{"id":497,"date":"2025-02-13T19:45:29","date_gmt":"2025-02-13T19:45:29","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/calculus-of-the-hyperbolic-functions-fresh-take\/"},"modified":"2025-02-13T19:45:29","modified_gmt":"2025-02-13T19:45:29","slug":"calculus-of-the-hyperbolic-functions-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/calculus-of-the-hyperbolic-functions-fresh-take\/","title":{"raw":"Calculus of the Hyperbolic Functions: Fresh Take","rendered":"Calculus of the Hyperbolic Functions: Fresh Take"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li>Differentiate and integrate hyperbolic functions and their inverse forms<\/li>\n\t<li>Understand the practical situations where the catenary curve appears<\/li>\n<\/ul>\n<\/section>\n<h2>Derivatives and Integrals of the Hyperbolic Functions<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n\t<li class=\"whitespace-normal break-words\">Definitions of Hyperbolic Functions:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]\\sinh x = \\frac{e^x - e^{-x}}{2}[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">[latex]\\cosh x = \\frac{e^x + e^{-x}}{2}[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Key Derivatives:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx}(\\sinh x) = \\cosh x[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx}(\\cosh x) = \\sinh x[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx}(\\tanh x) = \\text{sech}^2 x[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Key Integrals:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]\\int \\sinh x , dx = \\cosh x + C[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">[latex]\\int \\cosh x , dx = \\sinh x + C[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">[latex]\\int \\text{sech}^2 x , dx = \\tanh x + C[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Similarities with Trigonometric Functions:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Derivatives of sinh and sin are similar<\/li>\n\t<li class=\"whitespace-normal break-words\">But [latex]\\frac{d}{dx}(\\cos x) = -\\sin x[\/latex], while [latex]\\frac{d}{dx}(\\cosh x) = \\sinh x[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">U-substitution is often useful for integrals involving hyperbolic functions<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p id=\"fs-id1167794028505\">Evaluate the following derivatives:<\/p>\n<ol id=\"fs-id1167794043812\" style=\"list-style-type: lower-alpha;\">\n\t<li>[latex]\\frac{d}{dx}(\\text{tanh}({x}^{2}+3x))[\/latex]<\/li>\n\t<li>[latex]\\frac{d}{dx}(\\dfrac{1}{{(\\text{sinh}x)}^{2}})[\/latex]<\/li>\n<\/ol>\n\n[reveal-answer q=\"fs-id1167794187159\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167794187159\"]\n\n<ol style=\"list-style-type: lower-alpha;\">\n\t<li>[latex]\\frac{d}{dx}(\\text{tanh}({x}^{2}+3x))=({\\text{sech}}^{2}({x}^{2}+3x))(2x+3)[\/latex]<\/li>\n\t<li>[latex]\\frac{d}{dx}(\\frac{1}{{(\\text{sinh}x)}^{2}})=\\frac{d}{dx}{(\\text{sinh}x)}^{-2}=-2{(\\text{sinh}x)}^{-3}\\text{cosh}x[\/latex]<\/li>\n<\/ol>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ozRCEFk1tmA?controls=0&amp;start=134&amp;end=300&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.9CalculusOfHyperbolicFunctions134to300_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.9 Calculus of Hyperbolic Functions\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1167793819956\">Evaluate the following integrals:<\/p>\n<ol id=\"fs-id1167793985861\" style=\"list-style-type: lower-alpha;\">\n\t<li>[latex]\\displaystyle\\int {\\text{sinh}}^{3}x\\text{cosh}xdx[\/latex]<\/li>\n\t<li>[latex]\\displaystyle\\int {\\text{sech}}^{2}(3x)dx[\/latex]<\/li>\n<\/ol>\n\n[reveal-answer q=\"fs-id1167793373829\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793373829\"]\n\n<ol id=\"fs-id1167793373829\" style=\"list-style-type: lower-alpha;\">\n\t<li>[latex]\\displaystyle\\int {\\text{sinh}}^{3}x\\text{cosh}xdx=\\frac{{\\text{sinh}}^{4}x}{4}+C[\/latex]<\/li>\n\t<li>[latex]\\displaystyle\\int {\\text{sech}}^{2}(3x)dx=\\frac{\\text{tanh}(3x)}{3}+C[\/latex]<\/li>\n<\/ol>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ozRCEFk1tmA?controls=0&amp;start=478&amp;end=598&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.9CalculusOfHyperbolicFunctions478to598_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.9 Calculus of Hyperbolic Functions\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Evaluate the following integral:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\int x \\sinh(x^2) , dx[\/latex]<\/p>\n<p><br>\n[reveal-answer q=\"773910\"]Show Answer[\/reveal-answer]<br>\n[hidden-answer a=\"773910\"]<\/p>\n<p>This integral suggests a u-substitution. Let [latex]u = x^2[\/latex].