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Math Help - Derivative of sigmoid function query

  1. #1
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    Derivative of sigmoid function query

    Hi,
    As part of the differentiation of the sigmoid function the following step occurs.

    (1/(1+e^-z))^2 * (1/e^z) = 1/(1+e^-z) * (1-(1/(1+e^-z)))


    But I just cant see my way through the algebra. Really niggling me, so an elucidation would be greatly appreciated.

    Thanks in advance, Jon
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  2. #2
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    Hello, Mathsdog!

    As part of the differentiation of the sigmoid function the following step occurs:

    . . \displaystyle \frac{1}{(1+e^{-z})^2}\cdot\frac{1}{e^z} \;=\;\frac{1}{1+e^{-z}}\cdot \left(1-\frac{1}{1+e^{-z}}\right)

    But I just can't see my way through the algebra.

    A strange step . . . I wouldn't call it "simplifying".


    \displaystyle \text{We have: }\;\frac{1}{(1+e^{-z})^2} \cdot \frac{1}{e^x}

    . . . . . . \displaystyle =\;\frac{1}{1+e^{-z}}\cdot\frac{1}{1+e^{-z}} \cdot e^{-z}

    . . . . . . \displaystyle =\;\frac{1}{1 + e^{-z}} \cdot\frac{e^{-z}}{1+e^{-z}}


    In the second fraction, add and subtract 1 to the numerator:

    . . . . . . \displaystyle =\; \frac{1}{1+e^{-z}} \cdot \frac{{\bf1} + e^{-z} {\bf -1}}{1+e^{-z}}

    . . . . . . \displaystyle =\;\frac{1}{1+e^{-z}}\left(\frac{1+e^{-z}}{1+e^{-z}} - \frac{1}{1+e^{-z}}\right)

    . . . . . . \displaystyle =\;\frac{1}{1 + e^{-z}}\left(1 - \frac{1}{1+e^{-z}}\right)

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  3. #3
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    Hi Soroban. Thanks so much. That was really annoying. As you say, its not a simpification. The object was to show that where h(z) = 1/1+e^-z (i.e. the sigmoid function), then h'(z) = h(z)*(1-h(z)). As you see, thats the final form you so kindly demonstrated. Thanks again. J
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