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.)

If you have problems in computing derivatives, just do them slowly. Set $f(x)=\sinh x$ and $g(x)=\cosh x-\sinh x$; notice that $f'(x)=\cosh x$ and $g'(x)=\sinh x-\cosh x$. Then
$$
F(x)=\frac{\sinh x}{\cosh x-\sinh x}=\frac{f(x)}{g(x)}
$$
and therefore
\begin{align}
F'(x)
&=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2} \\[6px]
&=\frac{\cosh x(\cosh x-\sinh x)-\sinh x(\sinh x-\cosh x)}
       {(\cosh x-\sinh x)^2} \\[6px]
&=\frac{\cosh^2x-\cosh x\sinh x-\sinh^2x+\sinh x\cosh x}
       {(\cosh x-\sinh x)^2} \\[6px]
&=\frac{\cosh^2x-\sinh^2x}{(\cosh x-\sinh x)^2}\\[6px]
&=\frac{\cosh x+\sinh x}{\cosh x-\sinh x}
\end{align}
but this can be written
$$
\frac{1}{(\cosh x-\sinh x)^2}
$$
as well.
