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Study Guides > Calculus Volume 1

Calculus of the Hyperbolic Functions

Learning Objectives

  • Apply the formulas for derivatives and integrals of the hyperbolic functions.
  • Apply the formulas for the derivatives of the inverse hyperbolic functions and their associated integrals.
  • Describe the common applied conditions of a catenary curve.

We were introduced to hyperbolic functions in Introduction to Functions and Graphs, along with some of their basic properties. In this section, we look at differentiation and integration formulas for the hyperbolic functions and their inverses.

Derivatives and Integrals of the Hyperbolic Functions

Recall that the hyperbolic sine and hyperbolic cosine are defined as

sinhx=ex−e−x2 and coshx=ex+e−x2.\text{sinh}x=\frac{{e}^{x}-{e}^{\text{−}x}}{2}\text{ and }\text{cosh}x=\frac{{e}^{x}+{e}^{\text{−}x}}{2}.

The other hyperbolic functions are then defined in terms of sinhx\text{sinh}x and coshx.\text{cosh}x. 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 1. Graphs of the hyperbolic functions.

It is easy to develop differentiation formulas for the hyperbolic functions. For example, looking at sinhx\text{sinh}x we have

ddx(sinhx)=ddx(ex−e−x2)=12[ddx(ex)−ddx(e−x)]=12[ex+e−x]=coshx.\begin{array}{cc}\hfill \frac{d}{dx}(\text{sinh}x)& =\frac{d}{dx}(\frac{{e}^{x}-{e}^{\text{−}x}}{2})\hfill \\ & =\frac{1}{2}\left[\frac{d}{dx}({e}^{x})-\frac{d}{dx}({e}^{\text{−}x})\right]\hfill \\ & =\frac{1}{2}\left[{e}^{x}+{e}^{\text{−}x}\right]=\text{cosh}x.\hfill \end{array}

Similarly, (d/dx)coshx=sinhx.(d\text{/}dx)\text{cosh}x=\text{sinh}x. We summarize the differentiation formulas for the hyperbolic functions in the following table.

Derivatives of the Hyperbolic Functions
f(x)f(x) ddxf(x)\frac{d}{dx}f(x)
sinhx\text{sinh}x coshx\text{cosh}x
coshx\text{cosh}x sinhx\text{sinh}x
tanhx\text{tanh}x sech2x{\text{sech}}^{2}x
cothx\text{coth}x −csch2x\text{−}{\text{csch}}^{2}x
sechx\text{sech}x −sechxtanhx\text{−}\text{sech}x\text{tanh}x
cschx\text{csch}x −cschxcothx\text{−}\text{csch}x\text{coth}x

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)sin⁡x=cos⁡x(d\text{/}dx) \sin x= \cos x and (d/dx)sinhx=coshx.(d\text{/}dx)\text{sinh}x=\text{cosh}x. The derivatives of the cosine functions, however, differ in sign: (d/dx)cos⁡x=−sin⁡x,(d\text{/}dx) \cos x=\text{−} \sin x, but (d/dx)coshx=sinhx.(d\text{/}dx)\text{cosh}x=\text{sinh}x. 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\begin{array}{cccccccc}\hfill \int \text{sinh}udu& =\hfill & \text{cosh}u+C\hfill & & & \hfill \int {\text{csch}}^{2}udu& =\hfill & \text{−}\text{coth}u+C\hfill \\ \hfill \int \text{cosh}udu& =\hfill & \text{sinh}u+C\hfill & & & \hfill \int \text{sech}u\text{tanh}udu& =\hfill & \text{−}\text{sech}u+C\hfill \\ \hfill \int {\text{sech}}^{2}udu& =\hfill & \text{tanh}u+C\hfill & & & \hfill \int \text{csch}u\text{coth}udu& =\hfill & \text{−}\text{csch}u+C\hfill \end{array}

Differentiating Hyperbolic Functions

Evaluate the following derivatives:

  1. ddx(sinh(x2))\frac{d}{dx}(\text{sinh}({x}^{2}))
  2. ddx(coshx)2\frac{d}{dx}{(\text{cosh}x)}^{2}

Answer:

Using the formulas in (Figure) and the chain rule, we get

  1. ddx(sinh(x2))=cosh(x2)⋅2x\frac{d}{dx}(\text{sinh}({x}^{2}))=\text{cosh}({x}^{2})·2x
  2. ddx(coshx)2=2coshxsinhx\frac{d}{dx}{(\text{cosh}x)}^{2}=2\text{cosh}x\text{sinh}x

