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.)

Notice, $$y'=\frac{\cosh x}{\cosh x-\sinh x}-\frac{\sinh x(\sinh x-\cosh x)}{(\cosh x-\sinh x)^2}$$$$=\frac{\cosh^2x-\sinh x\cosh x-\sinh^2x+\sinh x\cosh x}{(\cosh x-\sinh x)^2}$$
$$=\frac{\cosh^2x-\sinh^2x}{(\cosh x-\sinh x)^2}$$
$$=\frac{(\cosh x+\sinh x)(\cosh x-\sinh x)}{(\cosh x-\sinh x)^2}$$
$$=\color{red}{\frac{\cosh x+\sinh x}{\cosh x-\sinh x}}$$
Now, substitute the values of $\cosh x$ & $\sinh x$ as follows 
$$y'=\frac{\frac{e^x+e^{-x}}{2}+\frac{e^x-e^{-x}}{2}}{\frac{e^x+e^{-x}}{2}-\frac{e^x-e^{-x}}{2}}=\frac{e^x}{e^{-x}}=\color{red}{e^{2x}}$$
