I'm following Jeff heatons book 'Introduction to Neural Networks with Java'. To get node deltas, we need to calculate $f'(sum)$. In the very first row for Training Element #1, we need to compute $f'(1.13) \cdot 0.25$ which Heaton evaluates to $0.045$.

Using my calculator, I get $f'(1.13)=-0.13$ (derivative of sigmoid activation function), then I multiply by $0.25$ to get $-0.0325$. I've been trying to figure out for days how heaton does his calculation but no success yet. Kindly assist.


The derivative of a sigmoid is always positive. Here's my math: Sigmoid(1.13) = 0.7558

I computed this on Wolfram Alpha

The derivative of a sigmoid y=S(x) is y * (1-y), so 0.7558 * 0.2442 = 0.0451

  • $\begingroup$ Hi,I think what you did is f '( Sigmoid(X) ).What the author did was f '(X)*0.25 and got 0.045 $\endgroup$ Apr 24 '14 at 5:04
  • $\begingroup$ @LeroyKayanda If your $f$ is the sigmoid, then MattD is correct. Certainly the fact that he got the same answer as the author is an indicator that his thinking parallels the book. I don't have that book and you haven't said enough to know. $\endgroup$
    – Gene
    Apr 24 '14 at 23:54
  • $\begingroup$ Hi. Doing F ' ( Sigmoid(X) ) works with training element #1 but not with #2 in ( heatonresearch.com/wiki/Backpropagation ) . Here's the video that explains the calculation process youtube.com/watch?v=p1-FiWjThs8 $\endgroup$ Apr 25 '14 at 0:33
  • $\begingroup$ Hey, I finall got it @MattD is right. The derivative of sigmoid $f'(x)=f(x)*(1-f(x))$ Equivalent to $f'(x)$=sigmoid(x)*(1-sigmoid(x)) . What I was doing was $f'(x)=x*(1-x)$ Thanks for the help $\endgroup$ Apr 25 '14 at 5:20

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