## Hyperbolic and Inverse Hyperbolic Functions

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On this page we discuss the derivative of hyperbolic and inverse 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.

 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.
 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.

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

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

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

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

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

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

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

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.

Inverse Hyperbolic Functions

We will give you just a taste of what the inverse hyperbolic trig functions look like and let you pursue the links in the additional additional topics panel for more details.

Since you know from the discussion above that the hyperbolic trig functions are based on the exponential function $$e^x$$, you would be correct if you thought that the inverse hyperbolic trig functions are based on the natural logarithm function $$\ln(x)$$. Here are just a few, including basic derivatives.

 $$\displaystyle{ \arcsinh(x) = \ln(x + \sqrt{x^2+1}) }$$ $$\displaystyle{ \frac{d}{dx}[ \arcsinh(x)] = \frac{1}{\sqrt{x^2+1}} }$$ $$\displaystyle{ \arccosh(x) = \ln(x + \sqrt{x+1}\sqrt{x-1}) }$$ $$\displaystyle{ \frac{d}{dx}[ \arccosh(x)] = \frac{1}{\sqrt{x^2-1}} }$$

As is common with inverse functions, sometimes you will see the notation $$\sinh^{-1}(x)$$ instead of $$\arcsinh(x)$$. Remember that $$\displaystyle{ \sinh^{-1}(x) \neq \frac{1}{\sinh(x)} }$$. This notation may be used by your instructor, in your textbook and in some of the practice problems on this site.

 Here is a video clip proving the derivative of $$\sinh^{-1}(x)$$, which shows you how to manipulate inverse hyperbolic functions.

### Hyperbolic Derivatives Resources

Practice Problems

Instructions - - Unless otherwise instructed, calculate the derivative of the following functions. Give your final answers in exact, completely factored form.
[ Click on the practice problem to reveal/hide the solution. ]

 Level A - Basic

 Level B - Intermediate
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