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Calculus of the Hyperbolic Functions
Apply the formulas for derivatives and integrals of the hyperbolic functions.
Derivatives and Integrals of the Hyperbolic Functions
Recall that the hyperbolic sine and hyperbolic cosine are defined as
\[\text{sinh}\ x=\frac{{e}^{x}-{e}^{\text{-}x}}{2}\ \text{and}\ \text{cosh}\ x=\frac{{e}^{x}+{e}^{\text{-}x}}{2}.\]The other hyperbolic functions are then defined in terms of \(\text{sinh}\ x\) and \(\text{cosh}\ x.\) 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 \(\text{sinh}\ x\) we have
\[\begin{array}{ll}\frac{d}{dx}(\text{sinh}\ x) & =\frac{d}{dx}(\frac{{e}^{x}-{e}^{\text{-}x}}{2}) \\ & =\frac{1}{2}[\frac{d}{dx}({e}^{x})-\frac{d}{dx}({e}^{\text{-}x})] \\ & =\frac{1}{2}[{e}^{x}+{e}^{\text{-}x}]=\text{cosh}\ x.\end{array}\]Similarly, \((d\text{/}dx)\text{cosh}\ x=\text{sinh}\ x.\) We summarize the differentiation formulas for the hyperbolic functions in the following table.
| \(f(x)\) | \(\frac{d}{dx}f(x)\) |
| \(\text{sinh}\ x\) | \(\text{cosh}\ x\) |
| \(\text{cosh}\ x\) | \(\text{sinh}\ x\) |
| \(\text{tanh}\ x\) | \({\text{sech}}^{2}\ x\) |
| \(\text{coth}\ x\) | \(\text{-}{\text{csch}}^{2}\ x\) |
| \(\text{sech}\ x\) | \(\text{-}\text{sech}\ x\ \text{tanh}\ x\) |
| \(\text{csch}\ x\) | \(\text{-}\text{csch}\ x\ \text{coth}\ x\) |
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: \((d\text{/}dx)\text{sin}\ x=\text{cos}\ x\) and \((d\text{/}dx)\text{sinh}\ x=\text{cosh}\ x.\) The derivatives of the cosine functions, however, differ in sign: \((d\text{/}dx)\text{cos}\ x=\text{-}\text{sin}\ x,\) but \((d\text{/}dx)\text{cosh}\ x=\text{sinh}\ x.\) 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.
\[\begin{array}{llllllll}\int \text{sinh}\ u\ du & = & \text{cosh}\ u+C & & & \int {\text{csch}}^{2}\ u\ du & = & \text{-}\text{coth}\ u+C \\ \int \text{cosh}\ u\ du & = & \text{sinh}\ u+C & & & \int \text{sech}\ u\ \text{tanh}\ u\ du & = & \text{-}\text{sech}\ u+C \\ \int {\text{sech}}^{2}u\ du & = & \text{tanh}\ u+C & & & \int \text{csch}\ u\ \text{coth}\ u\ du & = & \text{-}\text{csch}\ u+C\end{array}\]Example
Try it.
Evaluate the following derivatives:
- \(\frac{d}{dx}(\text{sinh}({x}^{2}))\)
- \(\frac{d}{dx}{(\text{cosh}\ x)}^{2}\)
Solution
Using the formulas in and the chain rule, we get
- \(\frac{d}{dx}(\text{sinh}({x}^{2}))=\text{cosh}({x}^{2})\cdot 2x\)
- \(\frac{d}{dx}{(\text{cosh}\ x)}^{2}=2\ \text{cosh}\ x\ \text{sinh}\ x\)
Condensed: the full section is in OpenStax Calculus Volume 2.
