From the hyperbolic functions page, the equation for hyperbolic sine is \[\displaystyle{ y = \sinh(x) = \frac{e^x-e^{-x}}{2} }\] We can use this definition to get the equation for inverse hyperbolic sine in terms of natural logarithms. Try this on your own before looking at the solution.

So you now know that the inverse hyperbolic sine function can be written as \(\displaystyle{ \arcsinh(x) = \ln(x + \sqrt{x^2+1}) }\). Now work this next practice problem to calculate the derivative of the inverse hyperbolic sine function using this function.

Okay, in summary, here are the definition of the inverse hyperbolic sine function and it's derivative.

As is common with inverse functions, sometimes you will see the notation \(\sinh^{-1}(x)\) instead of \( \arcsinh(x)\). Remember that \(\displaystyle{ \sinh^{-1}(x) \neq \frac{1}{\sinh(x)} }\). This notation may be used by your instructor, in your textbook and in some of the practice problems on this site.

Here is a video clip proving the derivative of \(\sinh^{-1}(x)\), which shows you how to manipulate inverse hyperbolic functions.

Figure 1 - \( y = \arcsinh(x) \)
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The plot of the inverse hyperbolic sine function (thicker green line) is shown in figure 1. The thin red line in the plot is \( y = \sinh(x) \). Since \( y = \arcsinh(x) \) is the inverse function of \( y = \sinh(x) \), it is of course a reflection about the line \( y = x \), also shown in this plot. Notice that the inverse hyperbolic sine function is defined for all x, i.e. the domain is \( -\infty < x < \infty \). The range is also the set of all real numbers.

Now let's work additional practice problems involving inverse hyperbolic sine. Unless otherwise instructed, calculate the derivative of these functions.

Now let's do the same thing with inverse hyperbolic cosine.

In summary, here are the definition of the inverse hyperbolic cosine function and it's derivative.

Figure 2 - \( y = \arccosh(x) \)
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Figure 2 shows the inverse hyperbolic cosine function (thicker green line), the hyperbolic cosine function (thin red line) and \(y=x\). Since \(y=\cosh(x)\) is not one-to-one (does not pass the horizontal line test), we need to restrict the domain to be able to define an inverse function. So the domain of \( y = \arccosh(x) \) is \( x \geq 1 \). The range is \( y \geq 0 \).

Before we go on, let's compare the first two inverse hyperbolic functions.

Figure 3 - Comparing \(\arcsinh(x)\) and \(\arccosh(x)\)
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The equations for these two functions are very similar, differing only by a minus sign, so it is interesting to compare them. Notice from Figure 3, that as x increases the plots seem to approach one another. This can be confirmed by the equations. Notice that as x increases it is squared in the square root term and the constant 1 becomes negligible when compared to it. So both plots approach \(y=\ln(2x)\), which is also shown in Figure 3 (red line).

An interesting observation is that the two functions are approaching one another very quickly. Figure 3 is plotted to only about \(x=5\) and already we can barely tell the difference between the three plots.

Now let's work additional practice problems involving inverse hyperbolic cosine. Unless otherwise instructed, calculate the derivative of these functions.

First, let's derive the equations for the functions themselves.

Figure 4 - \( y = \arctanh(x) \) [green], \( y = \arccoth(x) \) [orange]
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Let's look at the equations and plots in detail, shown in Figure 4.
First, notice that the domains of both functions are not all real numbers.
For inverse hyperbolic tangent the equation is \[ \displaystyle{ \arctanh(x) = \frac{1}{2}\ln\left[ \frac{1+x}{1-x} \right] } \] The natural logarithm function requires that \(\displaystyle{ \frac{1+x}{1-x} }\) must be positive. From the equation, it looks like \( x = \pm 1 \) are important points. So let's look at three interval, \( x < -1 \), \( -1 < x < 1 \) and \( x > 1 \).
Note - We can reject \( x=1 \) and \( x=-1\) immediately. [why?] Because when \(x=1\) the denominator of \(\displaystyle{ \frac{1+x}{1-x} }\) is zero. And when \(x=-1\) the fraction is zero and \(\ln(0)\) is not defined.

Let's look at the first interval, \( x < -1 \). In this case, the denominator of \(\displaystyle{ \frac{1+x}{1-x} }\) is positive but the numerator is negative. This means that the fraction is negative, which is not allowed by the natural logarithm function. So \( x < -1 \) is not part of the domain.
The interval \( -1 < x < 1 \) looks okay since the numerator and denominator are both positive for all values in this interval.
Finally, the interval \( x > 1 \) causes problems since the numerator is positive but the denominator is negative, thus the fraction is negative. Again, not allowed for the natural logarithm. (To check these conclusions, plot \(\displaystyle{ y=\frac{1+x}{1-x} }\) and determine the x-values where the \( y > 0 \).)
So our conclusion is that the domain of \(y = \arctanh(x)\) is \( -1 < x < 1 \), which can be written \( |x| < 1 \).

We can use similar logic to get the domain of \( y = \arccoth(x) \). However, let's take a shortcut. Notice that the fraction inside the inverse hyperbolic cotangent function is just the negative of the fraction inside the inverse hyperbolic tangent function. Let's compare the fractions.

The numerators are the same, the denominators are negatives of each other, which means that the fractions are negatives of each other. So the opposite conclusions can be made about the domain. This means that the domain of \( y = \arccoth(x) \) is \(|x|>1\).

Now let's calculate the derivatives.

Notice that the equations for the derivatives look the same but, the domains transfer directly from each function, which means that the derivatives are different. Here is a summary of the results.

Let's work some practice problems containing the inverse hyperbolic tangent and cotangent.

The inverse hyperbolic secant and cosecant can be analysed in a similar manner as we did for tangent and cotangent. So we will allow you to work through the practice problems and show the results.