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The context is these notes ( https://see.stanford.edu/materials/aimlcs229/cs229-notes1.pdf ) page 17.

From here:

$\frac{1}{{(}{1}\hspace{0.33em}{+}\hspace{0.33em}{e}^{{-}{z}}{)}^{2}}\hspace{0.33em}\cdot\hspace{0.33em}{(}{e}^{{-}{z}}{)}$

To Here:

$\frac{1}{{(}{1}\hspace{0.33em}{+}\hspace{0.33em}{e}^{{-}{z}}{)}}\hspace{0.33em}\cdot\hspace{0.33em}\left({{1}{-}\hspace{0.33em}\frac{1}{{(}{1}\hspace{0.33em}{+}\hspace{0.33em}{e}^{{-}{z}}{)}}}\right)$

Bit confused about the negative:

${-}\hspace{0.33em}\frac{1}{{(}{1}\hspace{0.33em}{+}\hspace{0.33em}{e}^{{-}{z}}{)}}$

Can someone please help by filling in some of the intemediate steps....

Thanks

markthekoala
  • 111
  • Thanks... do you mean rewriting the second expression as: $\frac{1}{{(}{1}{+}{e}^{{-}{z}}{)}}\hspace{0.33em}\bullet\hspace{0.33em}\left({\frac{{(}{1}{+}{e}^{{-}{z}}{)}}{{(}{1}{+}{e}^{{-}{z}}{)}}\hspace{0.33em}{-}\hspace{0.33em}\frac{1}{{(}{1}{+}{e}^{{-}{z}}{)}}}\right)$ – markthekoala Apr 05 '17 at 03:25
  • I think my question is a duplicate. I have found something really helpful here: http://math.stackexchange.com/questions/1115302/please-explain-the-algebra-in-the-last-part-of-derivative-of-the-sigmoid-functio?rq=1 – markthekoala Apr 05 '17 at 03:40

1 Answers1

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Hint: Find common denominators in the parentheses. As a general rule, every mysterious algebraic simplification is made easier by common denominators.

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