6.9 Calculus of the hyperbolic functions (Page 2/5)

Calculus volume 1 Page 2 / 5

Domains and ranges of the inverse hyperbolic functions
Function	Domain	Range
${sinh}^{−1} x$	$(− \infty, \infty)$	$(− \infty, \infty)$
${cosh}^{−1} x$	$(1, \infty)$	$[0, \infty)$
${tanh}^{−1} x$	$(−1, 1)$	$(− \infty, \infty)$
${coth}^{−1} x$	$(− \infty, −1) \cup (1, \infty)$	$(− \infty, 0) \cup (0, \infty)$
${sech}^{−1} x$	$(0, 1)$	$[0, \infty)$
${csch}^{−1} x$	$(− \infty, 0) \cup (0, \infty)$	$(− \infty, 0) \cup (0, \infty)$

The graphs of the inverse hyperbolic functions are shown in the following figure.

This figure has six graphs. The first graph labeled “a” is of the function y=sinh^-1(x). It is an increasing function from the 3rd quadrant, through the origin to the first quadrant. The second graph is labeled “b” and is of the function y=cosh^-1(x). It is in the first quadrant, beginning on the x-axis at 2 and increasing. The third graph labeled “c” is of the function y=tanh^-1(x). It is an increasing function from the third quadrant, through the origin, to the first quadrant. The fourth graph is labeled “d” and is of the function y=coth^-1(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis. The fifth graph is labeled “e” and is of the function y=sech^-1(x). It is a curve decreasing in the first quadrant and stopping on the x-axis at x=1. The sixth graph is labeled “f” and is of the function y=csch^-1(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis. — Graphs of the inverse hyperbolic functions.

To find the derivatives of the inverse functions, we use implicit differentiation. We have

\begin{matrix} y & = & {sinh}^{−1} x \\ sinh y & = & x \\ \frac{d}{d x} sinh y & = & \frac{d}{d x} x \\ cosh y \frac{d y}{d x} & = & 1 . \end{matrix}

Recall that ${cosh}^{2} y - {sinh}^{2} y = 1,$ so $cosh y = \sqrt{1 + {sinh}^{2} y} .$ Then,

\frac{d y}{d x} = \frac{1}{cosh y} = \frac{1}{\sqrt{1 + {sinh}^{2} y}} = \frac{1}{\sqrt{1 + x^{2}}} .

We can derive differentiation formulas for the other inverse hyperbolic functions in a similar fashion. These differentiation formulas are summarized in the following table.

Derivatives of the inverse hyperbolic functions
$f (x)$	$\frac{d}{d x} f (x)$
${sinh}^{−1} x$	$\frac{1}{\sqrt{1 + x^{2}}}$
${cosh}^{−1} x$	$\frac{1}{\sqrt{x^{2} - 1}}$
${tanh}^{−1} x$	$\frac{1}{1 - x^{2}}$
${coth}^{−1} x$	$\frac{1}{1 - x^{2}}$
${sech}^{−1} x$	$\frac{−1}{x \sqrt{1 - x^{2}}}$
${csch}^{−1} x$	$\frac{−1}{\| x \| \sqrt{1 + x^{2}}}$

Note that the derivatives of ${tanh}^{−1} x$ and ${coth}^{−1} x$ are the same. Thus, when we integrate $1 / (1 - x^{2}),$ we need to select the proper antiderivative based on the domain of the functions and the values of $x .$ Integration formulas involving the inverse hyperbolic functions are summarized as follows.

\begin{matrix} \int \frac{1}{\sqrt{1 + u^{2}}} d u & = & {sinh}^{−1} u + C & \int \frac{1}{u \sqrt{1 - u^{2}}} d u & = & − {sech}^{−1} | u | + C \\ \int \frac{1}{\sqrt{u^{2} - 1}} d u & = & {cosh}^{−1} u + C & \int \frac{1}{u \sqrt{1 + u^{2}}} d u & = & − {csch}^{−1} | u | + C \\ \int \frac{1}{1 - u^{2}} d u & = & {\begin{matrix} {tanh}^{−1} u + C if | u | < 1 \\ {coth}^{−1} u + C if | u | > 1 \end{matrix} \end{matrix}

Differentiating inverse hyperbolic functions

Evaluate the following derivatives:

$\frac{d}{d x} ({sinh}^{−1} (\frac{x}{3}))$
$\frac{d}{d x} {({tanh}^{−1} x)}^{2}$

Using the formulas in [link] and the chain rule, we obtain the following results:

$\frac{d}{d x} ({sinh}^{−1} (\frac{x}{3})) = \frac{1}{3 \sqrt{1 + \frac{x^{2}}{9}}} = \frac{1}{\sqrt{9 + x^{2}}}$
$\frac{d}{d x} {({tanh}^{−1} x)}^{2} = \frac{2 ({tanh}^{−1} x)}{1 - x^{2}}$

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Evaluate the following derivatives:

$\frac{d}{d x} ({cosh}^{−1} (3 x))$
$\frac{d}{d x} {({coth}^{−1} x)}^{3}$

$\frac{d}{d x} ({cosh}^{−1} (3 x)) = \frac{3}{\sqrt{9 x^{2} - 1}}$
$\frac{d}{d x} {({coth}^{−1} x)}^{3} = \frac{3 {({coth}^{−1} x)}^{2}}{1 - x^{2}}$

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Integrals involving inverse hyperbolic functions

Evaluate the following integrals:

$\int \frac{1}{\sqrt{4 x^{2} - 1}} d x$
$\int \frac{1}{2 x \sqrt{1 - 9 x^{2}}} d x$

We can use $u -substitution$ in both cases.

Let $u = 2 x .$ Then, $d u = 2 d x$ and we have

$\int \frac{1}{\sqrt{4 x^{2} - 1}} d x = \int \frac{1}{2 \sqrt{u^{2} - 1}} d u = \frac{1}{2} {cosh}^{−1} u + C = \frac{1}{2} {cosh}^{−1} (2 x) + C .$
Let $u = 3 x .$ Then, $d u = 3 d x$ and we obtain

$\int \frac{1}{2 x \sqrt{1 - 9 x^{2}}} d x = \frac{1}{2} \int \frac{1}{u \sqrt{1 - u^{2}}} d u = - \frac{1}{2} {sech}^{−1} | u | + C = - \frac{1}{2} {sech}^{−1} | 3 x | + C .$

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Evaluate the following integrals:

$\int \frac{1}{\sqrt{x^{2} - 4}} d x, x > 2$
$\int \frac{1}{\sqrt{1 - e^{2 x}}} d x$

$\int \frac{1}{\sqrt{x^{2} - 4}} d x = {cosh}^{−1} (\frac{x}{2}) + C$
$\int \frac{1}{\sqrt{1 - e^{2 x}}} d x = − {sech}^{−1} (e^{x}) + C$

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Applications

One physical application of hyperbolic functions involves hanging cables . If a cable of uniform density is suspended between two supports without any load other than its own weight, the cable forms a curve called a catenary . High-voltage power lines, chains hanging between two posts, and strands of a spider’s web all form catenaries. The following figure shows chains hanging from a row of posts.

An image of chains hanging between posts that all take the shape of a catenary. — Chains between these posts take the shape of a catenary. (credit: modification of work by OKFoundryCompany, Flickr)

Hyperbolic functions can be used to model catenaries. Specifically, functions of the form $y = a cosh (x / a)$ are catenaries. [link] shows the graph of $y = 2 cosh (x / 2) .$

This figure is a graph. It is of the function f(x)=2cosh(x/2). The curve decreases in the second quadrant to the y-axis. It intersects the y-axis at y=2. Then the curve becomes increasing. — A hyperbolic cosine function forms the shape of a catenary.

Using a catenary to find the length of a cable

Assume a hanging cable has the shape $10 cosh (x / 10)$ for $−15 \leq x \leq 15,$ where $x$ is measured in feet. Determine the length of the cable (in feet).

Recall from Section $6.4$ that the formula for arc length is

Arc Length = \int_{a}^{b} \sqrt{1 + {[f^{'} (x)]}^{2}} d x .

We have $f (x) = 10 cosh (x / 10),$ so $f^{'} (x) = sinh (x / 10) .$ Then

\begin{matrix} Arc Length & = \int_{a}^{b} \sqrt{1 + {[f^{'} (x)]}^{2}} d x \\ = \int_{−15}^{15} \sqrt{1 + {sinh}^{2} (\frac{x}{10})} d x . \end{matrix}

Now recall that $1 + {sinh}^{2} x = {cosh}^{2} x,$ so we have

\begin{matrix} Arc Length & = \int_{−15}^{15} \sqrt{1 + {sinh}^{2} (\frac{x}{10})} d x \\ = \int_{−15}^{15} cosh (\frac{x}{10}) d x \\ = 10 sinh {(\frac{x}{10}) |}_{−15}^{15} = 10 [sinh (\frac{3}{2}) - sinh (- \frac{3}{2})] = 20 sinh (\frac{3}{2}) \\ \approx 42.586 ft . \end{matrix}

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Questions & Answers

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Source: OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2

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