Evaluate the following derivatives:

  1. ddx(tanh(x2+3x))\frac{d}{dx}(\text{tanh}({x}^{2}+3x))
  2. ddx(1(sinhx)2)\frac{d}{dx}(\frac{1}{{(\text{sinh}x)}^{2}})

Answer:

  1. ddx(tanh(x2+3x))=(sech2(x2+3x))(2x+3)\frac{d}{dx}(\text{tanh}({x}^{2}+3x))=({\text{sech}}^{2}({x}^{2}+3x))(2x+3)
  2. ddx(1(sinhx)2)=ddx(sinhx)−2=−2(sinhx)−3coshx\frac{d}{dx}(\frac{1}{{(\text{sinh}x)}^{2}})=\frac{d}{dx}{(\text{sinh}x)}^{-2}=-2{(\text{sinh}x)}^{-3}\text{cosh}x

Hint

Use the formulas in (Figure) and apply the chain rule as necessary.

Integrals Involving Hyperbolic Functions

Evaluate the following integrals:

  1. ∫xcosh(x2)dx\int x\text{cosh}({x}^{2})dx
  2. ∫tanhxdx\int \text{tanh}xdx

Answer:

We can use uu-substitution in both cases.

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

Evaluate the following integrals:

  1. ∫sinh3xcoshxdx\int {\text{sinh}}^{3}x\text{cosh}xdx
  2. ∫sech2(3x)dx\int {\text{sech}}^{2}(3x)dx

Answer:

  1. ∫sinh3xcoshxdx=sinh4x4+C\int {\text{sinh}}^{3}x\text{cosh}xdx=\frac{{\text{sinh}}^{4}x}{4}+C
  2. ∫sech2(3x)dx=tanh(3x)3+C\int {\text{sech}}^{2}(3x)dx=\frac{\text{tanh}(3x)}{3}+C

Hint

Use the formulas above and apply uu-substitution as necessary.

Calculus of Inverse Hyperbolic Functions

Looking at the graphs of the hyperbolic functions, we see that with appropriate range restrictions, they all have inverses. Most of the necessary range restrictions can be discerned by close examination of the graphs. The domains and ranges of the inverse hyperbolic functions are summarized in the following table.
Domains and Ranges of the Inverse Hyperbolic Functions
Function Domain Range
sinh−1x{\text{sinh}}^{-1}x (−∞,∞)(\text{−}\infty ,\infty ) (−∞,∞)(\text{−}\infty ,\infty )
cosh−1x{\text{cosh}}^{-1}x (1,∞)(1,\infty ) [0,∞)[0,\infty )
tanh−1x{\text{tanh}}^{-1}x (−1,1)(-1,1) (−∞,∞)(\text{−}\infty ,\infty )
coth−1x{\text{coth}}^{-1}x (−∞,−1)∪(1,∞)(\text{−}\infty ,-1)\cup (1,\infty ) (−∞,0)∪(0,∞)(\text{−}\infty ,0)\cup (0,\infty )
sech−1x{\text{sech}}^{-1}x (0, 1)(0\text{, 1}) [0,∞)[0,\infty )
csch−1x{\text{csch}}^{-1}x (−∞,0)∪(0,∞)(\text{−}\infty ,0)\cup (0,\infty ) (−∞,0)∪(0,∞)(\text{−}\infty ,0)\cup (0,\infty )

The graphs of the inverse hyperbolic functions are shown in the following figure.

This figure has six graphs. The first graph labeled “a” is of the function y=sinh^-1(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^-1(x). It is in the first quadrant, beginning on the x-axis at 2 and increasing. The third graph labeled “c” is of the function y=tanh^-1(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^-1(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^-1(x). It is a curve decreasing in the first quadrant and stopping on the x-axis at x=1. The sixth graph is labeled “f” and is of the function y=csch^-1(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis.

