Hyperbolic Functions and Their Derivatives

The last set of functions that we’re going to be looking in this chapter at are the hyperbolic functions. In many physical situations combinations of $e^x$ and $e^{−x}$ arise fairly often. Because of this these combinations are given names. There are six hyperbolic functions and they are defined as follows.

Notice that the hyperbolic tangent, cotangent, secant, and cosecant are defined in a similar ways as with the standard trigonometric functions. This can make it easier to remember.

From the definitions, which rely on exponential functions, we can easily determine the domains and ranges of these functions.

Below are the graphs of the hyperbolic sine, cosine, and tangent funtions.

While these six new functions have many interesting relationships with one another, here are some identities that are similar, but not the same, to some of the more common trigonometric identities.

You should be able to verify these easily with the definitions of the functions, so we leave this as an exercise.

Derivatives of Hyperbolic Functions

Because the hyperbolic functions are defined in terms of exponential functions finding their derivatives is fairly simple. Just remember to use the chain rule when taking the derivative of $e^{-x}$. We’ll do the derivative for hyperbolic sine and leave the rest to you as an exercise.

Note that there was no need to use the quotient rule as the denominator was just a constant.

The other derivatives can be found in a similar way or by using the quotient rule where appropriate. Here are all six derivatives.

We see some similarities to the derivatives of the basic trigonometric functions, but be careful as they are not the same.

Let's do some practice.

Inverse Hyperbolic Functions

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From the graphs of the hyperbolic functions, we see that all of them are one-to-one except $\cosh(x)$ and $\mbox{sech}(x)$. If we restrict the domains of these two functions to the interval $[0,\infty)$, then all the hyperbolic functions are one-to-one, and we can define the inverse hyperbolic functions. Since the hyperbolic functions themselves involve exponential functions, the inverse hyperbolic functions involve logarithmic functions. Included are the alternate notations for these functions.

Before explaining how these come about, it is important to understand the domains and ranges of these functions. For example, the definition for the inverse hyperbolic tangent and cotangent look identical, but they are defined on entirely different domains.

Let's look at how to derive the formula for the inverse hyperbolic sine function. The others follow similarly.

Suppose $y=\sinh^{-1}(x)$. Then $x=\sinh(y)$ and, by the definition of the hyperbolic sine function,

Therefore,

Multiplying this equation by $e^y$, which is always positive, we obtain

This can be solved like a quadratic equation in terms of $e^y$, with the solutions

Since $e^y>0$, the only solution possible is the one with the addition. Finally, solving for $y$, we conclude that

We leave the remaining definitions as exercises.

Derivatives of Inverse Hyperbolic Functions

We can use the definitions of the inverse hyperbolic functions to easily compute their derivatives. We will need to remember how to take the derivative of the natural logarithm function as well as using the chain rule.

Here are the derivatives of all six inverse hyperbolic functions. Remember that the domains of the inverse hyperbolic tangent and cotangent are different, even though they look identical in their definitions.

As always, we should do lots of practice problems with these new derivatives.

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.

Hyperbolic functions can be used to model catenaries. Specifically, functions of the form $y=a\cosh\left(\frac{x}{a}\right)$ are catenaries.

A chain hangs from two posts 2 m apart to form a catenary described by the equation $y=3\cosh\left(\frac{x}{3}\right)−2$. Find the slope of the catenary at the left fence post.

Show Solution

First, let's take a quick look at the graph of this function.

Since the shape of the hanging chain will be symmetric through the center, and that the posts are located 2 meters apart, we will imagine that the posts are located at the $x=-1$ and $x=1$ locations.

So to find the slope of the caternary at the left post, we simply need to compute $\dfrac{dy}{dx}\biggr\vert_{x=-1}$. Let's start by finding the derivative.

$$\frac{dy}{dx}=3\sinh\left(\frac{x}{3}\right)\cdot\frac{1}{3}=\sinh\left(\frac{x}{3}\right)$$

Now we evaluate this at $x=-1$.

$$\dfrac{dy}{dx}\biggr\vert_{x=-1}=\sinh\left(\frac{-1}{3}\right)=\frac{1-e^{2/3}}{2\sqrt[3]{e}}\approx-0.3395$$

Practice Problems

For problems 1 – 7 differentiate the given function.

1. $f\left( x \right) = \sinh \left( x \right) + 2\cosh \left( x \right) - {\mathop{\rm sech}\nolimits} \left( x \right)$
Show Solution

$$f'\left( x \right) = \cosh \left( x \right) + 2\sinh \left( x \right) + {\mathop{\rm sech}\nolimits} \left( x \right)\tanh \left( x \right)$$


2. $R\left( t \right) = \tan \left( t \right) + {t^2}{\mathop{\rm csch}\nolimits} \left( t \right)$
Show Solution


Remember to use the product rule for the second term.

$$R'\left( t \right) = {{\sec }^2}\left( t \right) + 2t{\mathop{\rm csch}\nolimits} \left( t \right) - {t^2}{\mathop{\rm csch}\nolimits} \left( t \right)\coth \left( t \right)$$


3. $\displaystyle g\left( z \right) = \frac{{z + 1}}{{\tanh \left( z \right)}}$
Show Solution


Remember to use the quotient rule.

$$g'\left( z \right) = \frac{{\tanh \left( z \right) - \left( {z + 1} \right){{{\mathop{\rm sech}\nolimits} }^2}\left( z \right)}}{{{{\tanh }^2}\left( z \right)}}$$


4. $k(w)=\cot(3w)+\coth(3w)$
Show Solution


Not that the first term is the regular cotangent function.

$$k'(w)=-3\csc^2(3w)-3\,\mbox{csch}^2(3w)$$


5. $A(x)=\,\mbox{sech}\left(\ln(x)\right)$
Show Solution


We will use the chain rule.

$$A'(x)=\frac{-\,\mbox{sech}\left(\ln(x)\right)\tanh\left(\ln(x)\right)}{x}$$


6. $P(y)=\cosh^{-1}\left(\sinh(y)\right)$
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We will use the chain rule.

$$P'(y)=\frac{1}{\sqrt{\sinh^2(y)-1}}\cdot\cosh(y)=\frac{\cosh(y)}{\sqrt{\sinh^2(y)-1}}$$


7. $v(t)=t^2\tanh^{-1}\left(e^t\right)$
Show Solution


Make sure to use the product rule as well as the chain rule.

$$\begin{align*} v'(t) &= t^2\cdot\frac{1}{1-\left(e^t\right)^2}\cdot e^t+2t\tanh^{-1}\left(e^t\right)\\ &= \frac{t^2 e^t}{1-e^{2t}}+2t\tanh^{-1}\left(e^t\right) \end{align*}$$