<\/p>\n<p>Then [latex]du = 2x , dx[\/latex], or [latex]x , dx = \\frac{1}{2} du[\/latex].<\/p>\n<p>Rewrite the integral in terms of u:<\/p>\n<p style=\"text-align: center;\">[latex]\\int x \\sinh(x^2) , dx = \\int \\frac{1}{2} \\sinh(u) , du[\/latex]<\/p>\n<p>Now we can use the integration formula for sinh:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{2} \\int \\sinh(u) , du = \\frac{1}{2} \\cosh(u) + C[\/latex]<\/p>\n<p>Substitute back [latex]x^2[\/latex] for [latex]u[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{2} \\cosh(x^2) + C[\/latex]<\/p>\n<p>Therefore, [latex]\\int x \\sinh(x^2) , dx = \\frac{1}{2} \\cosh(x^2) + C[\/latex].<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<h2>Calculus of Inverse Hyperbolic Functions<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n\t<li class=\"whitespace-normal break-words\">All hyperbolic functions have inverses with appropriate range restrictions<\/li>\n\t<li class=\"whitespace-normal break-words\">Key Derivatives:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx}(\\sinh^{-1} x) = \\frac{1}{\\sqrt{1+x^2}}[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx}(\\cosh^{-1} x) = \\frac{1}{\\sqrt{x^2-1}}[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx}(\\tanh^{-1} x) = \\frac{1}{1-x^2}[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Key Integrals:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]\\int \\frac{1}{\\sqrt{1+u^2}} du = \\sinh^{-1} u + C[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">[latex]\\int \\frac{1}{\\sqrt{u^2-1}} du = \\cosh^{-1} u + C[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">[latex]\\int \\frac{1}{1-u^2} du = \\begin{cases} \\tanh^{-1} u + C &amp; \\text{if } |u| &lt; 1 \\ \\coth^{-1} u + C &amp; \\text{if } |u| &gt; 1 \\end{cases}[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Each inverse hyperbolic function has specific domain and range restrictions<\/li>\n\t<li class=\"whitespace-normal break-words\">Derivatives of inverse hyperbolic functions are found using implicit differentiation<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p id=\"fs-id1167794332395\">Evaluate the following derivatives:<\/p>\n<ol id=\"fs-id1167793590727\" style=\"list-style-type: lower-alpha;\">\n\t<li>[latex]\\frac{d}{dx}({\\text{cosh}}^{-1}(3x))[\/latex]<\/li>\n\t<li>[latex]\\frac{d}{dx}{({\\text{coth}}^{-1}x)}^{3}[\/latex]<\/li>\n<\/ol>\n\n[reveal-answer q=\"fs-id1167793498614\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793498614\"]\n\n<ol id=\"fs-id1167793498614\" style=\"list-style-type: lower-alpha;\">\n\t<li>[latex]\\frac{d}{dx}({\\text{cosh}}^{-1}(3x))=\\frac{3}{\\sqrt{9{x}^{2}-1}}[\/latex]<\/li>\n\t<li>[latex]\\frac{d}{dx}{({\\text{coth}}^{-1}x)}^{3}=\\frac{3{({\\text{coth}}^{-1}x)}^{2}}{1-{x}^{2}}[\/latex]<\/li>\n<\/ol>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ozRCEFk1tmA?controls=0&amp;start=827&amp;end=907&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.9CalculusOfHyperbolicFunctions827to907_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.9 Calculus of Hyperbolic Functions\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1167793871641\">Evaluate the following integrals:<\/p>\n<ol id=\"fs-id1167793937415\" style=\"list-style-type: lower-alpha;\">\n\t<li>[latex]\\displaystyle\\int \\frac{1}{\\sqrt{4{x}^{2}-1}}dx[\/latex]<\/li>\n\t<li>[latex]\\displaystyle\\int \\frac{1}{2x\\sqrt{1-9{x}^{2}}}dx[\/latex]<\/li>\n<\/ol>\n\n[reveal-answer q=\"fs-id1167793361818\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793361818\"]\n\n<p id=\"fs-id1167793361818\">We can use [latex]u\\text{-substitution}[\/latex] in both cases.<\/p>\n<ol id=\"fs-id1167794296597\" style=\"list-style-type: lower-alpha;\">\n\t<li>Let [latex]u=2x.[\/latex] Then, [latex]du=2dx[\/latex] and we have<br>\n<div id=\"fs-id1167793883719\" class=\"equation unnumbered\">[latex]\\displaystyle\\int \\frac{1}{\\sqrt{4{x}^{2}-1}}dx=\\displaystyle\\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.[\/latex]<\/div>\n<\/li>\n\t<li>Let [latex]u=3x.