To find the derivatives of the inverse functions, we use implicit differentiation. We have

y=sinh−1xsinhy=xddxsinhy=ddxxcoshydydx=1.\begin{array}{ccc}\hfill y& =\hfill & {\text{sinh}}^{-1}x\hfill \\ \hfill \text{sinh}y& =\hfill & x\hfill \\ \hfill \frac{d}{dx}\text{sinh}y& =\hfill & \frac{d}{dx}x\hfill \\ \hfill \text{cosh}y\frac{dy}{dx}& =\hfill & 1.\hfill \end{array}

Recall that cosh2y−sinh2y=1,{\text{cosh}}^{2}y-{\text{sinh}}^{2}y=1, so coshy=1+sinh2y.\text{cosh}y=\sqrt{1+{\text{sinh}}^{2}y}. Then,

dydx=1coshy=11+sinh2y=11+x2.\frac{dy}{dx}=\frac{1}{\text{cosh}y}=\frac{1}{\sqrt{1+{\text{sinh}}^{2}y}}=\frac{1}{\sqrt{1+{x}^{2}}}.

We can derive differentiation formulas for the other inverse hyperbolic functions in a similar fashion. These differentiation formulas are summarized in the following table.

Derivatives of the Inverse Hyperbolic Functions
f(x)f(x) ddxf(x)\frac{d}{dx}f(x)
sinh−1x{\text{sinh}}^{-1}x 11+x2\frac{1}{\sqrt{1+{x}^{2}}}
cosh−1x{\text{cosh}}^{-1}x 1x2−1\frac{1}{\sqrt{{x}^{2}-1}}
tanh−1x{\text{tanh}}^{-1}x 11−x2\frac{1}{1-{x}^{2}}
coth−1x{\text{coth}}^{-1}x 11−x2\frac{1}{1-{x}^{2}}
sech−1x{\text{sech}}^{-1}x −1x1−x2\frac{-1}{x\sqrt{1-{x}^{2}}}
csch−1x{\text{csch}}^{-1}x −1∣x∣1+x2\frac{-1}{|x|\sqrt{1+{x}^{2}}}

Note that the derivatives of tanh−1x{\text{tanh}}^{-1}x and coth−1x{\text{coth}}^{-1}x are the same. Thus, when we integrate 1/(1−x2),1\text{/}(1-{x}^{2}), we need to select the proper antiderivative based on the domain of the functions and the values of x.x. Integration formulas involving the inverse hyperbolic functions are summarized as follows.

∫11+u2du=sinh−1u+C∫1u1−u2du=−sech−1∣u∣+C∫1u2−1du=cosh−1u+C∫1u1+u2du=−csch−1∣u∣+C∫11−u2du={tanh−1u+C if ∣u∣<1coth−1u+C if ∣u∣>1\begin{array}{cccccccc}\hfill \int \frac{1}{\sqrt{1+{u}^{2}}}du& =\hfill & {\text{sinh}}^{-1}u+C\hfill & & & \hfill \int \frac{1}{u\sqrt{1-{u}^{2}}}du& =\hfill & \text{−}{\text{sech}}^{-1}|u|+C\hfill \\ \hfill \int \frac{1}{\sqrt{{u}^{2}-1}}du& =\hfill & {\text{cosh}}^{-1}u+C\hfill & & & \hfill \int \frac{1}{u\sqrt{1+{u}^{2}}}du& =\hfill & \text{−}{\text{csch}}^{-1}|u|+C\hfill \\ \hfill \int \frac{1}{1-{u}^{2}}du& =\hfill & \bigg\{\begin{array}{c}{\text{tanh}}^{-1}u+C\text{ if }|u|<1\hfill \\ {\text{coth}}^{-1}u+C\text{ if }|u|>1\hfill \end{array}\hfill & & & & & \end{array}

Differentiating Inverse Hyperbolic Functions

Evaluate the following derivatives:

  1. ddx(sinh−1(x3))\frac{d}{dx}({\text{sinh}}^{-1}(\frac{x}{3}))
  2. ddx(tanh−1x)2\frac{d}{dx}{({\text{tanh}}^{-1}x)}^{2}

Answer:

Using the formulas in (Figure) and the chain rule, we obtain the following results:

  1. ddx(sinh−1(x3))=131+x29=19+x2\frac{d}{dx}({\text{sinh}}^{-1}(\frac{x}{3}))=\frac{1}{3\sqrt{1+\frac{{x}^{2}}{9}}}=\frac{1}{\sqrt{9+{x}^{2}}}
  2. ddx(tanh−1x)2=2(tanh−1x)1−x2\frac{d}{dx}{({\text{tanh}}^{-1}x)}^{2}=\frac{2({\text{tanh}}^{-1}x)}{1-{x}^{2}}

Evaluate the following derivatives:

