Hyperbolic Function

Hyperbolic functions are like evil twins of trigonometric functions; just as trigonometric functions relate to circles, hyperbolic functions relate to hyperbolae. In the mathematical family tree, they are closely related to exponential functions as well.

Curious about where this all comes from?

Check out the incredibly written page on the Anatomy of a Hyperbola, which dives deeper than the Mariana Trench on all the nitty-gritty workings of hyperbolae and hyperbolic functions.

Each trigonometric function’s doppelganger can be obtained by slapping a fat $\text{h}$ after the function name:

Trigonometric	Hyperbolic
$\sin$	$\sinh$
$\cos$	$\cosh$
$\tan$	$\tanh$
$\sec$	$\text{sech}$
$\csc$	$\text{csch}$
$\cot$	$\coth$

Let’s take a closer look at each of these hyperbolic functions individually.

Function definitions

sinh

The hyperbolic sine function is defined as follows:

$$\sinh x=\frac{e^x-e^{-x}}{2}$$

Why? I dunno, go ask your mother. Check out Anatomy of a Hyperbola.¹

The graph of $y=\sinh x$ looks something like this:

Note how the function is asymptotic to both $\frac{e^x}{2}$ and $-\frac{e^{-x}}{2}$. This is because as $x$ gets really big (i.e. approaches $+\infty$), $e^{-x}$ gets really small and approaches zero, leaving behind just the other term - and inversely, when $x$ approaches $-\infty$ then $e^x$ gets really small.

cosh

The hyperbolic cosine function is defined very similarly, but with a sign flipped:

$$\cosh x=\frac{e^x+e^{-x}}{2}$$

I still don’t know why; go ask your father.

The graph of $y=\cosh x$ looks a bit like this:

Note how again, the function is asymptotic to its two constituent exponential parts being summed together, for the same reason as with the [[#Function Definitions#sinh|sinh function]].

tanh

Just like how $\tan =\frac{\sin }{\cos }$:

$$\begin{array}{rl}\tanh & =\frac{\sinh }{\cosh }\\ & =\frac{e^x-e^{-x}}{2}÷\frac{e^x+e^{-x}}{2}\\ & =\frac{e^x-e^{-x}}{\cancel{2}}\times \frac{\cancel{2}}{e^x+e^{-x}}\\ & =\frac{e^x-e^{-x}}{e^x+e^{-x}}\end{array}$$

The graph of $y=\tanh x$ looks a little like this:

Note how the function is asymptotic to both $y=1$ and $y=-1$. This property makes the function very useful for interpolation in applied mathematics, which is a fancy word which means smoothly filling in the space between two numbers.

sech, csch, coth

Just as there exist secant, cosecant and cotangent functions which are the reciprocals of the sine, cosine and tangent functions respectively, there are also hyperbolic equivalents:

$$\begin{array}{rl}\text{sech}x& =\frac{1}{\cosh x}=\frac{2}{e^x+e^{-x}}\\ \text{csch}x& =\frac{1}{\sinh x}=\frac{2}{e^x-e^{-x}}\\ \coth x& =\frac{1}{\tanh x}=\frac{\cosh x}{\sinh x}=\frac{e^x+e^{-x}}{e^x-e^{-x}}\end{array}$$

Here’s how the functions look when plotted:

Linking trigonometric and hyperbolic functions

Okay, so clearly there’s some sort of romantic tension between the trig and hyp families - from the similar naming and each function having a doppelganger, to the way in which their identities look so similar. But how do we actually, properly, relate the two?

Parametric plots of cos/cosh, sin/sinh

If we were to parametrically plot $(\cos t,\sin t)$, we would get the following:

A lovely circle! :D

Now, what if we used $(\cosh t,\sinh t)$ instead?

A-ha, there’s half of our unit hyperbola!

Question

I wonder what happens when you plug in a complex value for $t$? Would that allow me to reach the other, dashed arm of the hyperbola?

Using the imaginary unit

For some reason, bringing imaginary numbers into these functions bridges the gap between the circular trigonometric functions and the hyperbolic ones. It feels like a bridge from black magic!

Why does the following work? I have absolutely no clue, and wish I understood it better. There are proofs, but they don’t sit quite right in my head yet.

cos and cosh

The hyperbolic cosine is actually equivalent to just multiplying the inside of the cosine function by the imaginary unit:

$$\cosh x=\cos ix$$

Surprisingly, this works the other way too!

$$\cos x=\cosh ix$$

Question

I wonder if there’s any way to visualise this as “remapping” the real and imaginary axes as inputs to the $\cos$ and $\cosh$ functions, and if there’s any way to link that to conic sections.

