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Notes for Math 220, Calculus 2

Brenton LeMesurier

Section 2.9 Hyperbolic Functions

References.

Introduction.

The hyperbolic functions and their inverses are the last of the basic elementary functions. They arise in evaluating integrals (Chapter 3) and in solving differential equations (Chapter 4). The name comes from the fact that they are related to the hyperbola \(x^2-y^2=1\) in the same way that trigonometric functions (a.k.a. circular functions) are related to the circle \(x^2+y^2=1\text{.}\) Almost every trig. identity (except ones about periodicity) and all the derivative formulas have hyperbolic counterparts, with at most a sign change.

Subsection 2.9.1 Definitions

The first two definitions set things up so that everything else mimics trigonometry:
\begin{equation*} \cosh x = \frac{e^x + e^{-x}}{2}, \qquad \sinh x = \frac{e^x - e^{-x}}{2}. \end{equation*}
Then as you might guess, we define
\begin{align*} \tanh x = \frac{\sinh x}{\cosh x} \amp\amp \sech\,x = \frac{1}{\cosh x}\\ \csch\,x = \frac{1}{\sinh x} \amp\amp \coth x = \frac{1}{\tanh x} = \frac{\cosh x}{\sinh x} \end{align*}

Subsection 2.9.2 Identities

\begin{align*} \cosh(-x) = \cosh x \amp\amp \sinh(-x) = -\sinh x\\ \cosh^2 x - \sinh^2 x = 1 \amp\amp 1 - \tanh^2 x = \sech^2 x\\ \sinh(x+y) \amp=\amp \sinh x \cosh y + \cosh x \sinh y\\ \cosh(x+y) \amp=\amp \cosh x \cosh y + \sinh x \sinh y \end{align*}

Subsection 2.9.3 Derivatives of the Hyperbolic Functions

\begin{align*} \frac{d}{\,dx}(\cosh x) = \sinh x \amp\amp \frac{d}{\,dx}(\sinh x) = \cosh x\\ \frac{d}{\,dx}(\sech\, x) = -\sech\, x \tanh x \amp\amp \frac{d}{\,dx}(\csch\, x) = -\csch\, x \coth x\\ \frac{d}{\,dx}(\tanh x) = \sech^2 x \amp\amp \frac{d}{\,dx}(\coth x) = -\csch^2 x \end{align*}
For example, evaluate
\begin{equation*} \frac{d}{\,dx}(\cosh \sqrt{x}). \end{equation*}

Subsection 2.9.4 Domains, Ranges and Invertibility

sinh is defined everywhere and from the derivative above it is increasing, so it has an inverse. It can also be checked that it has the infinite limits
\begin{equation*} \lim_{x \to \pm\infty} \sinh x = \lim_{x \to \pm\infty} \frac{e^x - e^{-x}}{2} = \pm\infty. \end{equation*}
So its range is all of \(\reals\text{,}\) and it has inverse \(\sinh^{-1} x\) that is also defined everywhere.
cosh is also defined everywhere, is even, is increasing for positive \(x\text{,}\) goes to infinity as \(x \to \pm\infty\text{,}\) and \(\cosh(0) = 1\text{.}\) Thus its range is \([1,\infty)\text{.}\) It has no inverse on its whole domain (no even function can), but it has an inverse if the domain is restricted to \([0,\infty)\text{,}\) and that inverse \(\cosh^{-1} x\) has domain \([1,\infty)\text{,}\) range \([0,\infty)\text{.}\)
tanh is also defined everywhere because \(\cosh x\) is never zero, and is odd and increasing, with
\begin{equation*} \lim_{x \to \pm\infty} \tanh x = \lim_{x \to \pm\infty} \frac{e^x - e^{-x}}{e^x + e^{-x}} = \pm 1. \end{equation*}
Thus its range is \((-1,1)\text{,}\) and it has inverse \(\tanh^{-1} x\) with domain \((-1,1)\text{,}\) range \(\reals\text{.}\)
sech is even, defined everywhere, with range \((0,1]\text{,}\) is decreasing on \([0,\infty)\text{,}\) and its inverse is defined by restricting the domain to \([0,\infty)\text{.}\)
csch is odd, defined everywhere except at zero, decreasing, with disjointed range \((-\infty,0) \cup (0,\infty)\text{,}\) and has an inverse with the same disjointed range and domain \((-\infty,0) \cup (0,\infty)\text{.}\)
coth is odd, defined everywhere except at zero, with disjointed range \((-\infty,-1) \cup (1,\infty)\text{,}\) decreasing on each half of its domain, and has an inverse with disjointed domain \((-\infty,-1) \cup (1,\infty)\text{.}\)

Subsection 2.9.5 Derivatives of the Inverse Hyperbolic Functions

\begin{align*} \frac{d}{\,dx}(\cosh^{-1} x) = \frac{1}{\sqrt{x^2-1}} \amp\amp \frac{d}{\,dx}(\sinh^{-1} x) = \frac{1}{\sqrt{x^2+1}}\\ \frac{d}{\,dx}(\sech^{-1} x) = - \frac{1}{x\sqrt{1-x^2}} \amp\amp \frac{d}{\,dx}(\csch^{-1} x) = -\frac{1}{|x|\sqrt{x^2+1}}\\ \frac{d}{\,dx}(\tanh^{-1} x) = \frac{1}{1-x^2} \amp\amp \frac{d}{\,dx}(\coth^{-1} x) =\frac{1}{1-x^2} \end{align*}

Subsection 2.9.6 Some Useful Integrals

From the above derivative results, we get some useful indefinite integrals:
\begin{align*} \int \cosh x \,dx \amp= \sinh x + C \amp \int \sinh x \,dx \amp= \cosh x + C\\ \int \sech^2 x \,dx \amp= \tanh x + C \amp \int \csch^2 x \,dx \amp= -\coth x + C\\ \int \frac{\,dx}{\sqrt{x^2 + 1}} \amp= \sinh^{-1} x + C \amp \int \frac{\,dx}{\sqrt{x^2 - 1}} \amp= \cosh^{-1} x + C \end{align*}
The last two can be combined with \(\int \frac{\,dx}{\sqrt{1-x^2}} = \sin^{-1} x + C\) to handle many integrals involving the square roots of quadratics, as we will see in Section 3.3.
There are several more in Section 2.9 3  of OpenStax Calculus Volume 2, and in the Tables of Integrals 4  in that book, but the above are by far the ones that arise most often.

Study Guide.

Study the latter part of Calculus Volume 1, Section 1.5 5 , from Hyperbolic Functions onward, and Calculus Volume 2, Section 2.9 6 . In particular
  • note the derivative and integral formulas, and how they differ from those for their trigonometic cousins,
  • study Examples 47 to 50,
  • work Checkpoints 47 to 50,
  • and do one or several exercises from the following ranges: 385–390 and 405–409. (These are all derivatives; we will work with the integrals soon, in Chapter 3.)
openstax.org/books/calculus-volume-1/pages/1-5-exponential-and-logarithmic-functions
openstax.org/books/calculus-volume-2/pages/2-9-calculus-of-the-hyperbolic-functions
openstax.org/books/calculus-volume-2/pages/2-9-calculus-of-the-hyperbolic-functions
openstax.org/books/calculus-volume-2/pages/a-table-of-integrals
openstax.org/books/calculus-volume-1/pages/1-5-exponential-and-logarithmic-functions
openstax.org/books/calculus-volume-2/pages/2-9-calculus-of-the-hyperbolic-functions
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