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:

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