6.9 Calculus Of The Hyperbolic Functions - OpenStax

Derivatives and Integrals of the Hyperbolic Functions

Recall that the hyperbolic sine and hyperbolic cosine are defined as

sinhx=ex−e−x2andcoshx=ex+e−x2.sinhx=ex−e−x2andcoshx=ex+e−x2.

The other hyperbolic functions are then defined in terms of sinhxsinhx and coshx.coshx. The graphs of the hyperbolic functions are shown in the following figure.

Figure 6.81 Graphs of the hyperbolic functions.

It is easy to develop differentiation formulas for the hyperbolic functions. For example, looking at sinhxsinhx we have

ddx(sinhx)=ddx(ex−e−x2)=12[ddx(ex)−ddx(e−x)]=12[ex+e−x]=coshx.ddx(sinhx)=ddx(ex−e−x2)=12[ddx(ex)−ddx(e−x)]=12[ex+e−x]=coshx.

Similarly, (d/dx)coshx=sinhx.(d/dx)coshx=sinhx. We summarize the differentiation formulas for the hyperbolic functions in the following table.

f(x)f(x)	ddxf(x)ddxf(x)
sinhxsinhx	coshxcoshx
coshxcoshx	sinhxsinhx
tanhxtanhx	sech2xsech2x
cothxcothx	−csch2x−csch2x
sechxsechx	−sechxtanhx−sechxtanhx
cschxcschx	−cschxcothx−cschxcothx

Table 6.2 Derivatives of the Hyperbolic Functions

Let’s take a moment to compare the derivatives of the hyperbolic functions with the derivatives of the standard trigonometric functions. There are a lot of similarities, but differences as well. For example, the derivatives of the sine functions match: (d/dx)sinx=cosx(d/dx)sinx=cosx and (d/dx)sinhx=coshx.(d/dx)sinhx=coshx. The derivatives of the cosine functions, however, differ in sign: (d/dx)cosx=−sinx,(d/dx)cosx=−sinx, but (d/dx)coshx=sinhx.(d/dx)coshx=sinhx. As we continue our examination of the hyperbolic functions, we must be mindful of their similarities and differences to the standard trigonometric functions.

These differentiation formulas for the hyperbolic functions lead directly to the following integral formulas.

∫sinhudu=coshu+C∫csch2udu=−cothu+C∫coshudu=sinhu+C∫sechutanhudu=−sechu+C∫sech2udu=tanhu+C∫cschucothudu=−cschu+C∫sinhudu=coshu+C∫csch2udu=−cothu+C∫coshudu=sinhu+C∫sechutanhudu=−sechu+C∫sech2udu=tanhu+C∫cschucothudu=−cschu+C

Differentiating Hyperbolic Functions

Evaluate the following derivatives:

1. ddx(sinh(x2))ddx(sinh(x2))
2. ddx(coshx)2ddx(coshx)2

Solution

Using the formulas in Table 6.2 and the chain rule, we get

1. ddx(sinh(x2))=cosh(x2)·2xddx(sinh(x2))=cosh(x2)·2x
2. ddx(coshx)2=2coshxsinhxddx(coshx)2=2coshxsinhx

Evaluate the following derivatives:

1. ddx(tanh(x2+3x))ddx(tanh(x2+3x))
2. ddx(1(sinhx)2)ddx(1(sinhx)2)

Integrals Involving Hyperbolic Functions

Evaluate the following integrals:

1. ∫xcosh(x2)dx∫xcosh(x2)dx
2. ∫tanhxdx∫tanhxdx

Solution

We can use u-substitution in both cases.

1. Let u=x2.u=x2. Then, du=2xdxdu=2xdx and ∫xcosh(x2)dx=∫12coshudu=12sinhu+C=12sinh(x2)+C.∫xcosh(x2)dx=∫12coshudu=12sinhu+C=12sinh(x2)+C.
2. Let u=coshx.u=coshx. Then, du=sinhxdxdu=sinhxdx and ∫tanhxdx=∫sinhxcoshxdx=∫1udu=ln|u|+C=ln|coshx|+C.∫tanhxdx=∫sinhxcoshxdx=∫1udu=ln|u|+C=ln|coshx|+C. Note that coshx>0coshx>0 for all x,x, so we can eliminate the absolute value signs and obtain ∫tanhxdx=ln(coshx)+C.∫tanhxdx=ln(coshx)+C.

Evaluate the following integrals:

1. ∫sinh3xcoshxdx∫sinh3xcoshxdx
2. ∫sech2(3x)dx∫sech2(3x)dx