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.
| Function | Domain | Range |
| \({\text{sinh}}^{-1}x\) | \((\text{-}\infty ,\infty )\) | \((\text{-}\infty ,\infty )\) |
| \({\text{cosh}}^{-1}x\) | \([1,\infty )\) | \([0,\infty )\) |
| \({\text{tanh}}^{-1}x\) | \((-1,1)\) | \((\text{-}\infty ,\infty )\) |
| \({\text{coth}}^{-1}x\) | \((\text{-}\infty ,-1)\cup (1,\infty )\) | \((\text{-}\infty ,0)\cup (0,\infty )\) |
| \({\text{sech}}^{-1}x\) | \((0\text{, 1}]\) | \([0,\infty )\) |
| \({\text{csch}}^{-1}x\) | \((\text{-}\infty ,0)\cup (0,\infty )\) | \((\text{-}\infty ,0)\cup (0,\infty )\) |
The graphs of the inverse hyperbolic functions are shown in the following figure.
To find the derivatives of the inverse functions, we use implicit differentiation. We have
\[\begin{array}{lll}y & = & {\text{sinh}}^{-1}\ x \\ \text{sinh}\ y & = & x \\ \frac{d}{dx}\text{sinh}\ y & = & \frac{d}{dx}x \\ \text{cosh}\ y\frac{dy}{dx} & = & 1.\end{array}\]Recall that \({\text{cosh}}^{2}y-{\text{sinh}}^{2}y=1,\) so \(\text{cosh}\ y=\sqrt{1+{\text{sinh}}^{2}y}.\) Then,
\[\frac{dy}{dx}=\frac{1}{\text{cosh}\ y}=\frac{1}{\sqrt{1+{\text{sinh}}^{2}y}}=\frac{1}{\sqrt{1+{x}^{2}}}.\]We can derive differentiation formulas for the other inverse hyperbolic functions in a similar fashion. These differentiation formulas are summarized in the following table.
| \(f(x)\) | \(\frac{d}{dx}f(x)\) |
| \({\text{sinh}}^{-1}x\) | \(\frac{1}{\sqrt{1+{x}^{2}}}\) |
| \({\text{cosh}}^{-1}x\) | \(\frac{1}{\sqrt{{x}^{2}-1}}\) |
| \({\text{tanh}}^{-1}x\) | \(\frac{1}{1-{x}^{2}}\) |
| \({\text{coth}}^{-1}x\) | \(\frac{1}{1-{x}^{2}}\) |
| \({\text{sech}}^{-1}x\) | \(\frac{-1}{x\sqrt{1-{x}^{2}}}\) |
| \({\text{csch}}^{-1}x\) | \(\frac{-1}{|x|\sqrt{1+{x}^{2}}}\) |
Note that the derivatives of \({\text{tanh}}^{-1}\ x\) and \({\text{coth}}^{-1}\ x\) are the same. Thus, when we integrate \(1\text{/}(1-{x}^{2}),\) we need to select the proper antiderivative based on the domain of the functions and the values of \(x.\) Integration formulas involving the inverse hyperbolic functions are summarized as follows.
\[\begin{array}{llllllll}\int \frac{1}{\sqrt{1+{u}^{2}}}du & = & {\text{sinh}}^{-1}u+C & & & \int \frac{1}{u\sqrt{1-{u}^{2}}}du & = & \text{-}{\text{sech}}^{-1}|u|+C \\ \int \frac{1}{\sqrt{{u}^{2}-1}}du & = & {\text{cosh}}^{-1}u+C & & & \int \frac{1}{u\sqrt{1+{u}^{2}}}du & = & \text{-}{\text{csch}}^{-1}|u|+C \\ \int \frac{1}{1-{u}^{2}}du & = & \{\begin{array}{l}{\text{tanh}}^{-1}u+C\ \text{if}\ |u|<1 \\ {\text{coth}}^{-1}u+C\ \text{if}\ |u|>1\end{array} & & & & & \end{array}\]Condensed: the full section is in OpenStax Calculus Volume 2.
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.
Hyperbolic functions can be used to model catenaries. Specifically, functions of the form \(y=a\ \text{cosh}(x\text{/}a)\) are catenaries. shows the graph of \(y=2\ \text{cosh}(x\text{/}2).\)
Example
Try it.
Assume a hanging cable has the shape \(10\ \text{cosh}(x\text{/}10)\) for \(-15\le x\le 15,\) where \(x\) is measured in feet. Determine the length of the cable (in feet).
