Sigmoid function

 sigmoid function คือ

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

Derivative of sigmoid

$$\begin{array}{rl}{\sigma }^{\prime }(X)& =\frac{d(\frac{1}{1+e^{-X}})}{dX}\end{array}$$

จาก recipocal rule[1] ถ้ากำหนดให้ $u(X)=1+e^{-X}$

$$\begin{array}{rl}{\sigma }^{\prime }(X)& =\frac{d(\frac{1}{u(X)})}{dX}\\ {\sigma }^{\prime }(X)& =-\frac{1}{u(X{)}^2}\frac{du(X)}{dX}\\ {\sigma }^{\prime }(X)& =-\frac{1}{u(X{)}^2}{u}^{\prime }(X)\end{array}$$

จาก linearity rule[2]

$$\begin{array}{rl}{\sigma }^{\prime }(X)& =-\frac{1}{u(X{)}^2}{u}^{\prime }(X)\\ {\sigma }^{\prime }(X)& =-\frac{1}{u(X{)}^2}\frac{d1}{dX}\frac{d(e^{-X})}{dX}\\ {\sigma }^{\prime }(X)& =-\frac{1}{u(X{)}^2}\frac{(d{e}^{-X})}{dX}\end{array}$$

จาก exponential rule[3]

$$\begin{array}{rl}{\sigma }^{\prime }(X)& =-\frac{1}{u(X{)}^2}\frac{(d{e}^{-X})}{dX}\\ {\sigma }^{\prime }(X)& =-\frac{1}{u(X{)}^2}{e}^{-X}\frac{(d-X)}{dX}\\ {\sigma }^{\prime }(X)& =\frac{e^{-X}}{u(X{)}^2}\\ {\sigma }^{\prime }(X)& =\frac{e^{-X}}{(1+e^{-X}{)}^2}\end{array}$$

จะเห็นว่าสมการล่าสุด ลองพยายามทำให้อยู่ในรูปของ sigmoid function

$$\begin{array}{rl}{\sigma }^{\prime }(X)& =\frac{e^{-X}}{(1+e^{-X}{)}^2}\\ {\sigma }^{\prime }(X)& =\frac{1}{(1+e^{-X})}\frac{e^{-X}}{(1+e^{-X})}\\ {\sigma }^{\prime }(X)& =\sigma (X)\frac{e^{-X}}{(1+e^{-X})}\\ {\sigma }^{\prime }(X)& =\sigma (X)\frac{e^{-X}+1-1}{(1+e^{-X})}\\ {\sigma }^{\prime }(X)& =\sigma (X)[\frac{e^{-X}+1}{(1+e^{-X})}-\frac{1}{(1+e^{-X})}]\\ {\sigma }^{\prime }(X)& =\sigma (X)[1-\sigma (X)]\end{array}$$

นั่นคือ

$$\boxed{{\displaystyle {\sigma }^{\prime }(X)=\sigma (X)[1-\sigma (X)]}}$$

เอกสารอ้างอิง

[1] https://en.wikipedia.org/wiki/Reciprocal_rule

[2] https://en.wikipedia.org/wiki/Linearity_of_differentiation

[3] https://tutorial.math.lamar.edu/classes/calci/diffexplogfcns.aspx