  1. ddx(cosh−1(3x))\frac{d}{dx}({\text{cosh}}^{-1}(3x))
  2. ddx(coth−1x)3\frac{d}{dx}{({\text{coth}}^{-1}x)}^{3}

Answer:

  1. ddx(cosh−1(3x))=39x2−1\frac{d}{dx}({\text{cosh}}^{-1}(3x))=\frac{3}{\sqrt{9{x}^{2}-1}}
  2. ddx(coth−1x)3=3(coth−1x)21−x2\frac{d}{dx}{({\text{coth}}^{-1}x)}^{3}=\frac{3{({\text{coth}}^{-1}x)}^{2}}{1-{x}^{2}}

Hint

Use the formulas in (Figure) and apply the chain rule as necessary.

Integrals Involving Inverse Hyperbolic Functions

Evaluate the following integrals:

  1. ∫14x2−1dx\int \frac{1}{\sqrt{4{x}^{2}-1}}dx
  2. ∫12x1−9x2dx\int \frac{1}{2x\sqrt{1-9{x}^{2}}}dx

Answer:

We can use u-substitutionu\text{-substitution} in both cases.

  1. Let u=2x.u=2x. Then, du=2dxdu=2dx and we have
    ∫14x2−1dx=∫12u2−1du=12cosh−1u+C=12cosh−1(2x)+C.\int \frac{1}{\sqrt{4{x}^{2}-1}}dx=\int \frac{1}{2\sqrt{{u}^{2}-1}}du=\frac{1}{2}{\text{cosh}}^{-1}u+C=\frac{1}{2}{\text{cosh}}^{-1}(2x)+C.
  2. Let u=3x.u=3x. Then, du=3dxdu=3dx and we obtain
    ∫12x1−9x2dx=12∫1u1−u2du=−12sech−1∣u∣+C=−12sech−1∣3x∣+C.\int \frac{1}{2x\sqrt{1-9{x}^{2}}}dx=\frac{1}{2}\int \frac{1}{u\sqrt{1-{u}^{2}}}du=-\frac{1}{2}{\text{sech}}^{-1}|u|+C=-\frac{1}{2}{\text{sech}}^{-1}|3x|+C.

Evaluate the following integrals:

  1. ∫1x2−4dx,x>2\int \frac{1}{\sqrt{{x}^{2}-4}}dx,\text{}x>2
  2. ∫11−e2xdx\int \frac{1}{\sqrt{1-{e}^{2x}}}dx

Answer:

  1. ∫1x2−4dx=cosh−1(x2)+C\int \frac{1}{\sqrt{{x}^{2}-4}}dx={\text{cosh}}^{-1}(\frac{x}{2})+C
  2. ∫11−e2xdx=−sech−1(ex)+C\int \frac{1}{\sqrt{1-{e}^{2x}}}dx=\text{−}{\text{sech}}^{-1}({e}^{x})+C

Hint

Use the formulas above and apply u-substitutionu\text{-substitution} as necessary.

Applications

One physical application of hyperbolic functions involves hanging cables. If a cable of uniform density is suspended between two supports without any load other than its own weight, the cable forms a curve called a catenary. High-voltage power lines, chains hanging between two posts, and strands of a spider’s web all form catenaries. The following figure shows chains hanging from a row of posts.

An image of chains hanging between posts that all take the shape of a catenary. Figure 3. Chains between these posts take the shape of a catenary. (credit: modification of work by OKFoundryCompany, Flickr)

Hyperbolic functions can be used to model catenaries. Specifically, functions of the form y=acosh(x/a)y=a\text{cosh}(x\text{/}a) are catenaries. (Figure) shows the graph of y=2cosh(x/2).y=2\text{cosh}(x\text{/}2).

This figure is a graph. It is of the function f(x)=2cosh(x/2). The curve decreases in the second quadrant to the y-axis. It intersects the y-axis at y=2. Then the curve becomes increasing. Figure 4. A hyperbolic cosine function forms the shape of a catenary.

Using a Catenary to Find the Length of a Cable

Assume a hanging cable has the shape 10cosh(x/10)10\text{cosh}(x\text{/}10) for −15≤x≤15,-15\le x\le 15, where xx is measured in feet. Determine the length of the cable (in feet).

Answer:

Recall from Section 6.4 that the formula for arc length is

Arc Length=∫ab1+[f′(x)]2dx.\text{Arc Length}={\int }_{a}^{b}\sqrt{1+{\left[{f}^{\prime }(x)\right]}^{2}}dx.