Solution
Recall from Section \(2.4\) that the formula for arc length is
\[\text{Arc Length}={\int }_{a}^{b}\sqrt{1+{[{f}^{'}(x)]}^{2}}\ dx.\]We have \(f(x)=10\ \text{cosh}(x\text{/}10),\) so \({f}^{'}(x)=\text{sinh}(x\text{/}10).\) Then
\[\begin{array}{ll}\text{Arc Length} & ={\int }_{a}^{b}\sqrt{1+{[{f}^{'}(x)]}^{2}}\ dx \\ & ={\int }_{-15}^{15}\sqrt{1+{\text{sinh}}^{2}(\frac{x}{10})}\ dx.\end{array}\]Now recall that \(1+{\text{sinh}}^{2}x={\text{cosh}}^{2}x,\) so we have
\[\begin{array}{ll}\text{Arc Length} & ={\int }_{-15}^{15}\sqrt{1+{\text{sinh}}^{2}(\frac{x}{10})}\ dx \\ & ={\int }_{-15}^{15}\text{cosh}(\frac{x}{10})dx \\ & =10\ \text{sinh}{(\frac{x}{10})|}_{-15}^{15}=10[\text{sinh}(\frac{3}{2})-\text{sinh}(-\frac{3}{2})]=20\ \text{sinh}(\frac{3}{2}) \\ & \approx 42.586\ \text{ft}\text{.}\end{array}\]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.
Calculus of the Hyperbolic Functions
For the following exercises, find the derivatives of the given functions and graph along with the function to ensure your answer is correct.
For the following exercises, find the antiderivatives for the given functions.
For the following exercises, find the derivatives for the functions.
For the following exercises, find the antiderivatives for the functions.
For the following exercises, use the fact that a falling body with friction equal to velocity squared obeys the equation \(dv\text{/}dt=g-{v}^{2}.\)
For the following exercises, use this scenario: A cable hanging under its own weight has a slope \(S=dy\text{/}dx\) that satisfies \(dS\text{/}dx=c\sqrt{1+{S}^{2}}.\) The constant \(c\) is the ratio of cable density to tension.
For the following exercises, solve each problem.
Derivatives and Integrals of the Hyperbolic Functions
Recall that the hyperbolic sine and hyperbolic cosine are defined as
\[\text{sinh}\ x=\frac{{e}^{x}-{e}^{\text{-}x}}{2}\ \text{and}\ \text{cosh}\ x=\frac{{e}^{x}+{e}^{\text{-}x}}{2}.\]The other hyperbolic functions are then defined in terms of \(\text{sinh}\ x\) and \(\text{cosh}\ x.\) 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 \(\text{sinh}\ x\) we have
\[\begin{array}{ll}\frac{d}{dx}(\text{sinh}\ x) & =\frac{d}{dx}(\frac{{e}^{x}-{e}^{\text{-}x}}{2}) \\ & =\frac{1}{2}[\frac{d}{dx}({e}^{x})-\frac{d}{dx}({e}^{\text{-}x})] \\ & =\frac{1}{2}[{e}^{x}+{e}^{\text{-}x}]=\text{cosh}\ x.\end{array}\]Similarly, \((d\text{/}dx)\text{cosh}\ x=\text{sinh}\ x.\) We summarize the differentiation formulas for the hyperbolic functions in the following table.
| \(f(x)\) | \(\frac{d}{dx}f(x)\) |
| \(\text{sinh}\ x\) | \(\text{cosh}\ x\) |
| \(\text{cosh}\ x\) | \(\text{sinh}\ x\) |
| \(\text{tanh}\ x\) | \({\text{sech}}^{2}\ x\) |
| \(\text{coth}\ x\) | \(\text{-}{\text{csch}}^{2}\ x\) |
| \(\text{sech}\ x\) | \(\text{-}\text{sech}\ x\ \text{tanh}\ x\) |
| \(\text{csch}\ x\) | \(\text{-}\text{csch}\ x\ \text{coth}\ x\) |
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: \((d\text{/}dx)\text{sin}\ x=\text{cos}\ x\) and \((d\text{/}dx)\text{sinh}\ x=\text{cosh}\ x.\) The derivatives of the cosine functions, however, differ in sign: \((d\text{/}dx)\text{cos}\ x=\text{-}\text{sin}\ x,\) but \((d\text{/}dx)\text{cosh}\ x=\text{sinh}\ x.\) 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.
