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:

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

Got questions? Get instant answers now!

Evaluate the following derivatives:

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

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

Got questions? Get instant answers now!

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

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

Got questions? Get instant answers now!

Evaluate the following integrals:

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

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

Got questions? Get instant answers now!

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}

Got questions? Get instant answers now!

<< Chapter < Page Page > Chapter >>

Practice Key Terms 1

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

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

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 1' conversation and receive update notifications?

Ask

	27 AP 27 Reproductive System MCQ By OpenStax Start Quiz
	How much do you love him? By Zarina Chocolate Start Quiz
	26 Toxicology Gustafson- Biotoxins and Venoms By Brooke Delaney Start Exam
	Negotiations Conflict Management BUS403 By Charles Jumper Start Quiz
	Professional Writing MCQ By Mary Cohen Start Quiz
	23 AP 23 Digestive System Essay By OpenStax Start Flashcards
	Anatomy & Physiology By OpenStax Read Online Course
	2 AP Key Terms 02 Chemical Level of Organization By OpenStax Start Key Terms
	13 Neuroanatomy 13 The Hypothalamus By Stephen Voron Start Quiz
©flickr: Eduardo	Nursing Education NU589 Program Outcomes By Karen Gowdey Start Quiz