Calculus of the Hyperbolic Functions

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Evaluate the following derivatives:

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

   Using the formulas in and the chain rule, we get

   1. \(\frac{d}{dx}(\text{sinh}({x}^{2}))=\text{cosh}({x}^{2})\cdot 2x\)
   2. \(\frac{d}{dx}{(\text{cosh}\ x)}^{2}=2\ \text{cosh}\ x\ \text{sinh}\ x\)

2. Evaluate the following derivatives:

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

   1. \(\frac{d}{dx}(\text{tanh}({x}^{2}+3x))=({\text{sech}}^{2}({x}^{2}+3x))(2x+3)\)
   2. \(\frac{d}{dx}(\frac{1}{{(\text{sinh}\ x)}^{2}})=\frac{d}{dx}{(\text{sinh}\ x)}^{-2}=-2{(\text{sinh}\ x)}^{-3}\text{cosh}\ x\)

3. Evaluate the following integrals:

   1. \(\int x\ \text{cosh}({x}^{2})dx\)
   2. \(\int \text{tanh}\ x\ dx\)
   i

   We can use u-substitution in both cases.

   1. Let \(u={x}^{2}.\) Then, \(du=2x\ dx\) and
      \[\int x\ \text{cosh}({x}^{2})dx=\int \frac{1}{2}\text{cosh}\ u\ du=\frac{1}{2}\text{sinh}\ u+C=\frac{1}{2}\text{sinh}({x}^{2})+C.\]
   2. 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
      \[\int \text{tanh}\ x\ dx=\text{ln}(\text{cosh}\ x)+C.\]

4. Evaluate the following integrals:

   1. \(\int {\text{sinh}}^{3}x\ \text{cosh}\ x\ dx\)
   2. \(\int {\text{sech}}^{2}(3x)dx\)
   i

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

5. Evaluate the following derivatives:

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

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

   1. \(\frac{d}{dx}({\text{sinh}}^{-1}(\frac{x}{3}))=\frac{1}{3\sqrt{1+\frac{{x}^{2}}{9}}}=\frac{1}{\sqrt{9+{x}^{2}}}\)
   2. \(\frac{d}{dx}{({\text{tanh}}^{-1}x)}^{2}=\frac{2({\text{tanh}}^{-1}x)}{1-{x}^{2}}\)

6. Evaluate the following derivatives:

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

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

7. Evaluate the following integrals:

   1. \(\int \frac{1}{\sqrt{4{x}^{2}-1}}dx\)
   2. \(\int \frac{1}{2x\sqrt{1-9{x}^{2}}}dx\)
   i

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

   1. Let \(u=2x.\) Then, \(du=2dx\) and we have
      \[\int \frac{1}{\sqrt{4{x}^{2}-1}}dx=\int \frac{1}{2\sqrt{{u}^{2}-1}}du=\frac{1}{2}{\text{cosh}}^{-1}u+C=\frac{1}{2}{\text{cosh}}^{-1}(2x)+C.\]
   2. Let \(u=3x.\) Then, \(du=3dx\) and we obtain
      \[\int \frac{1}{2x\sqrt{1-9{x}^{2}}}dx=\frac{1}{2}\int \frac{1}{u\sqrt{1-{u}^{2}}}du=-\frac{1}{2}{\text{sech}}^{-1}|u|+C=-\frac{1}{2}{\text{sech}}^{-1}|3x|+C.\]

8. Evaluate the following integrals:

   1. \(\int \frac{1}{\sqrt{{x}^{2}-4}}dx,\ \ x>2\)
   2. \(\int \frac{1}{\sqrt{1-{e}^{2x}}}dx\)
   i

   1. \(\int \frac{1}{\sqrt{{x}^{2}-4}}dx={\text{cosh}}^{-1}(\frac{x}{2})+C\)
   2. \(\int \frac{1}{\sqrt{1-{e}^{2x}}}dx=\text{-}{\text{sech}}^{-1}({e}^{x})+C\)

9. 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).

   i

   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

   \[\begin{array}{ll}\text{Arc Length} & ={\int }_{-15}^{15}\sqrt{1+{\text{sinh}}^{2}(\frac{x}{10})}\ dx \\ & ={\int }_{-15}^{15}\text{cosh}(\frac{x}{10})dx \\ & =10\ \text{sinh}{(\frac{x}{10})|}_{-15}^{15}=10[\text{sinh}(\frac{3}{2})-\text{sinh}(-\frac{3}{2})]=20\ \text{sinh}(\frac{3}{2}) \\ & \approx 42.586\ \text{ft}\text{.}\end{array}\]

10. 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).

    i

    \(52.95\ \text{ft}\)

11. [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.

    i

    \({e}^{x}\ \text{and}\ {e}^{\text{-}x}\)

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

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

    i

    Answers may vary

14. Use the quotient rule to verify that \(\text{tanh}(x)'={\text{sech}}^{2}(x).\)

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

    i

    Answers may vary

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

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

    i

    Answers may vary

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

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

    i

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

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

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

    i

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

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

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

    i

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

24. [T] \({\text{cosh}}^{2}(x)-{\text{sinh}}^{2}(x)\)

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

    i

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

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

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

    i

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

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

29. \(\text{cosh}(2x+1)\)

    i

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

30. \(\text{tanh}(3x+2)\)

31. \(x\ \text{cosh}({x}^{2})\)

    i

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

32. \(3{x}^{3}\text{tanh}({x}^{4})\)

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

    i

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

34. \({\text{tanh}}^{2}(x){\text{sech}}^{2}(x)\)

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

    i

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

36. \(\text{coth}(x)\)

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

    i

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

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

39. \({\text{tanh}}^{-1}(4x)\)

    i

    \(\frac{4}{1-16{x}^{2}}\)

40. \({\text{sinh}}^{-1}({x}^{2})\)

Kuri Gukoresha

Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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