\[\begin{array}{llllllll}\int \text{sinh}\ u\ du & = & \text{cosh}\ u+C & & & \int {\text{csch}}^{2}\ u\ du & = & \text{-}\text{coth}\ u+C \\ \int \text{cosh}\ u\ du & = & \text{sinh}\ u+C & & & \int \text{sech}\ u\ \text{tanh}\ u\ du & = & \text{-}\text{sech}\ u+C \\ \int {\text{sech}}^{2}u\ du & = & \text{tanh}\ u+C & & & \int \text{csch}\ u\ \text{coth}\ u\ du & = & \text{-}\text{csch}\ u+C\end{array}\]Example
Try it.
Evaluate the following derivatives:
- \(\frac{d}{dx}(\text{sinh}({x}^{2}))\)
- \(\frac{d}{dx}{(\text{cosh}\ x)}^{2}\)
Solution
Using the formulas in and the chain rule, we get
- \(\frac{d}{dx}(\text{sinh}({x}^{2}))=\text{cosh}({x}^{2})\cdot 2x\)
- \(\frac{d}{dx}{(\text{cosh}\ x)}^{2}=2\ \text{cosh}\ x\ \text{sinh}\ x\)
Condensed: the full section is in OpenStax Calculus Volume 1.
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.
| Function | Domain | Range |
| \({\text{sinh}}^{-1}x\) | \((\text{-}\infty ,\infty )\) | \((\text{-}\infty ,\infty )\) |
| \({\text{cosh}}^{-1}x\) | \([1,\infty )\) | \([0,\infty )\) |
| \({\text{tanh}}^{-1}x\) | \((-1,1)\) | \((\text{-}\infty ,\infty )\) |
| \({\text{coth}}^{-1}x\) | \((\text{-}\infty ,-1)\cup (1,\infty )\) | \((\text{-}\infty ,0)\cup (0,\infty )\) |
| \({\text{sech}}^{-1}x\) | \((0\text{, 1}]\) | \([0,\infty )\) |
| \({\text{csch}}^{-1}x\) | \((\text{-}\infty ,0)\cup (0,\infty )\) | \((\text{-}\infty ,0)\cup (0,\infty )\) |
The graphs of the inverse hyperbolic functions are shown in the following figure.
To find the derivatives of the inverse functions, we use implicit differentiation. We have
\[\begin{array}{lll}y & = & {\text{sinh}}^{-1}\ x \\ \text{sinh}\ y & = & x \\ \frac{d}{dx}\text{sinh}\ y & = & \frac{d}{dx}x \\ \text{cosh}\ y\frac{dy}{dx} & = & 1.\end{array}\]Recall that \({\text{cosh}}^{2}y-{\text{sinh}}^{2}y=1,\) so \(\text{cosh}\ y=\sqrt{1+{\text{sinh}}^{2}y}.\) Then,
\[\frac{dy}{dx}=\frac{1}{\text{cosh}\ y}=\frac{1}{\sqrt{1+{\text{sinh}}^{2}y}}=\frac{1}{\sqrt{1+{x}^{2}}}.\]We can derive differentiation formulas for the other inverse hyperbolic functions in a similar fashion. These differentiation formulas are summarized in the following table.
| \(f(x)\) | \(\frac{d}{dx}f(x)\) |
| \({\text{sinh}}^{-1}x\) | \(\frac{1}{\sqrt{1+{x}^{2}}}\) |
| \({\text{cosh}}^{-1}x\) | \(\frac{1}{\sqrt{{x}^{2}-1}}\) |
| \({\text{tanh}}^{-1}x\) | \(\frac{1}{1-{x}^{2}}\) |
| \({\text{coth}}^{-1}x\) | \(\frac{1}{1-{x}^{2}}\) |
| \({\text{sech}}^{-1}x\) | \(\frac{-1}{x\sqrt{1-{x}^{2}}}\) |
| \({\text{csch}}^{-1}x\) | \(\frac{-1}{|x|\sqrt{1+{x}^{2}}}\) |
Note that the derivatives of \({\text{tanh}}^{-1}\ x\) and \({\text{coth}}^{-1}\ x\) are the same. Thus, when we integrate \(1\text{/}(1-{x}^{2}),\) we need to select the proper antiderivative based on the domain of the functions and the values of \(x.\) Integration formulas involving the inverse hyperbolic functions are summarized as follows.
