6.9 Calculus Of The Hyperbolic Functions - OpenStax

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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.

This figure has six graphs. The first graph labeled “a” is of the function y=sinh(x). It is an increasing function from the 3rd quadrant, through the origin to the first quadrant. The second graph is labeled “b” and is of the function y=cosh(x). It decreases in the second quadrant to the intercept y=1, then becomes an increasing function. The third graph labeled “c” is of the function y=tanh(x). It is an increasing function from the third quadrant, through the origin, to the first quadrant. The fourth graph is labeled “d” and is of the function y=coth(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis. The fifth graph is labeled “e” and is of the function y=sech(x). It is a curve above the x-axis, increasing in the second quadrant, to the y-axis at y=1 and then decreases. The sixth graph is labeled “f” and is of the function y=csch(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis.
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

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