<s>[INST] Trying to determine the Inverse function of $\sinh$ and $\cosh$ I'm trying to find out how to determine the inverse function in order to develop the  $$ \sinh(x).$$
I tried to expand to its exponential form $$\sinh(x) = \frac{1}{2} (e^x-e^{-x})  .$$
So I wrote
$$ \frac{1}{2}(e^x-e^{-x}) = y . $$
So  we find ourselves with this equation 
By substituting  $ e^x = t $, 
we have $$ t^2 -2yt +1 = 0 .$$
   Noting that  $\sinh$ changes its sign from negative to positive as shown below. 

What to do next because  $$ \delta = 4y^2-4 .$$
 [/INST] If you multiply both sides of the equation 
$$y = \frac{e^x - e^{-x}}{2}$$
by $2e^x$, you obtain
\begin{align*}
2e^{x}y & = e^{2x} - 1\\
0 & = e^{2x} - 2e^{x}y - 1
\end{align*}
so you should have obtained the equation $t^2 - 2ty - 1 = 0$ when you made the substitution $y = e^x$.
To find the inverse, we interchange $x$ and $y$ in the equation $e^{2x} - 2e^{x}y - 1 = 0$, then solve for $y$.
\begin{align*}
e^{2y} - 2xe^{y} - 1 & = 0\\
e^{2y} - 2xe^{y} & = 1\\
e^{2y} - 2xe^{y} + x^2 & = 1 + x^2 && \text{complete the square}\\
(e^y - x)^2 & = 1 + x^2\\
e^y - x & = \pm \sqrt{1 + x^2}\\
e^y & = x \pm \sqrt{1 + x^2}
\end{align*}
Since the range of $e^y$ is the set of all positive real numbers, we discard the negative root since $x - \sqrt{1 + x^2} < 0$.  Thus,
\begin{align*}
e^y & = x + \sqrt{1 + x^2}\\
y & = \log_e (x + \sqrt{1 + x^2})
\end{align*}
</s>