\[\begin{array}{llllllll}\int \frac{1}{\sqrt{1+{u}^{2}}}du & = & {\text{sinh}}^{-1}u+C & & & \int \frac{1}{u\sqrt{1-{u}^{2}}}du & = & \text{-}{\text{sech}}^{-1}|u|+C \\ \int \frac{1}{\sqrt{{u}^{2}-1}}du & = & {\text{cosh}}^{-1}u+C & & & \int \frac{1}{u\sqrt{1+{u}^{2}}}du & = & \text{-}{\text{csch}}^{-1}|u|+C \\ \int \frac{1}{1-{u}^{2}}du & = & \{\begin{array}{l}{\text{tanh}}^{-1}u+C\ \text{if}\ |u|<1 \\ {\text{coth}}^{-1}u+C\ \text{if}\ |u|>1\end{array} & & & & & \end{array}\]Condensed: the full section is in OpenStax Calculus Volume 1.
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.
Hyperbolic functions can be used to model catenaries. Specifically, functions of the form \(y=a\ \text{cosh}(x\text{/}a)\) are catenaries. shows the graph of \(y=2\ \text{cosh}(x\text{/}2).\)
Example
Try it.
Assume a hanging cable has the shape \(10\ \text{cosh}(x\text{/}10)\) for \(-15\le x\le 15,\) where \(x\) is measured in feet. Determine the length of the cable (in feet).
Solution
Recall from Section \(2.4\) that the formula for arc length is
\[\text{Arc Length}={\int }_{a}^{b}\sqrt{1+{[{f}^{'}(x)]}^{2}}\ dx.\]We have \(f(x)=10\ \text{cosh}(x\text{/}10),\) so \({f}^{'}(x)=\text{sinh}(x\text{/}10).\) Then
\[\begin{array}{ll}\text{Arc Length} & ={\int }_{a}^{b}\sqrt{1+{[{f}^{'}(x)]}^{2}}\ dx \\ & ={\int }_{-15}^{15}\sqrt{1+{\text{sinh}}^{2}(\frac{x}{10})}\ dx.\end{array}\]Now recall that \(1+{\text{sinh}}^{2}x={\text{cosh}}^{2}x,\) so we have
\[\begin{array}{ll}\text{Arc Length} & ={\int }_{-15}^{15}\sqrt{1+{\text{sinh}}^{2}(\frac{x}{10})}\ dx \\ & ={\int }_{-15}^{15}\text{cosh}(\frac{x}{10})dx \\ & =10\ \text{sinh}{(\frac{x}{10})|}_{-15}^{15}=10[\text{sinh}(\frac{3}{2})-\text{sinh}(-\frac{3}{2})]=20\ \text{sinh}(\frac{3}{2}) \\ & \approx 42.586\ \text{ft}\text{.}\end{array}\]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.
Calculus of the Hyperbolic Functions
For the following exercises, find the derivatives of the given functions and graph along with the function to ensure your answer is correct.
For the following exercises, find the antiderivatives for the given functions.
For the following exercises, find the derivatives for the functions.
For the following exercises, find the antiderivatives for the functions.
For the following exercises, use the fact that a falling body with friction equal to velocity squared obeys the equation \(dv\text{/}dt=g-{v}^{2}.\)
For the following exercises, use this scenario: A cable hanging under its own weight has a slope \(S=dy\text{/}dx\) that satisfies \(dS\text{/}dx=c\sqrt{1+{S}^{2}}.\) The constant \(c\) is the ratio of cable density to tension.
For the following exercises, solve each problem.