We have f(x)=10cosh(x/10),f(x)=10\text{cosh}(x\text{/}10), so f′(x)=sinh(x/10).{f}^{\prime }(x)=\text{sinh}(x\text{/}10). Then

Arc Length=∫ab1+[f′(x)]2dx=∫−15151+sinh2(x10)dx.\begin{array}{cc}\hfill \text{Arc Length}& ={\int }_{a}^{b}\sqrt{1+{\left[{f}^{\prime }(x)\right]}^{2}}dx\hfill \\ & ={\int }_{-15}^{15}\sqrt{1+{\text{sinh}}^{2}(\frac{x}{10})}dx.\hfill \end{array}

Now recall that 1+sinh2x=cosh2x,1+{\text{sinh}}^{2}x={\text{cosh}}^{2}x, so we have

Arc Length=∫−15151+sinh2(x10)dx=∫−1515cosh(x10)dx=10sinh(x10)∣−1515=10[sinh(32)−sinh(−32)]=20sinh(32)≈42.586ft.\begin{array}{cc}\hfill \text{Arc Length}& ={\int }_{-15}^{15}\sqrt{1+{\text{sinh}}^{2}(\frac{x}{10})}dx\hfill \\ & ={\int }_{-15}^{15}\text{cosh}(\frac{x}{10})dx\hfill \\ & =10\text{sinh}{(\frac{x}{10})|}_{-15}^{15}=10\left[\text{sinh}(\frac{3}{2})-\text{sinh}(-\frac{3}{2})\right]=20\text{sinh}(\frac{3}{2})\hfill \\ & \approx 42.586\text{ft}\text{.}\hfill \end{array}

Assume a hanging cable has the shape 15cosh(x/15)15\text{cosh}(x\text{/}15) for −20≤x≤20.-20\le x\le 20. Determine the length of the cable (in feet).

Answer: 52.95ft52.95\text{ft}

Hint

Use the procedure from the previous example.

Key Concepts

  • Hyperbolic functions are defined in terms of exponential functions.
  • Term-by-term differentiation yields differentiation formulas for the hyperbolic functions. These differentiation formulas give rise, in turn, to integration formulas.
  • With appropriate range restrictions, the hyperbolic functions all have inverses.
  • Implicit differentiation yields differentiation formulas for the inverse hyperbolic functions, which in turn give rise to integration formulas.
  • The most common physical applications of hyperbolic functions are calculations involving catenaries.

1. [T] Find expressions for coshx+sinhx\text{cosh}x+\text{sinh}x and coshx−sinhx.\text{cosh}x-\text{sinh}x. Use a calculator to graph these functions and ensure your expression is correct.

Answer:

ex and e−x{e}^{x}\text{ and }{e}^{\text{−}x}

2. From the definitions of cosh(x)\text{cosh}(x) and sinh(x),\text{sinh}(x), find their antiderivatives.

3. Show that cosh(x)\text{cosh}(x) and sinh(x)\text{sinh}(x) satisfy y″=y.y\text{″}=y.

Answer:

Answers may vary

4. Use the quotient rule to verify that tanh(x)′=sech2(x).\text{tanh}(x)\prime ={\text{sech}}^{2}(x).

5. Derive cosh2(x)+sinh2(x)=cosh(2x){\text{cosh}}^{2}(x)+{\text{sinh}}^{2}(x)=\text{cosh}(2x) from the definition.

Answer:

Answers may vary

6. Take the derivative of the previous expression to find an expression for sinh(2x).\text{sinh}(2x).

7. Prove sinh(x+y)=sinh(x)cosh(y)+cosh(x)sinh(y)\text{sinh}(x+y)=\text{sinh}(x)\text{cosh}(y)+\text{cosh}(x)\text{sinh}(y) by changing the expression to exponentials.

Answer:

Answers may vary

8. Take the derivative of the previous expression to find an expression for cosh(x+y).\text{cosh}(x+y).

For the following exercises, find the derivatives of the given functions and graph along with the function to ensure your answer is correct.

