Inverse Hyperbolic Trigonometric Functions
For background on inverse functions, refer to a previous post on the inverse standard trig functions. Looking at the graphs of the hyperbolic trig functions in the previous post, we can see that $\sinh$, $\tanh$, $\csch$, and $\coth$ are one-to-one, and therefore have a well-defined inverse. By contrast, $\cosh$ and $\sech$ are not one-to-one, so we will have to restrict their domain. Since we have exact formulas for the hyperbolic trig functions in terms of $e^x$, we will also get exact formulas for their inverse functions in terms of $\ln x$.
Derivations
Hyperbolic Sine
The standard method for solving for inverse functions is to swap $x$ and $y$ and then isolate for $y$.
\[y = \sinh(x) = \frac{e^x - e^{-x}}{2} \qquad\implies\qquad x = \frac{e^y - e^{-y}}{2}\]Now, we are going to isolate $e^y$, which will then let us isolate for $y$. Multiply both sides by $2e^{y}$.
\[2x e^y = e^{2y} - 1 \qquad\implies\qquad (e^y)^2 - 2x e^y - 1 = 0\]Use the quadratic formula.
\[e^y = \frac{2x \pm \sqrt{4x^2 + 4}}{2} \qquad\implies\qquad e^y = x \pm \sqrt{x^2 + 1}\]Since $x < \sqrt{x^2 + 1}$, the root $x - \sqrt{x^2 + 1}$ will result in a negative number and thus no solution for $y$.
\[e^y = x + \sqrt{x^2 + 1} \qquad\implies\qquad y = \arcsinh(x) = \ln( x + \sqrt{x^2 + 1} )\]Hyperbolic Cosine
The steps are almost identical to $\arcsinh$. I will skip a few steps to save space.
\[\begin{align} &y = \cosh(x) = \frac{e^x + e^{-x}}{2} \\[10pt] &x = \frac{e^y + e^{-y}}{2} \\[10pt] &(e^y)^2 - 2x e^y + 1 = 0 \\[10pt] &e^y = x \pm \sqrt{x^2 - 1} \\[10pt] &e^y = x + \sqrt{x^2 - 1} \\[10pt] &y = \arccosh(x) = \ln( x + \sqrt{x^2 - 1} ) \end{align}\]One thing to note is that the reason we neglect the root $e^y = x - \sqrt{x^2 - 1}$ is different than before. $x > \sqrt{x^2 - 1}$, therefore the root will be positive and produce a solution for $y$. Our choice of using the root $e^y = x + \sqrt{x^2 - 1}$ is due to our choice of domain restriction.
Hyperbolic Tangent
The methodology is very similar to the above. Again, I will skip some steps to save space.
\[\begin{align} &y = \tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \\[10pt] &x = \frac{e^y - e^{-y}}{e^y + e^{-y}} = \frac{e^{2y} - 1}{e^{2y} + 1} \\[10pt] &e^{2y} = \frac{1+x}{1-x} \\[10pt] &y = \arctanh(x) = \frac{1}{2} \ln \left ( \frac{1+x}{1-x} \right ) \end{align}\]Hyperbolic Secant
This is almost identical to the derivation of $\arccosh$.
\[\begin{align} &y = \sech(x) = \frac{2}{e^x + e^{-x}} \\[10pt] &x = \frac{2}{e^x + e^{-x}} \\[10pt] &(e^y)^2 - \frac{2}{x} \, e^y + 1 = 0 \\[10pt] &e^y = \frac{1}{x} \pm \sqrt{\frac{1}{x^2} - 1} \\[10pt] &e^y = \frac{1}{x} + \sqrt{\frac{1}{x^2} - 1} \\[10pt] &y = \arcsech(x) = \ln \left ( \frac{1}{x} + \sqrt{\frac{1}{x^2} - 1} \ \right ) \end{align}\]Hyperbolic Cosecant
This is almost identical to the derivation of $\arcsinh$.
\[\begin{align} &y = \csch(x) = \frac{2}{e^x - e^{-x}} \\[10pt] &x = \frac{2}{e^x - e^{-x}} \\[10pt] &(e^y)^2 + \frac{2}{x} \, e^y + 1 = 0 \\[10pt] &e^y = \frac{1}{x} \pm \sqrt{\frac{1}{x^2} + 1} \\[10pt] &e^y = \frac{1}{x} + \sqrt{\frac{1}{x^2} + 1} \\[10pt] &y = \arccsch(x) = \ln \left ( \frac{1}{x} + \sqrt{\frac{1}{x^2} + 1} \ \right ) \end{align}\]Hyperbolic Cotangent
This is almost identical to the derivation of $\arctanh$.
\[\begin{align} &y = \coth(x) = \frac{e^x + e^{-x}}{e^x - e^{-x}} \\[10pt] &x = \frac{e^y + e^{-y}}{e^y - e^{-y}} = \frac{e^{2y} + 1}{e^{2y} - 1} \\[10pt] &e^{2y} = \frac{x+1}{x-1} \\[10pt] &y = \arccoth(x) = \frac{1}{2} \ln \left ( \frac{x+1}{x-1} \right ) \end{align}\]