이제 너 계산기는 이것을 해결하지 않지만, 그 조각은 계산 가능합니다. 아래의 하나를 시도하거나 자신의 것을 입력하십시오.
연습 (40)
먼저 종이에 각각을 시도해 보세요. 확인하기 위해 답을 드러내세요. 확인된 답은 각 단계의 솔버에서 열 수 있습니다.
-
Evaluate the following derivatives:
- \(\frac{d}{dx}(\text{sinh}({x}^{2}))\)
- \(\frac{d}{dx}{(\text{cosh}\ x)}^{2}\)
답을 드러내세요
Using the formulas in and the chain rule, we get
- \(\frac{d}{dx}(\text{sinh}({x}^{2}))=\text{cosh}({x}^{2})\cdot 2x\)
- \(\frac{d}{dx}{(\text{cosh}\ x)}^{2}=2\ \text{cosh}\ x\ \text{sinh}\ x\)
-
Evaluate the following derivatives:
- \(\frac{d}{dx}(\text{tanh}({x}^{2}+3x))\)
- \(\frac{d}{dx}(\frac{1}{{(\text{sinh}\ x)}^{2}})\)
답을 드러내세요
- \(\frac{d}{dx}(\text{tanh}({x}^{2}+3x))=({\text{sech}}^{2}({x}^{2}+3x))(2x+3)\)
- \(\frac{d}{dx}(\frac{1}{{(\text{sinh}\ x)}^{2}})=\frac{d}{dx}{(\text{sinh}\ x)}^{-2}=-2{(\text{sinh}\ x)}^{-3}\text{cosh}\ x\)
-
Evaluate the following integrals:
- \(\int x\ \text{cosh}({x}^{2})dx\)
- \(\int \text{tanh}\ x\ dx\)
답을 드러내세요
We can use u-substitution in both cases.
- Let \(u={x}^{2}.\) Then, \(du=2x\ dx\) and
\[\int x\ \text{cosh}({x}^{2})dx=\int \frac{1}{2}\text{cosh}\ u\ du=\frac{1}{2}\text{sinh}\ u+C=\frac{1}{2}\text{sinh}({x}^{2})+C.\] - Let \(u=\text{cosh}\ x.\) Then, \(du=\text{sinh}\ x\ dx\) and
\[\int \text{tanh}\ x\ dx=\int \frac{\text{sinh}\ x}{\text{cosh}\ x}dx=\int \frac{1}{u}du=\text{ln}|u|+C=\text{ln}|\text{cosh}\ x|+C.\]
Note that \(\text{cosh}\ x>0\) for all \(x,\) so we can eliminate the absolute value signs and obtain
\[\int \text{tanh}\ x\ dx=\text{ln}(\text{cosh}\ x)+C.\]
-
Evaluate the following integrals:
- \(\int {\text{sinh}}^{3}x\ \text{cosh}\ x\ dx\)
- \(\int {\text{sech}}^{2}(3x)dx\)
답을 드러내세요
- \(\int {\text{sinh}}^{3}x\ \text{cosh}\ x\ dx=\frac{{\text{sinh}}^{4}x}{4}+C\)
- \(\int {\text{sech}}^{2}(3x)dx=\frac{\text{tanh}(3x)}{3}+C\)
-
Evaluate the following derivatives:
- \(\frac{d}{dx}({\text{sinh}}^{-1}(\frac{x}{3}))\)
- \(\frac{d}{dx}{({\text{tanh}}^{-1}x)}^{2}\)
답을 드러내세요
Using the formulas in and the chain rule, we obtain the following results:
- \(\frac{d}{dx}({\text{sinh}}^{-1}(\frac{x}{3}))=\frac{1}{3\sqrt{1+\frac{{x}^{2}}{9}}}=\frac{1}{\sqrt{9+{x}^{2}}}\)
- \(\frac{d}{dx}{({\text{tanh}}^{-1}x)}^{2}=\frac{2({\text{tanh}}^{-1}x)}{1-{x}^{2}}\)
-
Evaluate the following derivatives:
- \(\frac{d}{dx}({\text{cosh}}^{-1}(3x))\)
- \(\frac{d}{dx}{({\text{coth}}^{-1}x)}^{3}\)
답을 드러내세요
- \(\frac{d}{dx}({\text{cosh}}^{-1}(3x))=\frac{3}{\sqrt{9{x}^{2}-1}}\)
- \(\frac{d}{dx}{({\text{coth}}^{-1}x)}^{3}=\frac{3{({\text{coth}}^{-1}x)}^{2}}{1-{x}^{2}}\)
-
Evaluate the following integrals:
- \(\int \frac{1}{\sqrt{4{x}^{2}-1}}dx\)
- \(\int \frac{1}{2x\sqrt{1-9{x}^{2}}}dx\)
답을 드러내세요
We can use \(u\text{-substitution}\) in both cases.
