4.11: Hyperbolic Functions

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 \(\PageIndex{1}\): Hyperbolic Cosines and Sines

The hyperbolic cosine is the function

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

and the hyperbolic sine is the function

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

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 \(\PageIndex{2}\)

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

Proof

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

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

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 \(\cosh(x)\).

Definition \(\PageIndex{3}\): Hyperbolic Tangent and Cotangent

The other hyperbolic functions are

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

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 \(\PageIndex{1}\).

Figure \(\PageIndex{1}\): The hyperbolic functions: cosh, sinh, tanh, sech, csch, coth.

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

Theorem \(\PageIndex{4}\)

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

Proof

The proof is a straightforward computation:

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

This immediately gives two additional identities:

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

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 \(\PageIndex{2}\).

Figure \(\PageIndex{2}\): Geometric definitions of cosh, sinh, cos, sin: \(t\) is twice the shaded area in each figure.

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

Theorem \(\PageIndex{5}\)

\(\displaystyle {d\over dx}\cosh x=\sinh x\) and \(\displaystyle {d\over dx}\sinh x = \cosh x\).

Proof

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

and

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

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 \(\PageIndex{6}\)

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

Proof

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

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

and so

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

The other derivatives are left to the exercises.

Exercises \(\PageIndex{}\)

Exercise \(\PageIndex{1}\)

Show that the range of \(\sinh x\) is all real numbers. (Hint: show that if \(y=\sinh x,\) then \( x =\ln (y+\sqrt{y^2+1})\).)

Exercise \(\PageIndex{2}\)

Ex 4.11.2 Compute the following limits: \(\;\displaystyle\lim_{x\to\infty}\cosh x,\quad\lim_{x\to\infty}\sinh x,\quad\lim_{x\to\infty}\tanh x,\quad\lim_{x\to\infty}(\cosh x-\sinh x).\)

Answer

\(\infty,\quad\infty,\quad 1,\quad 0.\)

Exercise \(\PageIndex{3}\)

Show that the range of \(\tanh x\) is \((−1,1)\). What are the ranges of \(\coth, \hbox{sech},\) and \(\hbox{csch}\)? (Use the fact that they are reciprocal functions.)

Answer

Range of \(y=\coth(x)\) is \((-\infty,-1)\cup(1,\infty),\quad\) that is, \(y\not\in[-1,1].\)

Range of \(y=\hbox{sech}(x)\) is \((0,1],\quad\)that is, \(0\lt y\le 1.\)

Range of \(y=\hbox{csch}(x)\) is \(y\ne0.\)

Exercise \(\PageIndex{4}\)

Prove that for every pair of reals \(x,y\in\mathbb{R}\), \(\sinh (x+y) =\sinh x \cosh y + \cosh x \sinh y\). Obtain a similar identity for \(\sinh(x-y)\).

Answer

\(\sinh(x-y)= \sinh x \cosh y - \cosh x \sinh y\)

Exercise \(\PageIndex{5}\)

Prove that for every pair of reals \(x,y\in\mathbb{R}\), \(\cosh (x+y) =\cosh x \cosh y + \sinh x \sinh y\). Obtain a similar identity for \(\cosh(x-y)\).

Answer

\(\cosh(x-y)= \cosh x \cosh y - \sinh x \sinh y\)

Exercise \(\PageIndex{6}\)

Use exercises 4 and 5 to show that \(\sinh(2x)=2\sinh x \cosh x\) and \( \cosh(2x)=\cosh^2 x +\sinh^2 x\) for every \(x\).

Conclude also that \( (\cosh (2x) -1)/2 = \sinh ^2 x\).

Exercise \(\PageIndex{7}\)

Show that \(\dfrac{d}{dx} (\tanh x) =\hbox{sech}^2 x\).

Compute the derivatives of the remaining hyperbolic functions as well.

Answer

\(\dfrac{d}{dx} (\hbox{sech}\,x) =-\hbox{sech}\,x\tanh x;\quad\dfrac{d}{dx} (\hbox{csch}\,x) =-\hbox{csch}\,x\coth x;\quad\dfrac{d}{dx} (\coth x) =-\hbox{csch}^2\,x.\)

Exercise \(\PageIndex{8}\)

What are the domains of the six inverse hyperbolic functions?

Answer

Domain of \(\text{arccosh}\, x\quad=\quad[1,\infty)\)

Domain of \(\text{arcsinh}\, x\quad=\quad\mathbb{R}=(-\infty,\infty)\)

Domain of \(\text{arctanh}\, x\quad=\quad(-1,1)\)

Domain of \(\text{arcsech}\, x\quad=\quad(0,1]\)

Domain of \(\text{arccsch}\, x\quad=\quad x\ne 0\)

Domain of \(\text{arccoth}\, x\quad=\quad (-\infty,-1)\cup(1,\infty),\quad\) that is, \(x\not\in[-1,1].\)

Exercise \(\PageIndex{9}\)

Sketch the graphs of all six inverse hyperbolic functions.

Contributors

• David Guichard (Whitman College)

• Integrated by Justin Marshall.