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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 ?

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    $\begingroup$ Are you sure that that's the answer given in your book? $\endgroup$ – Zain Patel May 3 '17 at 11:22
  • $\begingroup$ @ZainPatel Let me check. $\endgroup$ – H G Sur May 3 '17 at 11:23
  • $\begingroup$ Just made sure, that was the answer given in the book. You can find it here on question 2.2.8(e) : archive.org/stream/ProblemsInCalculusOfOneVariableI.A.Maron/… $\endgroup$ – H G Sur May 3 '17 at 11:29
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    $\begingroup$ @H G Sur yes that's certainly a mistake. It looks like the author wanted to type $e^{\sinh ax}e^{bx}(a\cosh ax + b)$ as per my answer instead. $\endgroup$ – Zain Patel May 3 '17 at 11:36
  • $\begingroup$ @ZainPatel Thanks. $\endgroup$ – H G Sur May 3 '17 at 11:37
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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.

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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".

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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: $$\frac{e^{\sinh \left(ax\right)}\left(b+a\cosh \left(ax\right)\right)}{\sinh \left(bx\right)-\cosh \left(bx\right)} \color{red}{\ne} \:e^{\sinh \left(ax\right)}e^{bx\left(acosh(ax)+b\right)}$$

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  • $\begingroup$ Thank you. I got this answer too but how to simplify it to the form I mentioned in the question ? $\endgroup$ – H G Sur May 3 '17 at 11:12
  • $\begingroup$ @HGSur i updated my answer $\endgroup$ – Amarildo May 3 '17 at 11:19

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