9. [T]cosh(3x+1)\text{cosh}(3x+1)

Answer:

3sinh(3x+1)3\text{sinh}(3x+1)

10. [T]sinh(x2)\text{sinh}({x}^{2})

11. [T]1cosh(x)\frac{1}{\text{cosh}(x)}

Answer:

−tanh(x)sech(x)\text{−}\text{tanh}(x)\text{sech}(x)

12. [T]sinh(ln(x))\text{sinh}(\text{ln}(x))

13. [T]cosh2(x)+sinh2(x){\text{cosh}}^{2}(x)+{\text{sinh}}^{2}(x)

Answer:

4cosh(x)sinh(x)4\text{cosh}(x)\text{sinh}(x)

14. [T]cosh2(x)−sinh2(x){\text{cosh}}^{2}(x)-{\text{sinh}}^{2}(x)

15. [T]tanh(x2+1)\text{tanh}(\sqrt{{x}^{2}+1})

Answer:

xsech2(x2+1)x2+1\frac{x{\text{sech}}^{2}(\sqrt{{x}^{2}+1})}{\sqrt{{x}^{2}+1}}

16. [T]1+tanh(x)1−tanh(x)\frac{1+\text{tanh}(x)}{1-\text{tanh}(x)}

17. [T]sinh6(x){\text{sinh}}^{6}(x)

Answer:

6sinh5(x)cosh(x)6{\text{sinh}}^{5}(x)\text{cosh}(x)

18. [T]ln(sech(x)+tanh(x))\text{ln}(\text{sech}(x)+\text{tanh}(x))

For the following exercises, find the antiderivatives for the given functions.

19. cosh(2x+1)\text{cosh}(2x+1)

Answer:

12sinh(2x+1)+C\frac{1}{2}\text{sinh}(2x+1)+C

20. tanh(3x+2)\text{tanh}(3x+2)

21. xcosh(x2)x\text{cosh}({x}^{2})

Answer:

12sinh2(x2)+C\frac{1}{2}{\text{sinh}}^{2}({x}^{2})+C

23. 3x3tanh(x4)3{x}^{3}\text{tanh}({x}^{4})

24. cosh2(x)sinh(x){\text{cosh}}^{2}(x)\text{sinh}(x)

Answer:

13cosh3(x)+C\frac{1}{3}{\text{cosh}}^{3}(x)+C

25. tanh2(x)sech2(x){\text{tanh}}^{2}(x){\text{sech}}^{2}(x)

26. sinh(x)1+cosh(x)\frac{\text{sinh}(x)}{1+\text{cosh}(x)}

Answer: ln(1+cosh(x))+C\text{ln}(1+\text{cosh}(x))+C

27. coth(x)\text{coth}(x)

28. cosh(x)+sinh(x)\text{cosh}(x)+\text{sinh}(x)

Answer:

cosh(x)+sinh(x)+C\text{cosh}(x)+\text{sinh}(x)+C

29. (cosh(x)+sinh(x))n{(\text{cosh}(x)+\text{sinh}(x))}^{n}

For the following exercises, find the derivatives for the functions.

30. tanh−1(4x){\text{tanh}}^{-1}(4x)

Answer:

41−16x2\frac{4}{1-16{x}^{2}}

31. sinh−1(x2){\text{sinh}}^{-1}({x}^{2})

32. sinh−1(cosh(x)){\text{sinh}}^{-1}(\text{cosh}(x))

Answer:

sinh(x)cosh2(x)+1\frac{\text{sinh}(x)}{\sqrt{{\text{cosh}}^{2}(x)+1}}

33. cosh−1(x3){\text{cosh}}^{-1}({x}^{3})

34. tanh−1(cos⁡(x)){\text{tanh}}^{-1}( \cos (x))

Answer:

−csc⁡(x)\text{−} \csc (x)

35. esinh−1(x){e}^{{\text{sinh}}^{-1}(x)}

36. ln(tanh−1(x))\text{ln}({\text{tanh}}^{-1}(x))

Answer: −1(x2−1)tanh−1(x)-\frac{1}{({x}^{2}-1){\text{tanh}}^{-1}(x)}

For the following exercises, find the antiderivatives for the functions.

37. ∫dx4−x2\int \frac{dx}{4-{x}^{2}}

38. ∫dxa2−x2\int \frac{dx}{{a}^{2}-{x}^{2}}

Answer:

1atanh−1(xa)+C\frac{1}{a}{\text{tanh}}^{-1}(\frac{x}{a})+C

39. ∫dxx2+1\int \frac{dx}{\sqrt{{x}^{2}+1}}

40. ∫xdxx2+1\int \frac{xdx}{\sqrt{{x}^{2}+1}}

Answer:

x2+1+C\sqrt{{x}^{2}+1}+C

41. ∫−dxx1−x2\int -\frac{dx}{x\sqrt{1-{x}^{2}}}

42. ∫exe2x−1\int \frac{{e}^{x}}{\sqrt{{e}^{2x}-1}}

Answer: cosh−1(ex)+C{\text{cosh}}^{-1}({e}^{x})+C

43. ∫−2xx4−1\int -\frac{2x}{{x}^{4}-1}

For the following exercises, use the fact that a falling body with friction equal to velocity squared obeys the equation dv/dt=g−v2.dv\text{/}dt=g-{v}^{2}.

