\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - Derivatives of Hyperbolic Functions

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On this page we discuss the derivative of hyperbolic functions. First, we give a quick precalculus review then we discuss the basics of the derivatives.

Precalculus Review - What Are Hyperbolic Functions?

Hyperbolic functions are functions formed from exponentials. They appear so often that they are given the special name hyperbolic and they seem to work similar to trig functions, so they are also called hyperbolic trig functions. In trigonometry we have sine, cosine, tangent, etc. Hyperbolic functions are called hyperbolic sine, hyperbolic cosine, hyperbolic tangent and the abbreviations are written \(\sinh(x), \cosh(x), \tanh(x) . . . \), with an 'h' after the same name as the trig functions. Here are the definitions.

\(\displaystyle{ \sinh(x) = \frac{e^x-e^{-x}}{2} }\)

 

\(\displaystyle{ \cosh(x) = \frac{e^x+e^{-x}}{2} }\)

\(\displaystyle{ \csch(x) = \frac{1}{\sinh(x)} }\)

 

\(\displaystyle{ \sech(x) = \frac{1}{\cosh(x)} }\)

\(\displaystyle{ \tanh(x) = \frac{\sinh(x)}{\cosh(x)} }\)

 

\(\displaystyle{ \coth(x) = \frac{\cosh(x)}{\sinh(x)} }\)

So, you can see why we call them hyperbolic trig functions. Other than \(\sinh(x)\) and \(\cosh(x)\), we define the other hyperbolic functions similar to the corresponding definitions for trig functions. Now let's look at some identities and I think the relationship will become clearer.

\(\displaystyle{ \sinh(-x) = -\sinh(x) }\)

 

\(\displaystyle{ \cosh(-x) = \cosh(x) }\)

\(\displaystyle{ \cosh^2(x) - \sinh^2(x) = 1 }\)

 

\(\displaystyle{ 1-\tanh^2(x) = \sech^2(x) }\)

Be careful here. Notice that although they are similar, they are NOT always identical, usually with a change in sign or some other subtle difference. So don't gloss over this. Take a few minutes to notice the similarities AND the differences. Before we get started with the equations, let's take a few minutes to watch this video that explains hyperbolic functions using graphs. I think this will help you get a good feel for them.

Michel vanBiezen - What is a Hyperbolic Function? [8min-48secs]

video by Michel vanBiezen

Let's start with a very good video discussing some of these equations and working with them to prove some identities. This is a great video to get you started with hyperbolic functions.

PatrickJMT - Hyperbolic Functions [10min-0secs]

video by PatrickJMT

Okay, let's take a look at another quick video working with hyperbolic functions. Using the definitions for \(\cosh(x)\) and \(\sinh(x)\), this presenter proves the identity \(\displaystyle{ \cosh^2(x) - \sinh^2(x) = 1 }\) . Although the end result is not particularly significant, this video shows how you can work with hyperbolic trig functions using the definitions.

Krista King Math - Hyperbolic Functions [5min-14secs]

video by Krista King Math

Remember, there is nothing special or magic going on here. We just define \(\sinh(x)\) and \(\cosh(x)\) using exponentials and everything else builds from there.

Derivatives of Hyperbolic Functions

Okay, since nothing special is going on, you should be able to determine the derivatives of each hyperbolic function based only on exponentials. Doing so, produces the following formulas.

derivatives

\(\displaystyle{ \frac{d}{dx}[\sinh(u)] = \cosh(u)\frac{du}{dx} }\)

\(\displaystyle{ \frac{d}{dx}[\cosh(u)] = \sinh(u)\frac{du}{dx} }\)

\(\displaystyle{ \frac{d}{dx}[\tanh(u)] = \sech^2(u)\frac{du}{dx} }\)

\(\displaystyle{ \frac{d}{dx}[\coth(u)] = -\csch^2(u)\frac{du}{dx} }\)

\(\displaystyle{ \frac{d}{dx}[\sech(u)] = }\) \(\displaystyle{ -\sech(u)\tanh(u)\frac{du}{dx} }\)

\(\displaystyle{ \frac{d}{dx}[\csch(u)] = }\) \(\displaystyle{ -\csch(u)\coth(u)\frac{du}{dx} }\)

Again, notice similarities but also differences (especially minus signs) between these equations and the corresponding trig function derivatives.

Here is a video clip explaining some of these derivatives and showing how to use the exponential definitions to calculate the derivative.

PatrickJMT - Hyperbolics [7min-54secs]

video by PatrickJMT

Okay, let's try some practice problems.

Practice

Unless otherwise instructed, calculate the derivative of these functions. Give your final answers in exact, completely factored form.

\(f(x)=\tanh(4x)\)

Problem Statement

Calculate the derivative of the function \(f(x)=\tanh(4x)\). Give your final answer in exact, completely factored form.

Solution

1092 video

video by PatrickJMT

close solution

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\(f(x)=\ln(\sinh(x))\)

Problem Statement

Calculate the derivative of the function \(f(x)=\ln(\sinh(x))\). Give your final answer in exact, completely factored form.

Solution

1093 video

video by PatrickJMT

close solution

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\(f(x)=\sinh(x)\tanh(x)\)

Problem Statement

Calculate the derivative of the function \(f(x)=\sinh(x)\tanh(x)\). Give your final answer in exact, completely factored form.

Solution

1094 video

video by PatrickJMT

close solution

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\(f(x)=\text{sech}^2(e^x)\)

Problem Statement

Calculate the derivative of the function \(f(x)=\text{sech}^2(e^x)\). Give your final answer in exact, completely factored form.

Solution

1099 video

video by Krista King Math

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\(f(x)=\sin(\sinh(x))\)

Problem Statement

Calculate the derivative of the function \(f(x)=\sin(\sinh(x))\). Give your final answer in exact, completely factored form.

Solution

1098 video

video by Krista King Math

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\(f(x)=\sin\left[ \sinh(3x^2+2x) \right]\)

Problem Statement

Calculate the derivative of the function \(f(x)=\sin\left[ \sinh(3x^2+2x) \right]\). Give your final answer in exact, completely factored form.

Solution

2304 video

video by Krista King Math

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Topics You Need To Understand For This Page

Related Topics and Links

external links you may find helpful

Wikipedia - Inverse hyperbolic functions

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Unless otherwise instructed, calculate the derivative of these functions. Give your final answers in exact, completely factored form.

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