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.
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 |
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+CDifferentiating Hyperbolic Functions
Evaluate the following derivatives:
- ddx(sinh(x2))ddx(sinh(x2))
- ddx(coshx)2ddx(coshx)2
Solution
Using the formulas in Table 6.2 and the chain rule, we get
- ddx(sinh(x2))=cosh(x2)·2xddx(sinh(x2))=cosh(x2)·2x
- ddx(coshx)2=2coshxsinhxddx(coshx)2=2coshxsinhx
Evaluate the following derivatives:
- ddx(tanh(x2+3x))ddx(tanh(x2+3x))
- ddx(1(sinhx)2)ddx(1(sinhx)2)
Integrals Involving Hyperbolic Functions
Evaluate the following integrals:
- ∫xcosh(x2)dx∫xcosh(x2)dx
- ∫tanhxdx∫tanhxdx
Solution
We can use u-substitution in both cases.
- 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.
- 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:
- ∫sinh3xcoshxdx∫sinh3xcoshxdx
- ∫sech2(3x)dx∫sech2(3x)dx
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