# 7.3: Hyperbolic Functions

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The hyperbolic functions appear with some frequency in applications, and are quite similar in many respects to the trigonometric functions. This is a bit surprising given our initial definitions.

Definition 4.11.1: Hyperbolic Cosines and Sines

The **hyperbolic cosine** is the function

\[\cosh x ={e^x +e^{-x }\over2},\]

and the **hyperbolic sine** is the function

\[\sinh x ={e^x -e^{-x}\over 2}.\]

Notice that \(\cosh\) is even (that is, \(\cosh(-x)=\cosh(x)\)) while \(\sinh\) is odd (\(\sinh(-x)=-\sinh(x)\)), and \( \cosh x + \sinh x = e^x\). Also, for all \(x\), \(\cosh x >0\), while \(\sinh x=0\) if and only if \( e^x -e^{-x }=0\), which is true precisely when \(x=0\).

Lemma 4.11.2

The range of \(\cosh x\) is \([1,\infty)\).

Proof

Let \(y= \cosh x\). We solve for \(x\):

\[\eqalign{y&={e^x +e^{-x }\over 2}\cr 2y &= e^x + e^{-x }\cr 2ye^x &= e^{2x} + 1\cr 0 &= e^{2x}-2ye^x +1\cr e^{x} &= {2y \pm \sqrt{4y^2 -4}\over 2}\cr e^{x} &= y\pm \sqrt{y^2 -1}\cr} \]

From the last equation, we see \( y^2 \geq 1\), and since \(y\geq 0\), it follows that \(y\geq 1\).

Now suppose \(y\geq 1\), so \( y\pm \sqrt{y^2 -1}>0\). Then \( x = \ln(y\pm \sqrt{y^2 -1})\) is a real number, and \(y =\cosh x\), so \(y\) is in the range of \(\cosh(x)\).

\(\square\)

Definition 4.11.3: Hyperbolic Tangent and Cotangent

The other hyperbolic functions are

\[\eqalign{\tanh x &= {\sinh x\over\cosh x}\cr \coth x &= {\cosh x\over\sinh x}\cr \text{sech} x &= {1\over\cosh x}\cr \text{csch} x &= {1\over\sinh x}\cr} \]

The domain of \(\coth\) and \(\text{csch}\) is \(x\neq 0\) while the domain of the other hyperbolic functions is all real numbers. Graphs are shown in Figure \(\PageIndex{1}\)

Certainly the hyperbolic functions do not closely resemble the trigonometric functions graphically. But they do have analogous properties, beginning with the following identity.

Theorem 4.11.4

For all \(x\) in \(\mathbb{R}\), \( \cosh ^2 x -\sinh ^2 x = 1\).

Proof

The proof is a straightforward computation:

\[\cosh ^2 x -\sinh ^2 x = {(e^x +e^{-x} )^2\over 4} -{(e^x -e^{-x} )^2\over 4}= {e^{2x} + 2 + e^{-2x } - e^{2x } + 2 - e^{-2x}\over 4}= {4\over 4} = 1. \]

\(\square\)

This immediately gives two additional identities:

\[1-\tanh^2 x =\text{sech}^2 x\qquad\hbox{and}\qquad \coth^2 x - 1 =\text{csch}^2 x.\]

The identity of the theorem also helps to provide a geometric motivation. Recall that the graph of \( x^2 -y^2 =1\) is a hyperbola with asymptotes \(x=\pm y\) whose \(x\)-intercepts are \(\pm 1\). If \((x,y)\) is a point on the right half of the hyperbola, and if we let \(x=\cosh t\), then \( y=\pm\sqrt{x^2-1}=\pm\sqrt{\cosh^2x-1}=\pm\sinh t\). So for some suitable \(t\), \(\cosh t\) and \(\sinh t\) are the coordinates of a typical point on the hyperbola. In fact, it turns out that \(t\) is twice the area shown in the first graph of Figure \(\PageIndex{2}\). Even this is analogous to trigonometry; \(\cos t\) and \(\sin t\) are the coordinates of a typical point on the unit circle, and \(t\) is twice the area shown in the second graph of Figure \(\PageIndex{2}\).

Given the definitions of the hyperbolic functions, finding their derivatives is straightforward. Here again we see similarities to the trigonometric functions.

Theorem 4.11.5

\( {d\over dx}\cosh x=\sinh x\) and \thmrdef{thm:hyperbolic derivatives} \( {d\over dx}\sinh x = \cosh x\).

Proof

\[ {d\over dx}\cosh x= {d\over dx}{e^x +e^{-x}\over 2} = {e^x- e^{-x}\over 2} =\sinh x,\]

and

\[ {d\over dx}\sinh x = {d\over dx}{e^x -e^{-x}\over 2} = {e^x +e^{-x }\over 2} =\cosh x.\]

Since \(\cosh x > 0\), \(\sinh x\) is increasing and hence injective, so \(\sinh x\) has an inverse, \(\text{arcsinh} x\). Also, \(\sinh x > 0\) when \(x>0\), so \(\cosh x\) is injective on \([0,\infty)\) and has a (partial) inverse, \(\text{arccosh} x\). The other hyperbolic functions have inverses as well, though \(\text{arcsech} x\) is only a partial inverse. We may compute the derivatives of these functions as we have other inverse functions.

Theorem 4.11.6

\( {d\over dx}\text{arcsinh} x = {1\over\sqrt{1+x^2}}\).

Proof

Let \(y=\text{arcsinh} x\), so \(\sinh y=x\). Then

\[ {d\over dx}\sinh y = \cosh(y)\cdot y' = 1,\]

and so

\[ y' ={1\over\cosh y} ={1\over\sqrt{1 +\sinh^2 y}} = {1\over\sqrt{1+x^2}}.\]

The other derivatives are left to the exercises.