[\/latex] Then, [latex]du=3dx[\/latex] and we obtain<br>\n<div id=\"fs-id1167793444519\" class=\"equation unnumbered\">[latex]\\displaystyle\\int \\frac{1}{2x\\sqrt{1-9{x}^{2}}}dx=\\frac{1}{2}\\displaystyle\\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.[\/latex]<\/div>\n<\/li>\n<\/ol>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Evaluate the following integral:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\int \\frac{x}{\\sqrt{x^2+9}} dx[\/latex]<\/p>\n<p><br>\n[reveal-answer q=\"745283\"]Show Answer[\/reveal-answer]<br>\n[hidden-answer a=\"745283\"]<\/p>\n<p>This integral suggests a u-substitution. Let [latex]u = x^2 + 9[\/latex].<\/p>\n<p>Then [latex]du = 2x dx[\/latex], or [latex]x dx = \\frac{1}{2} du[\/latex].<\/p>\n<p>Rewrite the integral in terms of [latex]u[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\int \\frac{x}{\\sqrt{x^2+9}} dx = \\frac{1}{2} \\int \\frac{1}{\\sqrt{u}} du[\/latex]<\/p>\n<p>This is a standard integral. We get:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{2} \\int \\frac{1}{\\sqrt{u}} du = \\frac{1}{2} \\cdot 2\\sqrt{u} + C = \\sqrt{u} + C[\/latex]<\/p>\n<p>Substitute back [latex]x^2 + 9[\/latex] for [latex]u[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{x^2 + 9} + C[\/latex]<\/p>\n<p>However, we can express this result using an inverse hyperbolic function:<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{x^2 + 9} = 3\\sqrt{1 + (\\frac{x}{3})^2} = 3\\sinh(\\sinh^{-1}(\\frac{x}{3}))[\/latex]<\/p>\n<p>Therefore, [latex]\\int \\frac{x}{\\sqrt{x^2+9}} dx = 3\\sinh(\\sinh^{-1}(\\frac{x}{3})) + C[\/latex].<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<h2>Applications of Hyperbolic Functions<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n\t<li class=\"whitespace-normal break-words\">Catenary Curves:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">A catenary is the curve formed by a hanging cable or chain under uniform gravity<\/li>\n\t<li class=\"whitespace-normal break-words\">Mathematically modeled by: [latex]y = a \\cosh(\\frac{x}{a})[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Arc Length Formula:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Used to calculate the length of a catenary<\/li>\n\t<li class=\"whitespace-normal break-words\">[latex]\\text{Arc Length} = \\int_{a}^{b} \\sqrt{1 + [f'(x)]^2} dx[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Hyperbolic Function Properties:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]1 + \\sinh^2 x = \\cosh^2 x[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">This identity is useful in simplifying arc length calculations<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p>Assume a hanging cable has the shape [latex]15\\text{cosh}\\left(\\frac{x}{15}\\right)[\/latex] for [latex]-20\\le x\\le 20.[\/latex] Determine the length of the cable (in feet).<\/p>\n\n[reveal-answer q=\"fs-id1167793292342\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793292342\"] [latex]52.95\\text{ft}[\/latex]\n\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>A suspension bridge cable forms a catenary with the equation [latex]y = 50 \\cosh(\\frac{x}{50})[\/latex], where [latex]x[\/latex] and [latex]y[\/latex] are measured in meters. The cable is anchored at [latex]x = -100[\/latex] and [latex]x = 100[\/latex]. Calculate the length of the cable.<\/p>\n<p><br>\n[reveal-answer q=\"728741\"]Show Answer[\/reveal-answer]<br>\n[hidden-answer a=\"728741\"]<\/p>\n<p>We use the arc length formula: [latex]\\text{Arc Length} = \\int_{a}^{b} \\sqrt{1 + [f'(x)]^2} dx[\/latex]<\/p>\n<p>First, find [latex]f'(x)[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]f'(x) = 50 \\cdot \\frac{1}{50} \\sinh(\\frac{x}{50}) = \\sinh(\\frac{x}{50})[\/latex]<\/p>\n<p>Substitute into the arc length formula:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Arc Length} = \\int_{-100}^{100} \\sqrt{1 + [\\sinh(\\frac{x}{50})]^2} dx[\/latex]<\/p>\n<p>Use the identity [latex]1 + \\sinh^2 x = \\cosh^2 x[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Arc Length} = \\int_{-100}^{100} \\cosh(\\frac{x}{50}) dx[\/latex]<\/p>\n<p>Integrate:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br>\n\\text{Arc Length} &amp;=&amp; 50 \\sinh(\\frac{x}{50}) \\Big|_{-100}^{100} \\\\<br>\n&amp;=&amp; 50 [\\sinh(2) - \\sinh(-2)] \\\\<br>\n&amp;=&amp; 100 \\sinh(2) \\\\<br>\n&amp;\\approx&amp; 364.1 \\text{ meters}<br>\n\\end{array}[\/latex]<\/p>\n<p>Therefore, the length of the cable is approximately [latex]364.1[\/latex] meters.