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I am attempting to learn inverse hyperbolic functions but I can't even follow the examples in the book. The example is $y= \sinh y = \frac{e^y - e^{-y}}2$ they they make it into $e^y -2x - e^{-y} = 0$ I don't really understand what is going on, are inverses equal to x? I just don't follow what he did there at all. There just isn't much explanation here, or at least enough for me to figure it out.

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    ...and it's certainly unhealthy to be frequently, repeatedly calling yourself "stupid". Things like those, repeated enough, are self-fulfilling. You need to change that habit if you don't want to be forever stuck.2011-10-07

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The text of the question is not clear, and I am a few miles away from my copy of Stewart. But I imagine that the calculation goes something like this. Suppose that $x=\sinh y= \frac{e^y-e^{-y}}{2}.$ We want to solve for $y$ in terms of $x$. The first thing to do is to multiply both sides by $2$, so we won't have to carry fractions around. They are heavy.

So we get $2x=e^y-e^{-y}$. Now we can do one of several things. Maybe rewrite $e^{-y}$ as $1/e^y$. So now we have $2x=e^y -\frac{1}{e^y}.$ It will save typing, and be useful in other ways, to let $w=e^y$. We obtain $2x=w -\frac{1}{w}.$ Multiply both sides by $w$. We get $2xw=w^2-1$. Rearrange this equation a little. We get $w^2-2xw-1=0.$ This is a quadratic equation in $w$. The solutions are, by the Quadratic Formula, $w=\frac{2x+\sqrt{4x^2+4}}{2}=x \pm\sqrt{x^2+1}.$ Now we remember that $w=e^y$, so $e^y=x \pm\sqrt{x^2+1}.$ But note that $e^y$ is always positive, and $x-\sqrt{x^2+1}$ is negative. So the solution with the minus sign has to be rejected, and $e^y=x + \sqrt{x^2+1}.$ Take the natural logarithm ($\ln$) of both sides. We get $y=\ln\left(x+\sqrt{x^2+1}\right).$

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    You can actually just spot the answer the inverse hyperbolic cosine of $1$, without needing any algebra.2011-10-07
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Try writing $y = \dfrac{e^x - e^{-x}}{2}$, so $y = \sinh x$, not $y= \sinh y$.

$ \begin{align} y & = \frac{e^x + e^{-x}}{2} \\ \\ 2y & = e^x + e^{-x} \\ \\ 0 & = e^x + e^{-x} - 2y \end{align} $ If you multiply both sides of this by $e^x$, you get $ 0 = \left(e^x\right)^2 + 1 - 2ye^x, $ so $ 0 = u^2 + 1 - 2yu $ and you've got a quadratic equation in $u$ $ au^2 + bu + c = 0, $ where $a=1$, $b=-2y$, and $c=1$.

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    I believe you have the $y$ and $x$ backwards.2011-10-07