44. Show that v(t)=gtanh(gt)v(t)=\sqrt{g}\text{tanh}(\sqrt{gt}) satisfies this equation.

Answer:

Answers may vary

45. Derive the previous expression for v(t)v(t) by integrating dvg−v2=dt.\frac{dv}{g-{v}^{2}}=dt.

46. [T] Estimate how far a body has fallen in 12 seconds by finding the area underneath the curve of v(t).v(t).

47. 

Answer: 37.30

For the following exercises, use this scenario: A cable hanging under its own weight has a slope S=dy/dxS=dy\text{/}dx that satisfies dS/dx=c1+S2.dS\text{/}dx=c\sqrt{1+{S}^{2}}. The constant cc is the ratio of cable density to tension.

48. Show that S=sinh(cx)S=\text{sinh}(cx) satisfies this equation.

49. Integrate dy/dx=sinh(cx)dy\text{/}dx=\text{sinh}(cx) to find the cable height y(x)y(x) if y(0)=1/c.y(0)=1\text{/}c.

Answer:

y=1ccosh(cx)y=\frac{1}{c}\text{cosh}(cx)

50. Sketch the cable and determine how far down it sags at x=0.x=0.

For the following exercises, solve each problem.

51. [T] A chain hangs from two posts 2 m apart to form a catenary described by the equation y=2cosh(x/2)−1.y=2\text{cosh}(x\text{/}2)-1. Find the slope of the catenary at the left fence post.

Answer:

-0.521095

52. [T] A chain hangs from two posts four meters apart to form a catenary described by the equation y=4cosh(x/4)−3.y=4\text{cosh}(x\text{/}4)-3. Find the total length of the catenary (arc length).

53. [T] A high-voltage power line is a catenary described by y=10cosh(x/10).y=10\text{cosh}(x\text{/}10). Find the ratio of the area under the catenary to its arc length. What do you notice?

Answer: 10

54. A telephone line is a catenary described by y=acosh(x/a).y=a\text{cosh}(x\text{/}a). Find the ratio of the area under the catenary to its arc length. Does this confirm your answer for the previous question?

55. Prove the formula for the derivative of y=sinh−1(x)y={\text{sinh}}^{-1}(x) by differentiating x=sinh(y).x=\text{sinh}(y). (Hint: Use hyperbolic trigonometric identities.)

56. Prove the formula for the derivative of y=cosh−1(x)y={\text{cosh}}^{-1}(x) by differentiating x=cosh(y).x=\text{cosh}(y).

(Hint: Use hyperbolic trigonometric identities.)

57. Prove the formula for the derivative of y=sech−1(x)y={\text{sech}}^{-1}(x) by differentiating x=sech(y).x=\text{sech}(y). (Hint: Use hyperbolic trigonometric identities.)

58. Prove that (cosh(x)+sinh(x))n=cosh(nx)+sinh(nx).{(\text{cosh}(x)+\text{sinh}(x))}^{n}=\text{cosh}(nx)+\text{sinh}(nx).

59. Prove the expression for sinh−1(x).{\text{sinh}}^{-1}(x). Multiply x=sinh(y)=(1/2)(ey−e−y)x=\text{sinh}(y)=(1\text{/}2)({e}^{y}-{e}^{\text{−}y}) by 2ey2{e}^{y} and solve for y.y. Does your expression match the textbook?

60. Prove the expression for cosh−1(x).{\text{cosh}}^{-1}(x). Multiply x=cosh(y)=(1/2)(ey−e−y)x=\text{cosh}(y)=(1\text{/}2)({e}^{y}-{e}^{\text{−}y}) by 2ey2{e}^{y} and solve for y.y. Does your expression match the textbook?

Glossary

catenary
a curve in the shape of the function y=acosh(x/a)y=a\text{cosh}(x\text{/}a) is a catenary; a cable of uniform density suspended between two supports assumes the shape of a catenary

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