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.)
 A: The first thing to realize is that in your answer, you have $\sinh x - \cosh x$ in the numerator of the second term, and $\cosh x - \sinh x$ in the denominator.  In general, whenever you have $\frac{A-B}{B-A}$ you can reduce this to $-1$.
So do that, and then combine the two terms into a single fraction.  That will get you the first answer from the answer key.
To get the second answer, remember that $\cosh x = \frac{e^x + e^{-x}}{2}$ and $\sinh x = \frac{e^x - e^{-x}}{2}$.  Make those substitutions and simplify everything in sight.
Finally, you could also go back to the original problem, re-express everything in terms of exponentials, simplify, and then compute the derivative.  It turns out to be much simpler to differentiate if you attack it in that order.
A: Hint
Bringing back to exponential functions simplifies it.
A: 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.
A: 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 
$$y'=\frac{\frac{e^x+e^{-x}}{2}+\frac{e^x-e^{-x}}{2}}{\frac{e^x+e^{-x}}{2}-\frac{e^x-e^{-x}}{2}}=\frac{e^x}{e^{-x}}=\color{red}{e^{2x}}$$