If I had a plane intersecting a cone to create a circular intersection, I could rotate that plane by a right angle for it to intersect “vertically” with the cone instead, creating a hyperbolic intersection instead. Is that right-angle rotation possibly linked in any way to this multiplication by $i$?

sin and sinh

Unfortunately, bridging the gap between circular and hyperbolic sine functions isn’t quite as pretty, but it’s still doable. We just need to multiply the outside of the function by $-i$ as well:

$$\begin{array}{rl}\sinh x& =-i\sin ix\\ \sin x& =-i\sinh ix\end{array}$$

Osborn’s rule

Osborn’s rule states that you can take “any” existing trigonometric identity, substitute $\sin$ with $\sinh$, $\cos$ with $\cosh$, as long as you negate every product of sines (i.e. ${\sin }^{2}x$ becomes $-{\sinh }^{2}x$).

Warning

Keep in mind that Osborn’s rule is somewhat of a hand-wavey rule of thumb! Be careful, as equations can be rearranged and manipulated to disguise or spawn in ${\sin }^2$ terms.

For instance, if we were to take the trigonometric identity ${\cos }^{2}x+{\sin }^{2}x=1$ (derived from Pythagoras’ theorem), we get ${\cosh }^{2}x-{\sinh }^{2}x=1$.

Important

This identity is very useful - make sure you know it!

Here it is again for good measure:

${\cosh }^{2}x-{\sinh }^{2}x=1$

Inverse hyperbolic functions

So far, we’ve assembled a whole family of functions which map $a\to b$. But what if we wish to go the other way? We need to define some inverse functions which can map $b\to a$ instead.

However, since some of the hyperbolic functions don’t have a $1:1$ mapping from input to output, we need to be a little careful and restrict the domain to avoid any “mapping conflicts”.

sinh⁻¹

todo

cosh⁻¹

todo

tanh⁻¹

todo

Calculus with hyperbolic functions

All of these hyperbolic functions are integratable and differentiable similarly to our regular ol’ trigonometric functions, so we can do calculus with them!

Differentiation

Summary of derivatives of hyperbolic functions

You do not need to know this table off by heart, but you should be able to work out the derivative on paper where necessary.

$f(x)$ Function	$f^{\prime }(x)$ Derivative	Domain (if applicable)
$\sinh x$	$\cosh x$
$\cosh x$	$\sinh x$
$\tanh x$	${\text{sech}}^{2}x$
$\text{sech}x$	$-\tanh x\text{sech}x$
$\text{csch}x$	$-\coth x\text{csch}x$
$\coth x$	$-{\text{csch}}^{2}x$
${\sinh }^{-1}x$	$\frac{1}{\sqrt{x^2+1}}$	$x\in \mathbb{R}$
${\cosh }^{-1}x$	$\frac{1}{\sqrt{x^2-1}}$	$x\in [1,\infty )$
${\tanh }^{-1}x$	$\frac{1}{1-x^2}$	$x\in (-1,1)$

sinh

Given we [[#Function Definitions#sinh|know how to express]] $\sinh x$ exponentially, let’s try differentiating that:

$$\begin{array}{rl}\frac{d}{dx}(\sinh x)& =\frac{d}{dx}\left(\frac{e^x-e^{-x}}{2}\right)\\ & =\frac{d}{dx}\left(\frac{1}{2}{e}^x-\frac{1}{2}{e}^{-x}\right)\\ & =\frac{1}{2}{e}^x+\frac{1}{2}{e}^{-x}\\ & =\frac{e^x+e^{-x}}{2}\\ & =\cosh x\end{array}$$

Woah, isn’t that pretty! :O

The derivative of $\sinh$ is $\cosh$, just as the derivative of $\sin$ is $\cos$:

$$\frac{d}{dx}\sinh x=\cosh x$$

Maybe $e$ isn’t such a scary monster after all. When working with calculus, $e$ is my friend. I like you now, $e$.

cosh

Just as with $\sinh$ [[#Differentiation#sinh|above]], let’s try differentiating $\cosh$‘s [[#Function Definitions#cosh|exponential form]] to see what we get:

$$\begin{array}{rl}\frac{d}{dx}(\cosh x)& =\frac{d}{dx}\left(\frac{e^x+e^{-x}}{2}\right)\\ & =\frac{d}{dx}\left(\frac{1}{2}{e}^x+\frac{1}{2}{e}^{-x}\right)\\ & =\frac{1}{2}{e}^x-\frac{1}{2}{e}^{-x}\\ & =\frac{e^x-e^{-x}}{2}\\ & =\sinh x\end{array}$$

Woah, we’ve come full circle! The derivative of $\cosh$ is $\sinh$, not to be confused with the derivative of $\cos$ which gets its sign flipped to $-\sin$.

$$\frac{d}{dx}\cosh x=\sinh x$$

This completes the following “differentiation cycle” between $\sinh$ and $\cosh$:

Question

$\sinh x$ and $\cosh x$ both differentiate to each other, forming a “differentiation cycle” of two elements. $\sin x\to \cos x\to -\sin x\to -\cos x$ would form a cycle of four elements, and $e^x$ would be a cycle of one element as it differentiates to itself.

