## Calculus (3rd Edition)

$$h'(y)= \frac{2e^y}{(1-e^y)^2}.$$
Recall that $(e^x)'=e^x$ Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Since $h(y)=\frac{1+e^y}{1-e^y}$, then the derivative, using the quotient rule, is given by $$h'(y)=\frac{(1-e^y)(e^y)-(1+e^y)(-e^y)}{(1-e^y)^2}=\frac{2e^y}{(1-e^y)^2}.$$