- Let \(u=2x.\) Then, \(du=2dx\) and we have
\[\int \frac{1}{\sqrt{4{x}^{2}-1}}dx=\int \frac{1}{2\sqrt{{u}^{2}-1}}du=\frac{1}{2}{\text{cosh}}^{-1}u+C=\frac{1}{2}{\text{cosh}}^{-1}(2x)+C.\] - Let \(u=3x.\) Then, \(du=3dx\) and we obtain
\[\int \frac{1}{2x\sqrt{1-9{x}^{2}}}dx=\frac{1}{2}\int \frac{1}{u\sqrt{1-{u}^{2}}}du=-\frac{1}{2}{\text{sech}}^{-1}|u|+C=-\frac{1}{2}{\text{sech}}^{-1}|3x|+C.\]
-
Evaluate the following integrals:
- \(\int \frac{1}{\sqrt{{x}^{2}-4}}dx,\ \ x>2\)
- \(\int \frac{1}{\sqrt{1-{e}^{2x}}}dx\)
답을 드러내세요
- \(\int \frac{1}{\sqrt{{x}^{2}-4}}dx={\text{cosh}}^{-1}(\frac{x}{2})+C\)
- \(\int \frac{1}{\sqrt{1-{e}^{2x}}}dx=\text{-}{\text{sech}}^{-1}({e}^{x})+C\)
-
Assume a hanging cable has the shape \(10\ \text{cosh}(x\text{/}10)\) for \(-15\le x\le 15,\) where \(x\) is measured in feet. Determine the length of the cable (in feet).
답을 드러내세요
Recall from Section \(2.4\) that the formula for arc length is
\[\text{Arc Length}={\int }_{a}^{b}\sqrt{1+{[{f}^{'}(x)]}^{2}}\ dx.\]We have \(f(x)=10\ \text{cosh}(x\text{/}10),\) so \({f}^{'}(x)=\text{sinh}(x\text{/}10).\) Then
\[\begin{array}{ll}\text{Arc Length} & ={\int }_{a}^{b}\sqrt{1+{[{f}^{'}(x)]}^{2}}\ dx \\ & ={\int }_{-15}^{15}\sqrt{1+{\text{sinh}}^{2}(\frac{x}{10})}\ dx.\end{array}\]Now recall that \(1+{\text{sinh}}^{2}x={\text{cosh}}^{2}x,\) so we have
\[\begin{array}{ll}\text{Arc Length} & ={\int }_{-15}^{15}\sqrt{1+{\text{sinh}}^{2}(\frac{x}{10})}\ dx \\ & ={\int }_{-15}^{15}\text{cosh}(\frac{x}{10})dx \\ & =10\ \text{sinh}{(\frac{x}{10})|}_{-15}^{15}=10[\text{sinh}(\frac{3}{2})-\text{sinh}(-\frac{3}{2})]=20\ \text{sinh}(\frac{3}{2}) \\ & \approx 42.586\ \text{ft}\text{.}\end{array}\] -
Assume a hanging cable has the shape \(15\ \text{cosh}(x\text{/}15)\) for \(-20\le x\le 20.\) Determine the length of the cable (in feet).
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\(52.95\ \text{ft}\)
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[T] Find expressions for \(\text{cosh}\ x+\text{sinh}\ x\) and \(\text{cosh}\ x-\text{sinh}\ x.\) Use a calculator to graph these functions and ensure your expression is correct.
