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I have the following question in a course:

An example of the logistic function is defined by $$\varphi(v)=\frac{1}{1+e^{-av}}$$ whose limiting values are $0$ and $1$. Show that the derivative of $\varphi(v)$ with respect to $v$ is given by $$\frac{\mathrm{d}\varphi}{\mathrm{d}v}=a\varphi(v)[1-\varphi(v)]$$ What is the value of this derivative at the origin?

The first part is solved, but I couldn't understand the second (What is the value of this derivative at the origin?).

Can anyone help me? Thanks in advance.

BLAZE
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EFK10003
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3 Answers3

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Since your function $$\varphi(v)=\frac{1}{1+e^{-av}}$$ is a function of only one variable (which is $v$), the value of the derivative at the origin is given by setting $v=0$ in your equation for $$\frac{\mathrm{d}\varphi}{\mathrm{d}v}=\frac{a}{1+e^{-av}}\left[1-\frac{1}{1+e^{-av}}\right]=\frac{a}{1+e^{-a\cdot 0}}\left[1-\frac{1}{1+e^{-a\cdot 0}}\right]=\frac{a}{1+1}\left[1-\frac{1}{1+1}\right]=\frac{a}{4}$$

BLAZE
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  • Okay thank you @BLAZE . Just one more question, we always set v to 0 or just in this case since the limiting values are 0 and 1 ? – EFK10003 Oct 18 '15 at 12:27
  • @EFK Explain what you mean by limiting? – BLAZE Oct 18 '15 at 12:30
  • In the question, the author wrote that the logistic function whose limiting values are 0 and 1. My question is v set to 0 is related to limiting values? – EFK10003 Oct 18 '15 at 12:31
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$$\varphi(v)=\frac{1}{1+e^{-av}}\Longrightarrow$$

$$\frac{\partial\varphi(v)}{\partial v}=\frac{\partial}{\partial v}\left(\frac{1}{1+e^{-av}}\right)=\frac{a\cdot e^{av}}{\left(1+e^{av}\right)^2}$$

Now if $v=0$:

$$\varphi'(0)=\frac{a\cdot e^{a\cdot 0}}{\left(1+e^{a\cdot 0}\right)^2}=\frac{a\cdot e^{0}}{\left(1+e^{0}\right)^2}=\frac{a\cdot 1}{(1+1)^2}=\frac{a}{2^2}=\frac{a}{4}$$

Jan Eerland
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The limiting values are the limits the function is approaching towards positive or negative infinity. Graph the function with any $\alpha$ and you will see the result.

user281498
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