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 9.6.1 The hyperbolic cosine is the function $$\cosh x ={e^x +e^{-x }\over2},$$ and the hyperbolic sine is the function $$\sinh x ={e^x -e^{-x}\over 2}.$$ $\square$

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

Lemma 9.6.2 The range of $\cosh x$ is $[1,\infty)$. $\square$

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 $\ds y^2 \geq 1$, and since $y\geq 0$, it follows that $y\geq 1$.

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

Definition 9.6.3 The other hyperbolic functions are \eqalign{\tanh x &= {\sinh x\over\cosh x}\cr \coth x &= {\cosh x\over\sinh x}\cr \sech x &= {1\over\cosh x}\cr \csch x &= {1\over\sinh x}\cr} The domain of $\coth$ and $\csch$ is $x\neq 0$ while the domain of the other hyperbolic functions is all real numbers. Graphs are shown in figure 9.6.1 $\square$

 $\cosh$ $\sinh$ $\tanh$ $\sech$ $\csch$ $\coth$
Figure 9.6.1. The hyperbolic functions.

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

Theorem 9.6.4 For all $x$ in $\R$, $\ds \cosh ^2 x -\sinh ^2 x = 1$.

Proof. The proof is a straightforward computation: $$\cosh ^2 x -\sinh ^2 x = {(e^x +e^{-x} )^2\over 4} -{(e^x -e^{-x} )^2\over 4}= {e^{2x} + 2 + e^{-2x } - e^{2x } + 2 - e^{-2x}\over 4}= {4\over 4} = 1.$$ $\qed$

This immediately gives two additional identities: $$1-\tanh^2 x =\sech^2 x\qquad\hbox{and}\qquad \coth^2 x - 1 =\csch^2 x.$$

The identity of the theorem also helps to provide a geometric motivation. Recall that the graph of $\ds x^2 -y^2 =1$ 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 $\ds y=\pm\sqrt{x^2-1}=\pm\sqrt{\cosh^2 t-1}=\pm\sinh t$. 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 9.6.2. 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 9.6.2.

Figure 9.6.2. Geometric definitions of sin, cos, sinh, cosh: $t$ 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 9.6.5 $\ds{d\over dx}\cosh x=\sinh x$ and $\ds{d\over dx}\sinh x = \cosh x$.

Proof. $\ds{d\over dx}\cosh x= {d\over dx}{e^x +e^{-x}\over 2} = {e^x- e^{-x}\over 2} =\sinh x$, and $\ds\ds{d\over dx}\sinh x = {d\over dx}{e^x -e^{-x}\over 2} = {e^x +e^{-x }\over 2} =\cosh x$. $\qed$

Of course, this immediately gives us two anti-derivatives as well.

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

Theorem 9.6.6 $\ds{d\over dx}\arcsinh x = {1\over\sqrt{1+x^2}}$.

Proof. Let $y=\arcsinh x$, so $\sinh y=x$. Then $\ds{d\over dx}\sinh y = \cosh(y)\cdot y' = 1$, and so $\ds y' ={1\over\cosh y} ={1\over\sqrt{1 +\sinh^2 y}} = {1\over\sqrt{1+x^2}}$. $\qed$

The other derivatives are left to the exercises.

## Exercises 9.6

Ex 9.6.1 Show that the range of $\sinh x$ is all real numbers. (Hint: show that if $y=\sinh x$ then $\ds x =\ln (y+\sqrt{y^2+1})$.)

Ex 9.6.2 Compute the following limits:

a. $\ds \lim_{x\to \infty } \cosh x$

b. $\ds \lim_{x\to \infty } \sinh x$

c. $\ds \lim_{x\to \infty } \tanh x$

d. $\ds \lim_{x\to \infty } (\cosh x -\sinh x)$

Ex 9.6.3 Show that the range of $\tanh x$ is $(-1,1)$. What are the ranges of $\coth$, $\sech$, and $\csch$? (Use the fact that they are reciprocal functions.)

Ex 9.6.4 Prove that for every $x,y\in\R$, $\sinh (x+y) =\sinh x \cosh y + \cosh x \sinh y$. Obtain a similar identity for $\sinh(x-y)$.

Ex 9.6.5 Prove that for every $x,y\in\R$, $\cosh (x+y) =\cosh x \cosh y + \sinh x \sinh y$. Obtain a similar identity for $\cosh(x-y)$.

Ex 9.6.6 Use exercises 4 and 5 to show that $\sinh(2x)=2\sinh x \cosh x$ and $\ds \cosh(2x)=\cosh^2 x +\sinh^2 x$ for every $x$. Conclude also that $\ds (\cosh (2x) -1)/2 = \sinh ^2 x$ and $\ds(\cosh (2x)+1)/2 = \cosh^2 x$.

Ex 9.6.7 Show that $\ds {d\over dx} (\tanh x) =\sech^2 x$. Compute the derivatives of the remaining hyperbolic functions as well.

Ex 9.6.8 What are the domains of the six inverse hyperbolic functions?

Ex 9.6.9 Sketch the graphs of all six inverse hyperbolic functions.

Ex 9.6.10 Compute $\ds \int \sqrt{x^2 + 1}\,dx$. (Hint: make the substitution $x=\sinh(u)$ and then use exercise 6.)

Ex 9.6.11 Fix $t>0$. The shaded region in the left-hand graph in figure 9.6.2 is bounded by $y=x\tanh t$, $y=0$, and $\ds x^2-y^2 =1$. Prove that twice the area of this region is $t$.