<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Differentiate and integrate hyperbolic functions and their inverse forms<\/li>\n<li>Understand the practical situations where the catenary curve appears<\/li>\n<\/ul>\n<\/section>\n<h2>Derivatives and Integrals of the Hyperbolic Functions<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Definitions of Hyperbolic Functions:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]\\sinh x = \\frac{e^x - e^{-x}}{2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\cosh x = \\frac{e^x + e^{-x}}{2}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Key Derivatives:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx}(\\sinh x) = \\cosh x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx}(\\cosh x) = \\sinh x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx}(\\tanh x) = \\text{sech}^2 x[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Key Integrals:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]\\int \\sinh x , dx = \\cosh x + C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\int \\cosh x , dx = \\sinh x + C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\int \\text{sech}^2 x , dx = \\tanh x + C[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Similarities with Trigonometric Functions:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Derivatives of sinh and sin are similar<\/li>\n<li class=\"whitespace-normal break-words\">But [latex]\\frac{d}{dx}(\\cos x) = -\\sin x[\/latex], while [latex]\\frac{d}{dx}(\\cosh x) = \\sinh x[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">U-substitution is often useful for integrals involving hyperbolic functions<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p id=\"fs-id1167794028505\">Evaluate the following derivatives:<\/p>\n<ol id=\"fs-id1167794043812\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dx}(\\text{tanh}({x}^{2}+3x))[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}(\\dfrac{1}{{(\\text{sinh}x)}^{2}})[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794187159\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167794187159\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dx}(\\text{tanh}({x}^{2}+3x))=({\\text{sech}}^{2}({x}^{2}+3x))(2x+3)[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}(\\frac{1}{{(\\text{sinh}x)}^{2}})=\\frac{d}{dx}{(\\text{sinh}x)}^{-2}=-2{(\\text{sinh}x)}^{-3}\\text{cosh}x[\/latex]<\/li>\n<\/ol>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ozRCEFk1tmA?controls=0&amp;start=134&amp;end=300&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.9CalculusOfHyperbolicFunctions134to300_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.9 Calculus of Hyperbolic Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1167793819956\">Evaluate the following integrals:<\/p>\n<ol id=\"fs-id1167793985861\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\displaystyle\\int {\\text{sinh}}^{3}x\\text{cosh}xdx[\/latex]<\/li>\n<li>[latex]\\displaystyle\\int {\\text{sech}}^{2}(3x)dx[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793373829\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793373829\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1167793373829\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\displaystyle\\int {\\text{sinh}}^{3}x\\text{cosh}xdx=\\frac{{\\text{sinh}}^{4}x}{4}+C[\/latex]<\/li>\n<li>[latex]\\displaystyle\\int {\\text{sech}}^{2}(3x)dx=\\frac{\\text{tanh}(3x)}{3}+C[\/latex]<\/li>\n<\/ol>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ozRCEFk1tmA?controls=0&amp;start=478&amp;end=598&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.9CalculusOfHyperbolicFunctions478to598_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.9 Calculus of Hyperbolic Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Evaluate the following integral:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\int x \\sinh(x^2) , dx[\/latex]<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q773910\">Show Answer<\/button><\/p>\n<div id=\"q773910\" class=\"hidden-answer\" style=\"display: none\">\n<p>This integral suggests a u-substitution. Let [latex]u = x^2[\/latex].<\/p>\n<p>Then [latex]du = 2x , dx[\/latex], or [latex]x , dx = \\frac{1}{2} du[\/latex].