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.
$\endgroup$ 63 Answers
$\begingroup$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 $$\frac{\mathrm{d}\varphi}{\mathrm{d}v}=\frac{a}{1+e^{-av}}\left[1-\frac{1}{1+e^{-av}}\right]=\frac{a}{1+e^{-a\cdot 0}}\left[1-\frac{1}{1+e^{-a\cdot 0}}\right]=\frac{a}{1+1}\left[1-\frac{1}{1+1}\right]=\frac{a}{4}$$
$\endgroup$ 3 $\begingroup$$$\varphi(v)=\frac{1}{1+e^{-av}}\Longrightarrow$$
$$\frac{\partial\varphi(v)}{\partial v}=\frac{\partial}{\partial v}\left(\frac{1}{1+e^{-av}}\right)=\frac{a\cdot e^{av}}{\left(1+e^{av}\right)^2}$$
Now if $v=0$:
$$\varphi'(0)=\frac{a\cdot e^{a\cdot 0}}{\left(1+e^{a\cdot 0}\right)^2}=\frac{a\cdot e^{0}}{\left(1+e^{0}\right)^2}=\frac{a\cdot 1}{(1+1)^2}=\frac{a}{2^2}=\frac{a}{4}$$
$\endgroup$ $\begingroup$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.
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