Providing a function is one to one, it is possible to find an inverse function. This module discusses inverse hyperbolic functions, which are used in advanced calculus for integration and in the solution of differential equations.
The inverse sinh function is \(\sinh^{-1}\) and is sometimes denoted
as \(\text{asinh or $\text{arcsinh.}$ }\)The latter is pronounced
“arc shine”.
The inverse cosh function is \(\cosh^{-1}\) and is sometimes denoted
by \(\text{acosh}\left(x\right)\) or \(\text{arccosh$\left(x\right)$ .}\)
The latter is pronounced “arc cosh”. Note that \(\cosh\left(x\right)\)
is not one to one. Restricting its domain to \([0,\infty)\), it is
one to one and the inverse function is defined for this restricted
function.
Finally, the inverse \(\tanh\) function is \(\tanh^{-1}\) or \(\text{atanh}\)
or \(\text{arctanh.}\) The latter is pronounced “arc than”.
The inverse hyperbolic functions may be written in terms of the natural
log function as \[\begin{align*}
\sinh^{-1}\left(x\right) & =\ln\left(x+\sqrt{x^{2}+1}\right)\\
\cosh^{-1}\left(x\right) & =\ln\left(x+\sqrt{x^{2}-1}\right),\ x\geq1\\
\tanh^{-1}\left(x\right) & =\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right),\ \left|x\right|<1\\
\text{csch$^{-1}\left(x\right)$ } & =\ln\left(\sqrt{\frac{1}{x^{2}}+1}+\frac{1}{x}\right),\ -\infty<x<\infty,\ x\neq0\\
\text{sech$^{-1}\left(x\right)$ } & =\ln\left(\sqrt{\frac{1}{x^{2}}-1}+\frac{1}{x}\right),\ 0<x\leq1\\
\text{coth$^{-1}$ }\left(x\right) & =\frac{1}{2}\ln\left(\frac{x+1}{x-1}\right),\ -\infty<x<-1\text{ or $1<x<\infty.$ }
\end{align*}\]
Derivative of inverse hyperbolic functions
The derivatives of the inverse hyperbolic functions are: \[\begin{align*}
\frac{d}{dx}\text{sinh$^{-1}\left(x\right)$ } & =\frac{1}{\sqrt{x^{2}+1}},\text{ for all real }x\\
\frac{d}{dx}\text{cosh$^{-1}$ $\left(x\right)$ } & =\frac{1}{\sqrt{x^{2}-1}},\text{for all real }x>1\\
\frac{d}{dx}\tanh^{-1}\left(x\right) & =\frac{1}{1-x^{2}},\text{ for all real }|x|<1\\
\frac{d}{dx}\text{csch$^{-1}\left(x\right)$ } & =\frac{-1}{|x|\sqrt{1+x^{2}}},\text{ for all real }x,x\neq0\\
\frac{d}{dx}\text{sech$^{-1}\left(x\right)$ } & =\frac{-1}{x\sqrt{1-x^{2}}},\text{ for all real }x\in(0,1)\\
\frac{d}{dx}\text{coth$^{-1}$ }\left(x\right) & =\frac{1}{1-x^{2}},\text{ for all real }|x|>1
\end{align*}\]
Example 1
What is the derivative of \(\sinh^{-1}\left(4x\right)\)?
Solution
Since \(\frac{d}{dx}\text{sinh$^{-1}\left(x\right)$ }=\frac{1}{\sqrt{x^{2}+1}}\),
we use the chain rule with \(u=4x.\) So \[\begin{align*}
\frac{du}{dx} & =4
\end{align*}\]
and \[\begin{align*}
\frac{d}{dx} & \sinh^{-1}\left(4x\right)=\\
\frac{d}{dx}\sinh^{-1}\left(4x\right) & =\frac{d}{du}\sinh^{-1}\left(u\right)\frac{du}{dx}\\
& =\frac{1}{\sqrt{u^{2}+1}}\cdot4\\
& =\frac{4}{\sqrt{\left(4x\right)^{2}+1}}\\
& =\frac{4}{\sqrt{16x^{2}+1}}.
\end{align*}\]
Alternative solution
Using the definition of sinh\(^{-1}\left(x\right)\) in terms of the
natural log function, \[\begin{align*}
\sinh^{-1}\left(4x\right) & =\ln\left(4x+\sqrt{\left(4x\right)^{2}+1}\right)\\
& =\ln\left(4x+\sqrt{16x^{2}+1}\right).
\end{align*}\]
Setting \(u=4x+\sqrt{16x^{2}+1}\), \[\begin{align*}
\frac{du}{dx} & =4+\frac{1}{2}\left(16x^{2}+1\right)^{-\frac{1}{2}}\left(32x\right)\\
& =4+16x\left(16x^{2}+1\right)^{-\frac{1}{2}}.
\end{align*}\]
Using the chain rule, \[\begin{align*}
\frac{d}{dx}\left(\sinh^{-1}\left(4x\right)\right) & =\frac{d}{dx}\left(\ln\left(4x+\sqrt{16x^{2}+1}\right)\right)\\
& =\frac{d}{du}\left(\ln\left(u\right)\right)\frac{du}{dx}\\
& =\frac{1}{u}\left(4+16x\left(16x^{2}+1\right)^{-\frac{1}{2}}\right)\\
& =\frac{4+16x\left(16x^{2}+1\right)^{-\frac{1}{2}}}{4x+\sqrt{16x^{2}+1}}.
\end{align*}\]
Now, rationalise the denominator by multiplying the top and bottom
by \(4x-\sqrt{16x^{2}+1}\), \[\begin{align*}
\frac{d}{dx}\left(\sinh^{-1}\left(4x\right)\right) & =\frac{4+16x\left(16x^{2}+1\right)^{-\frac{1}{2}}}{4x+\sqrt{16x^{2}+1}}\times\frac{4x-\sqrt{16x^{2}+1}}{4x-\sqrt{16x^{2}+1}}\\
& =\frac{16x-4\sqrt{16x^{2}+1}+64x^{2}\left(16x^{2}+1\right)^{-\frac{1}{2}}-16x}{16x^{2}-16x^{2}-1}\\
& =4\sqrt{16x^{2}+1}-64x^{2}\left(16x^{2}+1\right)^{-\frac{1}{2}}\\
& =4\left(\sqrt{16x^{2}+1}-\frac{16x^{2}}{\sqrt{16x^{2}+1}}\right)\\
& =4\left(\frac{16x^{2}+1}{\sqrt{16x^{2}+1}}-\frac{16x^{2}}{\sqrt{16x^{2}+1}}\right)\\
& =\frac{4}{\sqrt{16x^{2}+1}}.
\end{align*}\]
Exercises
Find the derivative, with respect to \(x,\) of \(\quad\)a) \(y=6\cosh^{-1}\left(x/3\right)\) \(\quad\)b) \(y=\frac{1}{2}\sinh^{-1}\left(2x+1\right)\).
Express \(\sinh^{-1}\left(-\frac{5}{12}\right)\) in terms
of the natural log function.
