Calculus of the Hyperbolic Functions: Learn It 2

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Calculus of the Hyperbolic Functions: Learn It 2

Calculus of Inverse Hyperbolic Functions

Looking at the graphs of the hyperbolic functions, we see that with appropriate range restrictions, they all have inverses. The inverse hyperbolic functions have specific domain and range restrictions, summarized in the table below:

Domains and Ranges of the Inverse Hyperbolic Functions
Function	Domain	Range
${\text{sinh}}^{-1}x$	$(\text{−}\infty ,\infty )$	$(\text{−}\infty ,\infty )$
${\text{cosh}}^{-1}x$	$(1,\infty )$	$[0,\infty )$
${\text{tanh}}^{-1}x$	$(-1,1)$	$(\text{−}\infty ,\infty )$
${\text{coth}}^{-1}x$	$(\text{−}\infty ,-1)\cup (1,\infty )$	$(\text{−}\infty ,0)\cup (0,\infty )$
${\text{sech}}^{-1}x$	$(0\text{, 1})$	$[0,\infty )$
${\text{csch}}^{-1}x$	$(\text{−}\infty ,0)\cup (0,\infty )$	$(\text{−}\infty ,0)\cup (0,\infty )$

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. — Figure 2. Graphs of the inverse hyperbolic functions.

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

\begin{array}{ccc} 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{array}

$\begin{array}{ccc}\hfill y& =\hfill & {\text{sinh}}^{-1}x\hfill \\ \hfill \text{sinh}y& =\hfill & x\hfill \\ \hfill \frac{d}{dx}\text{sinh}y& =\hfill & \frac{d}{dx}x\hfill \\ \hfill \text{cosh}y\frac{dy}{dx}& =\hfill & 1.\hfill \end{array}$

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

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

$\frac{dy}{dx}=\frac{1}{\text{cosh}y}=\frac{1}{\sqrt{1+{\text{sinh}}^{2}y}}=\frac{1}{\sqrt{1+{x}^{2}}}.$

We can derive differentiation formulas for the other inverse hyperbolic functions in a similar fashion.

derivatives of inverse hyperbolic functions

Derivatives of the Inverse Hyperbolic Functions
$f(x)$	$\frac{d}{dx}f(x)$
${\text{sinh}}^{-1}x$	$\frac{1}{\sqrt{1+{x}^{2}}}$
${\text{cosh}}^{-1}x$	$\frac{1}{\sqrt{{x}^{2}-1}}$
${\text{tanh}}^{-1}x$	$\frac{1}{1-{x}^{2}}$
${\text{coth}}^{-1}x$	$\frac{1}{1-{x}^{2}}$
${\text{sech}}^{-1}x$	$\frac{-1}{x\sqrt{1-{x}^{2}}}$
${\text{csch}}^{-1}x$	$\frac{-1}{\|x\|\sqrt{1+{x}^{2}}}$

Note that the derivatives of ${\text{tanh}}^{-1}x$ and ${\text{coth}}^{-1}x$ are the same. Thus, when we integrate $1\text{/}(1-{x}^{2}),$ we need to select the proper antiderivative based on the domain of the functions and the values of $x.$

integral formulas for inverse hyperbolic functions

$\begin{array}{cccccccc}\hfill \displaystyle\int \frac{1}{\sqrt{1+{u}^{2}}}du& =\hfill & {\text{sinh}}^{-1}u+C\hfill & & & \hfill \displaystyle\int \frac{1}{u\sqrt{1-{u}^{2}}}du& =\hfill & \text{−}{\text{sech}}^{-1}|u|+C\hfill \\ \hfill \displaystyle\int \frac{1}{\sqrt{{u}^{2}-1}}du& =\hfill & {\text{cosh}}^{-1}u+C\hfill & & & \hfill \displaystyle\int \frac{1}{u\sqrt{1+{u}^{2}}}du& =\hfill & \text{−}{\text{csch}}^{-1}|u|+C\hfill \\ \hfill \displaystyle\int \frac{1}{1-{u}^{2}}du& =\hfill & \bigg\{\begin{array}{c}{\text{tanh}}^{-1}u+C\text{ if }|u|<1\hfill \\ {\text{coth}}^{-1}u+C\text{ if }|u|>1\hfill \end{array}\hfill & & & & & \end{array}$

Evaluate the following derivatives:

$\frac{d}{dx}({\text{sinh}}^{-1}(\frac{x}{3}))$
$\frac{d}{dx}{({\text{tanh}}^{-1}x)}^{2}$

Using the formulas in the table on derivatives of the inverse hyperbolic functions and the chain rule, we obtain the following results:

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

Evaluate the following integrals:

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

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

Watch the following video to see the worked solution to this example.

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