4.9: Hyperbolic Functions

Last updated
Save as PDF

Page ID: 525

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$ \newcommand{\dsum}{\displaystyle\sum\limits} $

$ \newcommand{\dint}{\displaystyle\int\limits} $

$ \newcommand{\dlim}{\displaystyle\lim\limits} $

$ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\id}{\mathrm{id}}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\kernel}{\mathrm{null}\,}$

$ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$

$ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$

$ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$

$ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vectorC}[1]{\textbf{#1}} $

$ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $

$ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $

$ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$\newcommand{\longvect}{\overrightarrow}$

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$\newcommand{\ket}[1]{\left| #1 \right>}$
$\newcommand{\bra}[1]{\left< #1 \right|}$
$\newcommand{\braket}[2]{\left< #1 \vphantom{#2} \right| \left. #2 \vphantom{#1} \right>}$
$\newcommand{\braopket}[3]{\left< #1 \vphantom{#2}\vphantom{#3} \right| #2 \vphantom{#1}\vphantom{#3} \left| #3 \vphantom{#1}\vphantom{#2} \right>}$
$\newcommand{\qmvec}[1]{\mathbf{\vec{#1}}}$
$\newcommand{\op}[1]{\hat{\mathbf{#1}}}$
$\newcommand{\expect}[1]{\langle #1 \rangle}$
$\newcommand{\dfn}[1]{\emph{\textbf{#1}}}$

$\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y}$ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real}{\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var}{\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left|#1\right|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$

Definition of the Hyperbolic Functions

We define the hyperbolic functions as follows:

\[ \sinh x = \dfrac{e^x - e^{-x}}{2}, \nonumber \]

\[ \cosh x = \dfrac{e^x + e^{-x}}{2}, \nonumber \]

\[ \tanh x = \dfrac{\sinh x}{\cosh x}. \nonumber \]

$hyperb8a.gif$ $hyperb8b.gif$

Properties of Hyperbolic Functions

$ \cosh^2 x - \sinh^2 x = 1 $,
$ \dfrac{d}{dx} \sinh x = \cosh x$,
$\dfrac{d}{dx} \cosh x = \sinh x$.

Proof of Property A

We find

\[\begin{align*} \cosh^2 x - \sinh^2 x &= \left( \dfrac{e^x+e^{-x}}{2} \right)^2 - \left( \dfrac{e^x-e^{-x}}{2} \right)^2 \\[4pt] &= \dfrac{e^{2x}+2+e^{-2x}}{4} - \dfrac{e^{2x}-2+e^{-2x}}{4} \\[4pt] &= \dfrac{4}{4} =1. \end{align*} \nonumber \]

$\square$

The Derivative of the Inverse Hyperbolic Trig Functions

\[ \dfrac{d}{dx} \sinh^{-1} x = \dfrac{1}{\sqrt{1+x^2}}, \nonumber \]

\[ \dfrac{d}{dx} \cosh^{-1} x = \dfrac{1}{\sqrt{x^2-1}}, \nonumber \]

\[ \dfrac{d}{dx} \tanh^{-1} x = \dfrac{d}{dx} \coth^{-1} x = \dfrac{1}{1-x^2}, \nonumber \]

\[ \dfrac{d}{dx} \text{sech}^{-1} x = \dfrac{1}{x\sqrt{1-x^2}}, \nonumber \]

\[ \dfrac{d}{dx} \text{csch}^{-1} x = \dfrac{1}{x\sqrt{1+x^2}}. \nonumber \]

Proof of the third identity

We have

\[ \tanh (\tanh^{-1} x) = x. \nonumber \]

Taking derivatives implicitly, we have

\[ \dfrac{d}{dx} \text{sech}^2 (\tanh^{-1} x = \tanh^{-1} x = 1. \nonumber \]

Dividing gives

\[ \dfrac{d}{dx} \tanh^{-1} x = \dfrac{1}{\text{sech}^2 (\tan^{-1} x)}. \nonumber \]

Since

\[ \cosh^2(x) - \sinh^2(x) = 1, \nonumber \]

dividing by $\cosh^2(x)$, we get

\[1 - \tanh^2(x) = \text{sech}^2(x) \nonumber \]

so that

\[\begin{align*} \dfrac{d}{dx} \tan^{-1} x &= \dfrac{1}{1-\tanh^2 (\tanh^{-1} x)} \\[4pt] &= \dfrac{1}{1-x^2}. \end{align*}

$\square$

For the derivative of the $\text{sech}^{-1} (x)$ click here.

Integration and Hyperbolic Functions

Now we are ready to use the arc hyperbolic functions for integration.

Example $\PageIndex{1}$

Evaluate

\[ \int \dfrac{dx}{4-x^2} \nonumber \]

Solution

\[ \int \dfrac{dx}{4-x^2} = \int \dfrac{1}{4} \int \dfrac{dx}{1-(2/3)^2}\nonumber \]

let $ u = \dfrac{x}{2}$, then $du = \dfrac{1}{2}dx$

\[ \dfrac{1}{2} \int \dfrac{du}{1-u^2}= \dfrac{1}{2}\tanh^{-1} u +C = \dfrac{1}{2} \tanh^{-1} \left(\dfrac{x}{2}\right) + C.\nonumber \]

Example $\PageIndex{2}$

Evaluate

\[ \int \dfrac{x}{1-x^4} dx. \nonumber \]

Solution

Although this is not directly a derivative of a hyperbolic trig function, we can use the substitution $ u = x^2 $ and $du = 2x\, dx$.

To change the integral to

\[\begin{align*} \dfrac{1}{2} \int \dfrac{du}{1-u^2} &= \dfrac{1}{2} \tanh^{-1} u + C \\[4pt] &= \dfrac{1}{2} \tanh^{-1} (x^2) + C. \end{align*} \]

Contributors and Attributions

Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.

Search

Text Color

Text Size

Margin Size

Font Type

Properties of Hyperbolic Functions

Proof of Property A

Proof of the third identity

Example \(\PageIndex{1}\)

Solution

Example \(\PageIndex{2}\)

Solution