Do there exist any other such cycles?

If yes, how many are there? Is it possible to have multiple cycles with the same number of elements in it?

If no, how can we prove that there can be no more such cycles that exist?

tanh

We can use the quotient rule and our newfound knowledge of the derivatives of [[#Differentiation#sinh|sinh]] and [[#Differentiation#cosh|cosh]] to find the derivative of $\tanh x$:

$$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{(g(x){)}^2}$$ $$\begin{array}{rl}\frac{d}{dx}(\tanh x)& =\frac{d}{dx}\left(\frac{\sinh x}{\cosh x}\right)\\ & =\frac{\cosh x\cosh x+\sinh x\sinh x}{(\cosh x{)}^2}\\ & =\frac{{\cosh }^{2}x-{\sinh }^{2}x}{{\cosh }^{2}x}\end{array}$$

Since we know from Osborn’s Rule that ${\cosh }^{2}x-{\sin }^{2}x=1$, we can use this identity to simplify further:

$$\begin{array}{rl}& \frac{{\cosh }^{2}x-{\sinh }^{2}x}{{\cosh }^{2}x}\\ =& \frac{1}{{\cosh }^{2}x}\\ =& {\text{sech}}^{2}x\end{array}$$

And there we have it!

$$\frac{d}{dx}(\tanh x)={\text{sech}}^{2}x$$

sech, csch, coth

The derivatives of the [[#Function Definitions#sech, csch, coth|hyperbolic secants, cosecants and cotangents]] can be worked out on the spot by rewriting them in terms of [[#Differentiation#sinh|sinh]], [[#Differentiation#cosh|cosh]] and [[#Differentiation#tanh|tanh]] and using the quotient rule:

Derivative of $\text{sech}x$

$$\begin{array}{rl}\frac{d}{dx}(\text{sech}x)& =\frac{d}{dx}\left(\frac{1}{\cosh x}\right)\\ & =\frac{-\frac{d}{dx}(\cosh x)}{{\cosh }^{2}x}\\ & =\frac{-\sinh x}{{\cosh }^{2}x}\\ & =-\left(\frac{\sinh x}{\cosh x}\right)\left(\frac{1}{\cosh x}\right)\\ & =-\tanh x\text{sech}x\end{array}$$

Derivative of $\text{csch}x$

$$\begin{array}{rl}\frac{d}{dx}(\text{csch}x)& =\frac{d}{dx}\left(\frac{1}{\sinh x}\right)\\ & =\frac{-\frac{d}{dx}(\sinh x)}{{\sinh }^{2}x}\\ & =\frac{-\cosh x}{{\sinh }^{2}x}\\ & =-\left(\frac{\cosh x}{\sinh x}\right)\left(\frac{1}{\sinh x}\right)\\ & =-\coth x\text{csch}x\end{array}$$

Derivative of $\coth x$

$$\begin{array}{rl}\frac{d}{dx}(\coth x)& =\frac{d}{dx}\left(\frac{1}{\tanh x}\right)\\ & =\frac{-\frac{d}{dx}(\tanh x)}{{\tanh }^{2}x}\\ & =\frac{-{\text{sech}}^{2}x}{{\tanh }^{2}x}\\ & =-\left(\frac{1}{{\cosh }^{2}x}\right)\left(\frac{{\cosh }^{2}x}{{\sinh }^{2}x}\right)\\ & =-\frac{1}{{\sinh }^{2}x}\\ & =-{\text{csch}}^{2}x\end{array}$$

sinh⁻¹

todo

cosh⁻¹

todo

tanh⁻¹

todo

Integration

sinh

todo

cosh

todo

tanh

todo

Footnotes

1. Originally, this snarky response was because I had not the slightest of clues as to where this definition came from, and this footnote contained a link to the BetterExplained article on hyperbolic functions. However, now there’s an awesome page touching on the topic in this very wiki! Does this imply the author of the page is now our collective mothers? Let’s not think about that! ↩