Differentiate $y=\frac{e^{\sinh ax}}{\sinh bx-\cosh bx}$ 
Differentiate $$y=\frac{e^{\sinh ax}}{\sinh bx-\cosh bx}$$

I tried the quotient rule and got this :
$$y'=\frac{a\cosh ax(\sinh bx - \cosh bx)e^{\sinh ax}-e^{\sinh ax}(b\cosh bx + sinh bx)}{\sinh bx - \cosh bx}$$
I don't find a way to simplify this into the answer given in the book :
$$e^{\sinh ax}e^{bx(acosh ax + b)}$$
Where am I doing wrong ?
 A: Note that $\displaystyle \sinh bx - \cosh bx = \frac{e^{bx} - e^{-bx}}{2} - \frac{e^{bx} + e^{-bx}}{2} = -e^{-bx}$. 
Now: $$\frac{d}{dx}{e^{bx + \sinh ax}} = (b +a\cosh ax)e^{bx+\sinh ax}.$$
So your derivative is simply $-(b+a\cosh ax)e^{bx+\sinh ax}$. The 'form' you're looking for seems to be incorrect. 
A: By applying the quotient rule:
$$
\begin{aligned}
\frac{d}{dx}\left(\frac{e^{\sinh \left(ax\right)}}{\sinh \left(bx\right)-\cosh \left(bx\right)}\right)
& =\frac{\frac{d}{dx}\left(e^{\sinh \left(ax\right)}\right)\left(\sinh \left(bx\right)-\cosh \left(bx\right)\right)-\frac{d}{dx}\left(\sinh \left(bx\right)-\cosh \left(bx\right)\right)e^{\sinh \left(ax\right)}}{\left(\sinh \left(bx\right)-\cosh \left(bx\right)\right)^2}
\\ & = \color{red}{\frac{e^{\sinh \left(ax\right)}\left(b+a\cosh \left(ax\right)\right)}{\sinh \left(bx\right)-\cosh \left(bx\right)}}
\end{aligned}$$
NOTE:
$$\frac{e^{\sinh \left(ax\right)}\left(b+a\cosh \left(ax\right)\right)}{\sinh \left(bx\right)-\cosh \left(bx\right)} \color{red}{\ne} \:e^{\sinh \left(ax\right)}e^{bx\left(acosh(ax)+b\right)}$$
A: Remember that
$$
\sinh x = \frac{e^x - e^{-x}}{2},\quad \cosh x = \frac{e^x + e^{-x}}{2}.
$$
Then
$$
\sinh bx -\cosh bx =-e^{-bx}.
$$
Then from your result(correcting typos):
\begin{align}
y'&=\frac{a\cosh ax(\sinh bx - \cosh bx)e^{\sinh ax}-e^{\sinh ax}(b\cosh bx -b\sinh bx)}{(\sinh bx - \cosh bx)^2}\\
&=(a\cosh ax \cdot (-e^{-bx})e^{\sinh ax} - be^{\sinh ax}\cdot (e^{-bx}))(e^{2bx})\\
&=e^{\sinh ax}\left(-a\cdot \cosh ax\cdot e^{bx} -be^{bx}\right)\\
&=e^{\sinh ax}e^{bx}\left(-a\cdot \cosh ax -b\right)\\
&=-e^{\sinh ax}e^{bx}(a\cosh ax+b),
\end{align}
which is different from the "answer".