<\/p>\n<p>Rewrite the integral in terms of u:<\/p>\n<p style=\"text-align: center;\">[latex]\\int x \\sinh(x^2) , dx = \\int \\frac{1}{2} \\sinh(u) , du[\/latex]<\/p>\n<p>Now we can use the integration formula for sinh:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{2} \\int \\sinh(u) , du = \\frac{1}{2} \\cosh(u) + C[\/latex]<\/p>\n<p>Substitute back [latex]x^2[\/latex] for [latex]u[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{2} \\cosh(x^2) + C[\/latex]<\/p>\n<p>Therefore, [latex]\\int x \\sinh(x^2) , dx = \\frac{1}{2} \\cosh(x^2) + C[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h2>Calculus of Inverse Hyperbolic Functions<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">All hyperbolic functions have inverses with appropriate range restrictions<\/li>\n<li class=\"whitespace-normal break-words\">Key Derivatives:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx}(\\sinh^{-1} x) = \\frac{1}{\\sqrt{1+x^2}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx}(\\cosh^{-1} x) = \\frac{1}{\\sqrt{x^2-1}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\frac{d}{dx}(\\tanh^{-1} x) = \\frac{1}{1-x^2}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Key Integrals:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]\\int \\frac{1}{\\sqrt{1+u^2}} du = \\sinh^{-1} u + C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\int \\frac{1}{\\sqrt{u^2-1}} du = \\cosh^{-1} u + C[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\int \\frac{1}{1-u^2} du = \\begin{cases} \\tanh^{-1} u + C & \\text{if } |u| < 1 \\ \\coth^{-1} u + C & \\text{if } |u| > 1 \\end{cases}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Each inverse hyperbolic function has specific domain and range restrictions<\/li>\n<li class=\"whitespace-normal break-words\">Derivatives of inverse hyperbolic functions are found using implicit differentiation<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p id=\"fs-id1167794332395\">Evaluate the following derivatives:<\/p>\n<ol id=\"fs-id1167793590727\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dx}({\\text{cosh}}^{-1}(3x))[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}{({\\text{coth}}^{-1}x)}^{3}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793498614\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793498614\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1167793498614\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\frac{d}{dx}({\\text{cosh}}^{-1}(3x))=\\frac{3}{\\sqrt{9{x}^{2}-1}}[\/latex]<\/li>\n<li>[latex]\\frac{d}{dx}{({\\text{coth}}^{-1}x)}^{3}=\\frac{3{({\\text{coth}}^{-1}x)}^{2}}{1-{x}^{2}}[\/latex]<\/li>\n<\/ol>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ozRCEFk1tmA?controls=0&amp;start=827&amp;end=907&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.9CalculusOfHyperbolicFunctions827to907_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.9 Calculus of Hyperbolic Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1167793871641\">Evaluate the following integrals:<\/p>\n<ol id=\"fs-id1167793937415\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\displaystyle\\int \\frac{1}{\\sqrt{4{x}^{2}-1}}dx[\/latex]<\/li>\n<li>[latex]\\displaystyle\\int \\frac{1}{2x\\sqrt{1-9{x}^{2}}}dx[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793361818\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793361818\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793361818\">We can use [latex]u\\text{-substitution}[\/latex] in both cases.<\/p>\n<ol id=\"fs-id1167794296597\" style=\"list-style-type: lower-alpha;\">\n<li>Let [latex]u=2x.[\/latex] Then, [latex]du=2dx[\/latex] and we have\n<div id=\"fs-id1167793883719\" class=\"equation unnumbered\">[latex]\\displaystyle\\int \\frac{1}{\\sqrt{4{x}^{2}-1}}dx=\\displaystyle\\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.[\/latex]<\/div>\n<\/li>\n<li>Let [latex]u=3x.[\/latex] Then, [latex]du=3dx[\/latex] and we obtain\n<div id=\"fs-id1167793444519\" class=\"equation unnumbered\">[latex]\\displaystyle\\int \\frac{1}{2x\\sqrt{1-9{x}^{2}}}dx=\\frac{1}{2}\\displaystyle\\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.[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Evaluate the following integral:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\int \\frac{x}{\\sqrt{x^2+9}} dx[\/latex]<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q745283\">Show Answer<\/button><\/p>\n<div id=\"q745283\" class=\"hidden-answer\" style=\"display: none\">\n<p>This integral suggests a u-substitution. Let [latex]u = x^2 + 9[\/latex].<\/p>\n<p>Then [latex]du = 2x dx[\/latex], or [latex]x dx = \\frac{1}{2} du[\/latex].