Differentiate the Function: $y=\frac{e^x}{1-e^x}$ $y=\frac{e^x}{1-e^x}$
Utilize the quotient rule: 
$y'=\frac{(1-e^x)\cdot\ e^x\ - [e^x(-e^x) ]}{(1-e^x)^2}$
$y'=\frac{(1-e^x)\cdot\ e^x\ - e^x(+e^x) }{(1-e^x)^2}$
I am confused how to solve the problem from this point. Can I factor out the negative sign in the numerator? If so why? 

It would probably be easier to write the function as $$y=\frac{-(1-e^x)}{1-e^x}-\frac{1}{1-e^x}$$
$$y=-1-\frac{1}{1-e^x}$$
Now use the chain rule:
$$y'=-(1-e^x)^{-2}\cdot-e^x=e^x(1-e^x)^{-2}=\frac{e^x}{(1-e^x)^{2}}$$
