Simplifying Dervatives of Hyperbolic functions Last minute Calc I reviews have me stumbling on this question
$$D_x\left[\frac {\sinh x}{\cosh x-\sinh x}\right] $$
I've solved the derivative as 
$$ y' = \frac{\cosh x}{\cosh x-\sinh x} -\frac{\sinh x(\sinh x-\cosh x)}{(\cosh x-\sinh x)^2} $$
which is consistent with an online derivative calculator I've been using to check my answers. However, the answer sheet my professor handed out has the following as the answer:
$$ \frac{\cosh x + \sinh x}{\cosh x - \sinh x}$$
or even 
$$e^{2x}$$
I haven't the foggiest how she got either of those from the derivative. Can anyone help me simplify it? (This is not a graded assignment, it's for review purposes and she already gave us the answers.)

Hint
Bringing back to exponential functions simplifies it.
