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? 

$y=-1+(1-e^x)^{-1} \implies y'=e^x(1-e^x)^{-2}$.
