4.11: Hyperbolic Functions
( \newcommand{\kernel}{\mathrm{null}\,}\)
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.
The hyperbolic cosine is the function
coshx=ex+e−x2,
and the hyperbolic sine is the function
sinhx=ex−e−x2.
Notice that cosh is even (that is, cosh(−x)=cosh(x)) while sinh is odd (sinh(−x)=−sinh(x)), and coshx+sinhx=ex. Also, for all x, coshx>0, while sinhx=0 if and only if ex−e−x=0, which is true precisely when x=0.
The range of coshx is [1,∞).
Proof
Let y=coshx. We solve for x:
y=ex+e−x22y=ex+e−x2yex=e2x+10=e2x−2yex+1ex=2y±√4y2−42ex=y±√y2−1
From the last equation, we see y2≥1, and since y≥0, it follows that y≥1.
Now suppose y≥1, so y±√y2−1>0. Then x=ln(y±√y2−1) is a real number, and y=coshx, so y is in the range of cosh(x).
The other hyperbolic functions are
tanhx=sinhxcoshxcothx=coshxsinhxsechx=1coshxcschx=1sinhx
The domain of coth and csch is x≠0 while the domain of the other hyperbolic functions is all real numbers. Graphs are shown in Figure 4.11.1.
Certainly the hyperbolic functions do not closely resemble the trigonometric functions graphically. But they do have analogous properties, beginning with the following identity.
For all x in R, cosh2x−sinh2x=1.
Proof
The proof is a straightforward computation:
cosh2x−sinh2x=(ex+e−x)24−(ex−e−x)24=e2x+2+e−2x−e2x+2−e−2x4=44=1.
This immediately gives two additional identities:
1−tanh2x=sech2xandcoth2x−1=csch2x.
The identity of the theorem also helps to provide a geometric motivation. Recall that the graph of x2−y2=1 is a hyperbola with asymptotes x=±y whose x-intercepts are ±1. If (x,y) is a point on the right half of the hyperbola, and if we let x=cosht, then y=±√x2−1=±√cosh2x−1=±sinht. So for some suitable t, cosht and sinht 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 4.11.2. Even this is analogous to trigonometry; cost and sint are the coordinates of a typical point on the unit circle, and t is twice the area shown in the second graph of Figure 4.11.2.
Given the definitions of the hyperbolic functions, finding their derivatives is straightforward. Here again we see similarities to the trigonometric functions.
ddxcoshx=sinhx and ddxsinhx=coshx.
Proof
ddxcoshx=ddxex+e−x2=ex−e−x2=sinhx,
and
ddxsinhx=ddxex−e−x2=ex+e−x2=coshx.
Since coshx>0, sinhx is increasing and hence injective, so sinhx has an inverse, arcsinhx. Also, sinhx>0 when x>0, so coshx is injective on [0,∞) and has a (partial) inverse, arccoshx. The other hyperbolic functions have inverses as well, though arcsechx is only a partial inverse. We may compute the derivatives of these functions as we have other inverse functions.
ddxarcsinhx=1√1+x2.
Proof
Let y=arcsinhx, so sinhy=x. Then
ddxsinhy=cosh(y)⋅y′=1,
and so
y′=1coshy=1√1+sinh2y=1√1+x2.
The other derivatives are left to the exercises.
Exercises 4.11.
Show that the range of sinhx is all real numbers. (Hint: show that if y=sinhx, then x=ln(y+√y2+1).)
Ex 4.11.2 Compute the following limits: limx→∞coshx,limx→∞sinhx,limx→∞tanhx,limx→∞(coshx−sinhx).
- Answer
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∞,∞,1,0.
Show that the range of tanhx is (−1,1). What are the ranges of coth,sech, and csch? (Use the fact that they are reciprocal functions.)
- Answer
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Range of y=coth(x) is (−∞,−1)∪(1,∞), that is, y∉[−1,1].
Range of y=sech(x) is (0,1],that is, 0<y≤1.
Range of y=csch(x) is y≠0.
Prove that for every pair of reals x,y∈R, sinh(x+y)=sinhxcoshy+coshxsinhy. Obtain a similar identity for sinh(x−y).
- Answer
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sinh(x−y)=sinhxcoshy−coshxsinhy
Prove that for every pair of reals x,y∈R, cosh(x+y)=coshxcoshy+sinhxsinhy. Obtain a similar identity for cosh(x−y).
- Answer
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cosh(x−y)=coshxcoshy−sinhxsinhy
Use exercises 4 and 5 to show that sinh(2x)=2sinhxcoshx and cosh(2x)=cosh2x+sinh2x for every x.
Conclude also that (cosh(2x)−1)/2=sinh2x.
Show that ddx(tanhx)=sech2x.
Compute the derivatives of the remaining hyperbolic functions as well.
- Answer
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ddx(sechx)=−sechxtanhx;ddx(cschx)=−cschxcothx;ddx(cothx)=−csch2x.
What are the domains of the six inverse hyperbolic functions?
- Answer
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Domain of arccoshx=[1,∞)
Domain of arcsinhx=R=(−∞,∞)
Domain of arctanhx=(−1,1)
Domain of arcsechx=(0,1]
Domain of arccschx=x≠0
Domain of arccothx=(−∞,−1)∪(1,∞), that is, x∉[−1,1].
Sketch the graphs of all six inverse hyperbolic functions.
Contributors
Integrated by Justin Marshall.