# why are we ignoring universal constant function in derivative of sigmoid function?

Please take a look at my case:

$$t(x)=1/(1+exp^{-x})\\therefore\ we\ can\ create\ three\ functions: \\f(x)=exp^{-x}\\g(f(x)) = 1+f(x)\\c(g(f(x)) = 1/g(f(x))\\$$

we know

$$t'(x) = t(x)*(1-t(x))$$

and i think it should be:

$$t'(x) = t(x)*(1-t(x)*exp^{-x})$$

why?

chain rule says, find derivative of each function and multiply together, therefore we should get out this:

$$t'(x) = -1(1/g(f(x)))^-2*exp^{-x}*exp^{-x}$$

or on different case:

$$\\t(x) = ((x^2)^2)^2\\ \\f(x) = x^2\\ \\g(f(x)) = f(x)^2\\ \\c(g(f(x)) = g(f(x))^2\\$$

therefore derivative is:

$$t'(x) = 8x^7$$

• MathJax hints: if you put backslashes before common functions you get the proper font and spacing, so \exp x gives $\exp x$. When you want multicharacter exponents, put them in braces, so e^{-x} gives $e^{-x}$ instead of e^-x which gives $e^-x$. This works everywhere-thinks in braces are considered a unit. – Ross Millikan Mar 29 '18 at 16:20
• @RossMillikan Thanks I fixed it – filtertips Mar 29 '18 at 16:22
• Introducing a different $t(x)$ makes your argument hard to follow. You should start from $t(x)=c(g(f(x)))$ and apply the chain rule to that. It should also be either $\exp(-x)$ or $e^{-x}$, not $exp^{-x}$ – Ross Millikan Mar 29 '18 at 16:23
• – Hans Lundmark Mar 29 '18 at 18:28

Since $g(y) = 1+y$, we have $g'(y) = 1$ for any $y$, so the middle step of your chain rule is wrong. It should be $t'(x) = -\frac{1}{g(f(x))^2} \cdot 1 \cdot (-e^{-x})$.

• Thank you. I have read on wikipedia derivative of e{x} is always e{x}. By the way, where this you get - (minuses) in t'(x). I would appreciate it. – filtertips Mar 29 '18 at 16:38