<\/p>\n<p>Rewrite the integral in terms of [latex]u[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\int \\frac{x}{\\sqrt{x^2+9}} dx = \\frac{1}{2} \\int \\frac{1}{\\sqrt{u}} du[\/latex]<\/p>\n<p>This is a standard integral. We get:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{2} \\int \\frac{1}{\\sqrt{u}} du = \\frac{1}{2} \\cdot 2\\sqrt{u} + C = \\sqrt{u} + C[\/latex]<\/p>\n<p>Substitute back [latex]x^2 + 9[\/latex] for [latex]u[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{x^2 + 9} + C[\/latex]<\/p>\n<p>However, we can express this result using an inverse hyperbolic function:<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{x^2 + 9} = 3\\sqrt{1 + (\\frac{x}{3})^2} = 3\\sinh(\\sinh^{-1}(\\frac{x}{3}))[\/latex]<\/p>\n<p>Therefore, [latex]\\int \\frac{x}{\\sqrt{x^2+9}} dx = 3\\sinh(\\sinh^{-1}(\\frac{x}{3})) + C[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h2>Applications of Hyperbolic Functions<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Catenary Curves:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">A catenary is the curve formed by a hanging cable or chain under uniform gravity<\/li>\n<li class=\"whitespace-normal break-words\">Mathematically modeled by: [latex]y = a \\cosh(\\frac{x}{a})[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Arc Length Formula:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Used to calculate the length of a catenary<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{Arc Length} = \\int_{a}^{b} \\sqrt{1 + [f'(x)]^2} dx[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Hyperbolic Function Properties:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]1 + \\sinh^2 x = \\cosh^2 x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">This identity is useful in simplifying arc length calculations<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p>Assume a hanging cable has the shape [latex]15\\text{cosh}\\left(\\frac{x}{15}\\right)[\/latex] for [latex]-20\\le x\\le 20.[\/latex] Determine the length of the cable (in feet).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793292342\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793292342\" class=\"hidden-answer\" style=\"display: none\"> [latex]52.95\\text{ft}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>A suspension bridge cable forms a catenary with the equation [latex]y = 50 \\cosh(\\frac{x}{50})[\/latex], where [latex]x[\/latex] and [latex]y[\/latex] are measured in meters. The cable is anchored at [latex]x = -100[\/latex] and [latex]x = 100[\/latex]. Calculate the length of the cable.<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q728741\">Show Answer<\/button><\/p>\n<div id=\"q728741\" class=\"hidden-answer\" style=\"display: none\">\n<p>We use the arc length formula: [latex]\\text{Arc Length} = \\int_{a}^{b} \\sqrt{1 + [f'(x)]^2} dx[\/latex]<\/p>\n<p>First, find [latex]f'(x)[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]f'(x) = 50 \\cdot \\frac{1}{50} \\sinh(\\frac{x}{50}) = \\sinh(\\frac{x}{50})[\/latex]<\/p>\n<p>Substitute into the arc length formula:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Arc Length} = \\int_{-100}^{100} \\sqrt{1 + [\\sinh(\\frac{x}{50})]^2} dx[\/latex]<\/p>\n<p>Use the identity [latex]1 + \\sinh^2 x = \\cosh^2 x[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Arc Length} = \\int_{-100}^{100} \\cosh(\\frac{x}{50}) dx[\/latex]<\/p>\n<p>Integrate:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/> \\text{Arc Length} &=& 50 \\sinh(\\frac{x}{50}) \\Big|_{-100}^{100} \\\\<br \/> &=& 50 [\\sinh(2) - \\sinh(-2)] \\\\<br \/> &=& 100 \\sinh(2) \\\\<br \/> &\\approx& 364.1 \\text{ meters}<br \/> \\end{array}[\/latex]<\/p>\n<p>Therefore, the length of the cable is approximately [latex]364.1[\/latex] meters.<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":6,"menu_order":18,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":479,"module-header":"","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/497"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":0,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/497\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/479"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/497\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=497"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=497"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=497"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=497"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}