Chapter 6 - Inverse Functions - Review - Exercises - Page 505: 22

$$g'(t)=\frac{e^t}{\left(1+e^t\right)^2}$$

Given $$g(t)=\frac{e^t}{1+e^t}$$ Then \begin{align*} g'(t)&= \frac{\frac{d}{dt}\left(e^t\right)\left(1+e^t\right)-\frac{d}{dt}\left(1+e^t\right)e^t}{\left(1+e^t\right)^2}\\ &=\frac{e^t\left(1+e^t\right)-e^te^t}{\left(1+e^t\right)^2}\\ &=\frac{e^t}{\left(1+e^t\right)^2} \end{align*}