Learning Lab

Definitions

The basic hyperbolic functions are sinh and cosh and are defined as follows.

The hyperbolic sine function is

\[\begin{align*} \sinh\left(x\right) & =\frac{e^{x}-e^{-x}}{2}. \end{align*}\] It is pronounced as “shine \(x\)”.

The hyperbolic cosine function is defined as \[\begin{align*} \cosh\left(x\right) & =\frac{e^{x}+e^{-x}}{2}. \end{align*}\] It is pronounced as “cosh \(x\)”.

In addition to these we also define \[\begin{align*} \tanh\left(x\right) & =\frac{\sinh\left(x\right)}{\cosh\left(x\right)}\\ & =\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}. \end{align*}\] It is pronounced as “than \(x\)” where the “than” is pronounced as in “thank”.

Just as for the circular functions, there are reciprocal hyperbolic functions. They are:

\[\begin{align*} \text{$\text{coshec$\left(x\right)$ }$ } & =\frac{1}{\sinh\left(x\right)}\\ \text{sech$\left(x\right)$ } & =\frac{1}{\cosh\left(x\right)}\\ \coth\left(x\right) & =\frac{1}{\tanh\left(x\right)}. \end{align*}\]

Derivatives of Hyperbolic Functions

The derivatives of the hyperbolic functions may be found using their definitions.

For \(\sinh,\) let \(a\) be a constant, then: \[\begin{align*} \frac{d}{dx}\sinh\left(ax\right) & =\frac{d}{dx}\left(\frac{e^{ax}-e^{-ax}}{2}\right)\\ & =\frac{ae^{ax}+ae^{-ax}}{2}\\ & =a\cosh\left(ax\right). \end{align*}\] For cosh, \[\begin{align*} \frac{d}{dx}\cosh\left(ax\right) & =\frac{d}{dx}\left(\frac{e^{ax}+e^{-ax}}{2}\right)\\ & =\frac{ae^{ax}-ae^{-ax}}{2}\\ & =a\sinh\left(ax\right). \end{align*}\]

The quotient, product and chain rules can be applied to functions involving hyperbolic functions. For example, using the quotient rule, \[\begin{align*} \frac{d}{dx}\tanh\left(ax\right) & =\frac{d}{dx}\left(\frac{\sinh\left(ax\right)}{\cosh\left(ax\right)}\right)\\ & =\frac{d}{dx}\left(\frac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}\right)\\ & =\frac{\left(e^{ax}+e^{-ax}\right)\frac{d}{dx}\left(e^{ax}-e^{-ax}\right)-\left(e^{ax}-e^{-ax}\right)\frac{d}{dx}\left(e^{ax}+e^{-ax}\right)}{\left(e^{ax}+e^{-ax}\right)^{2}}\\ & =\frac{\left(e^{ax}+e^{-ax}\right)a\left(e^{ax}+e^{-ax}\right)-a\left(e^{ax}-e^{-ax}\right)\left(e^{ax}-e^{-ax}\right)}{\left(e^{ax}+e^{-ax}\right)^{2}}\\ & =\frac{a\left(e^{ax}+e^{-ax}\right)^{2}-a\left(e^{ax}-e^{-ax}\right)^{2}}{\left(e^{ax}+e^{-ax}\right)^{2}}\\ & =a-a\frac{\left(e^{ax}-e^{-ax}\right)^{2}}{\left(e^{ax}+e^{-ax}\right)^{2}}\\ & =a\left(1-\tanh^{2}\left(ax\right)\right)\\ & =a\text{sech$^{2}\left(ax\right).$ } \end{align*}\]

In summary, \[\begin{align*} \frac{d}{dx}\sinh\left(ax\right) & =a\cosh\left(ax\right)\\ \frac{d}{dx}\cosh\left(ax\right) & =a\sinh\left(ax\right)\\ \frac{d}{dx}\tanh\left(ax\right) & =a\text{sech$^{2}$ $\left(ax\right)$ .} \end{align*}\]

Integrals of Hyperbolic Functions

Let \(a\) be a constant. Using the definitions, \[\begin{align*} \int\sinh\left(ax\right)dx & =\int\left(\frac{e^{ax}-e^{-ax}}{2}\right)dx\\ & =\frac{\frac{1}{a}e^{ax}+\frac{1}{a}e^{-ax}}{2}+c,\quad c\in\mathbb{R}\\ & =\frac{1}{a}\cosh\left(ax\right)+c. \end{align*}\] Similarly, \[\begin{align*} \int\cosh\left(ax\right)dx & =\int\left(\frac{e^{ax}+e^{-ax}}{2}\right)dx\\ & =\frac{\frac{1}{a}e^{ax}-\frac{1}{a}e^{-ax}}{2}+c,\quad c\in\mathbb{R}\\ & =\frac{1}{a}\sinh\left(ax\right)+c. \end{align*}\] For the integral of \(\tanh,\) we use integration by substitution. Let \(u=\cosh\left(ax\right),\) then \(du/dx=a\,\sinh\left(x\right)\) and \[\begin{align*} \int\tanh\left(ax\right)dx & =\int\frac{\sinh\left(ax\right)}{\cosh\left(ax\right)}dx\\ & =\frac{1}{a}\int\frac{1}{\cosh\left(ax\right)}\frac{du}{dx}dx\\ & =\frac{1}{a}\int\frac{1}{u}du\\ & =\frac{1}{a}\ln\left(\cosh\left(ax\right)\right)+c,\quad c\in\mathbb{R}. \end{align*}\]

Summarising \[\begin{align*} \int\sinh\left(ax\right)dx & =\frac{1}{a}\cosh\left(ax\right)+c\\ \int\cosh\left(ax\right)dx & =\frac{1}{a}\sinh\left(ax\right)+c\\ \int\tanh\left(ax\right)dx & =\frac{1}{a}\ln\left(\cosh\left(ax\right)\right)+c. \end{align*}\]

Exercises

1. Find the derivative, with respect to \(x,\) of
   \(\quad\)a) \(y=6\cosh\left(x/3\right)\)
   \(\quad\)b) \(y=\frac{1}{2}\sinh\left(2x+1\right)\)

2. Evaluate:
   \(\quad\)a) \(\int\cosh\left(3x\right)dx\)
   \(\quad\)b) \(\int_{1}^{2}\frac{\cosh\left(\ln\left(t\right)\right)}{t}dt.\)