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\({e}^{x}\ \text{and}\ {e}^{\text{-}x}\)
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From the definitions of \(\text{cosh}(x)\) and \(\text{sinh}(x),\) find their antiderivatives.
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Show that \(\text{cosh}(x)\) and \(\text{sinh}(x)\) satisfy \({y}^{″}=y.\)
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Answers may vary
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Use the quotient rule to verify that \(\text{tanh}(x)'={\text{sech}}^{2}(x).\)
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Derive \({\text{cosh}}^{2}(x)+{\text{sinh}}^{2}(x)=\text{cosh}(2x)\) from the definition.
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Answers may vary
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Take the derivative of the previous expression to find an expression for \(\text{sinh}(2x).\)
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Prove \(\text{sinh}(x+y)=\text{sinh}(x)\text{cosh}(y)+\text{cosh}(x)\text{sinh}(y)\) by changing the expression to exponentials.
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Answers may vary
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Take the derivative of the previous expression to find an expression for \(\text{cosh}(x+y).\)
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[T] \(\text{cosh}(3x+1)\)
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\(3\ \text{sinh}(3x+1)\)
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[T] \(\text{sinh}({x}^{2})\)
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[T] \(\frac{1}{\text{cosh}(x)}\)
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\(\text{-}\text{tanh}(x)\text{sech}(x)\)
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[T] \(\text{sinh}(\text{ln}(x))\)
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[T] \({\text{cosh}}^{2}(x)+{\text{sinh}}^{2}(x)\)
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\(4\ \text{cosh}(x)\text{sinh}(x)\)
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[T] \({\text{cosh}}^{2}(x)-{\text{sinh}}^{2}(x)\)
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[T] \(\text{tanh}(\sqrt{{x}^{2}+1})\)
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\(\frac{x\ {\text{sech}}^{2}(\sqrt{{x}^{2}+1})}{\sqrt{{x}^{2}+1}}\)
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[T] \(\frac{1+\text{tanh}(x)}{1-\text{tanh}(x)}\)
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[T] \({\text{sinh}}^{6}(x)\)
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\(6\ {\text{sinh}}^{5}(x)\text{cosh}(x)\)
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[T] \(\text{ln}(\text{sech}(x)+\text{tanh}(x))\)
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\(\text{cosh}(2x+1)\)
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\(\frac{1}{2}\text{sinh}(2x+1)+C\)
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\(\text{tanh}(3x+2)\)
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\(x\ \text{cosh}({x}^{2})\)
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\(\frac{1}{2}\text{sinh}({x}^{2})+C\)
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\(3{x}^{3}\text{tanh}({x}^{4})\)
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\({\text{cosh}}^{2}(x)\text{sinh}(x)\)
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\(\frac{1}{3}{\text{cosh}}^{3}(x)+C\)
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\({\text{tanh}}^{2}(x){\text{sech}}^{2}(x)\)
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\(\frac{\text{sinh}(x)}{1+\text{cosh}(x)}\)
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\(\text{ln}(1+\text{cosh}(x))+C\)
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\(\text{coth}(x)\)
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\(\text{cosh}(x)+\text{sinh}(x)\)
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\(\text{cosh}(x)+\text{sinh}(x)+C\)
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\({(\text{cosh}(x)+\text{sinh}(x))}^{n}\)
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\({\text{tanh}}^{-1}(4x)\)
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\(\frac{4}{1-16{x}^{2}}\)
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\({\text{sinh}}^{-1}({x}^{2})\)
여기서 사용된 기호
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어떻게: Calculus of the Hyperbolic Functions
- 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.
- Let
- Let
- Let
- Let
- Hyperbolic functions are defined in terms of exponential functions.
사람들이 묻는 질문
What is a derivative in one sentence?
The slope of the graph at a point, the rate at which the output is changing there. Speed is the derivative of position.
What is an integral in one sentence?
The accumulated total of a rate: the area under the curve. Distance travelled is the integral of speed.
Why are derivatives and integrals opposites?
That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.
When do I use substitution and when integration by parts?
Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function, a polynomial times an exponential, log or trig function.
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에 더 Calculus
LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests