向传播技术创建一个用于学习的神经网络.

I'm creating a neural network using the backpropagation technique for learning.

我了解我们需要找到所用激活函数的导数.我正在使用标准的S型函数

I understand we need to find the derivative of the activation function used. I'm using the standard sigmoid function

f(x) = 1 / (1 + e^(-x))

我已经知道它的派生词是

and I've seen that its derivative is

dy/dx = f(x)' = f(x) * (1 - f(x))

这可能是一个愚蠢的问题，但这是否意味着我们必须在方程式中将x通过S型函数传递两次，因此它将扩展为

This may be a daft question, but does this mean that we have to pass x through the sigmoid function twice during the equation, so it would expand to

dy/dx = f(x)' = 1 / (1 + e^(-x)) * (1 - (1 / (1 + e^(-x))))

或者只是获取已经计算出的 f(x)输出(即神经元的输出)，然后将该值替换为 f(x)即可.代码>?

or is it simply a matter of taking the already calculated output of f(x), which is the output of the neuron, and replace that value for f(x)?

推荐答案

这两种方法是等效的(因为数学函数没有副作用，并且对于给定的输出总是返回相同的输入)，所以您也许还可以(更快)做第二种方式.

The two ways of doing it are equivalent (since mathematical functions don't have side-effects and always return the same input for a given output), so you might as well do it the (faster) second way.

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