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:
$$ t'(x) = 8x^7 $$
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 $t'(x) = -\frac{1}{g(f(x))^2} \cdot 1 \cdot (-e^{-x})$.