{"id":433,"date":"2023-07-03T10:10:06","date_gmt":"2023-07-03T10:10:06","guid":{"rendered":"https:\/\/mathority.org\/nl\/hyperbolisch-arccosinederivaat\/"},"modified":"2023-07-03T10:10:06","modified_gmt":"2023-07-03T10:10:06","slug":"hyperbolisch-arccosinederivaat","status":"publish","type":"post","link":"https:\/\/mathority.org\/nl\/hyperbolisch-arccosinederivaat\/","title":{"rendered":"Afgeleide van hyperbolische boogcosinus"},"content":{"rendered":"<p>Op deze pagina ziet u wat de afgeleide is van de hyperbolische boogcosinus (formule). Je vindt ook oefeningen die stap voor stap worden opgelost voor afgeleiden van de hyperbolische boogcosinus van een functie. En ten slotte vind je de demonstratie van de formule voor de afgeleide van dit type trigonometrische functie. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"formula-de-la-derivada-del-arcocoseno-hiperbolico\"><\/span> Formule voor de afgeleide van de hyperbolische boogcosinus<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> <strong>De afgeleide van de hyperbolische arccosinus van x is \u00e9\u00e9n gedeeld door de wortel van x in het kwadraat min 1.<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-cf6cb0ef7aae071322695ae7c8455d1a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(x)=\\text{arccosh}(x) \\quad\\color{orange}\\bm{\\longrightarrow}\\quad\\color{black} f'(x)=\\cfrac{1}{\\sqrt{x^2-1}}\" title=\"Rendered by QuickLaTeX.com\" height=\"42\" width=\"426\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p> Daarom is de <strong>afgeleide van de hyperbolische boogcosinus van een functie<\/strong> gelijk aan het quoti\u00ebnt van de afgeleide van die functie gedeeld door de vierkantswortel van die functie in het kwadraat min \u00e9\u00e9n.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-675bee296952a25c6048af071e7ce4e6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(x)=\\text{arccosh}(u) \\quad\\color{orange}\\bm{\\longrightarrow}\\quad\\color{black} f'(x)=\\cfrac{u'}{\\sqrt{u^2-1}}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"427\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p> De tweede formule omvat de kettingregel en kan daarom worden gebruikt om elke hyperbolische arccosinus af te leiden. Als we x vervangen door de u, krijgen we in feite de eerste formule. In plaats daarvan werkt de eerste formule alleen voor de hyperbolische arccosinusderivaat van x. <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/derivee-de-larccosine-hyperbolique.webp\" alt=\"afgeleide van de hyperbolische boogcosinus\" class=\"wp-image-2337\" width=\"403\" height=\"305\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<p> Hyperbolische arccosinus is de inverse functie van hyperbolische cosinus, en daarom zijn de twee functies gerelateerd. U kunt de formule voor de afgeleide van deze trigonometrische functie bekijken door hier te klikken:<\/p>\n<p> <span style=\"color:#ff951b\">\u27a4<\/span> <strong>Zie:<\/strong> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/nl\/hyperbolische-cosinusderivaat\/\">formule voor de afgeleide van de hyperbolische cosinus<\/a><\/span> <span style=\"text-decoration: underline;\"><\/span><\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"ejemplos-de-la-derivada-del-arcoseno-hiperbolico\"><\/span> Voorbeelden van het hyperbolische arcsinusderivaat <span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\" id=\"block-46cfc7df-b680-41c2-ad53-bd8a19834b32\"> voorbeeld 1<\/h3>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-83158beb1fa0dda8f8a6469cc6f87cd0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(x)=\\text{arccosh}(5x)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"147\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p id=\"block-2a112ce1-0dbe-43d5-95b3-4d8506c1a246\"> Om de afgeleide van de hyperbolische arccosinus te vinden, moeten we de overeenkomstige formule gebruiken, namelijk:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-675bee296952a25c6048af071e7ce4e6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(x)=\\text{arccosh}(u) \\quad\\color{orange}\\bm{\\longrightarrow}\\quad\\color{black} f'(x)=\\cfrac{u'}{\\sqrt{u^2-1}}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"427\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p id=\"block-a4fe1876-f662-49c1-8d09-6a6c4b5528dd\"> Daarom moeten we in de teller van de breuk de afgeleide van 5x plaatsen, wat 5 is. En in de noemer hoeven we alleen maar de vierkantswortel van de argumentfunctie in het kwadraat min 1 te plaatsen: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8704c0fb8ce82efc60e67bbca363b205_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(x)=\\text{arccosh}(5x) \\quad\\color{orange}\\bm{\\longrightarrow}\\quad\\color{black} f'(x)=\\cfrac{5}{\\sqrt{(5x)^2-1}}=\\cfrac{5}{\\sqrt{25x^2-1}}\" title=\"Rendered by QuickLaTeX.com\" height=\"47\" width=\"567\" style=\"vertical-align: -20px;\"><\/p>\n<\/p>\n<h3 class=\"wp-block-heading\" id=\"block-1446420a-0d61-44d3-9e31-8c5935a432a7\"> Voorbeeld 2<\/h3>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-07f0e806641b4f8ddc3441faa6059bc0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(x)=\\text{arccosh}(x^4-5x^2)\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"194\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p id=\"block-0514a2ea-d85a-4b25-a7db-9c27533e7436\"> De uit deze oefening af te leiden functie is een hyperbolische arccosinus, dus gebruiken we de volgende formule om deze af te leiden:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-675bee296952a25c6048af071e7ce4e6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(x)=\\text{arccosh}(u) \\quad\\color{orange}\\bm{\\longrightarrow}\\quad\\color{black} f'(x)=\\cfrac{u'}{\\sqrt{u^2-1}}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"427\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p id=\"block-6abf3c5a-c400-48c6-8375-c05fcb255b20\"> Dus schrijven we in de teller de afgeleide van het argument van de functie en in de noemer de vierkantswortel van de functie van het argument verhoogd tot 2 min 1: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-376cce9ffe78b04c056d4aa90af1be3d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(x)=\\text{arccosh}(x^4-5x^2) \\quad\\color{orange}\\bm{\\longrightarrow}\\quad\\color{black} f'(x)=\\cfrac{4x^3-10x}{\\sqrt{\\left(x^4-5x^2\\right)^2-1}}\" title=\"Rendered by QuickLaTeX.com\" height=\"59\" width=\"555\" style=\"vertical-align: -30px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"demostracion-de-la-derivada-del-arcocoseno-hiperbolico\"><\/span> Bewijs van de afgeleide van de hyperbolische boogcosinus<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Ten slotte zullen we de formule demonstreren voor de afgeleide van de hyperbolische boogcosinus.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-48f91a3e46d7a7275e54619efd3ede1a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y=\\text{arccosh}(x)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"113\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Eerst transformeren we de hyperbolische boogcosinus in een hyperbolische cosinus:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7032f56de52ca8602783ca5f4bbb0767_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x=\\text{cosh}(y)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"90\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Van beide kanten van de gelijkheid leiden we het volgende af:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-56122b4a5ee02b2c4c6f62751b5b21b6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"1=\\text{senh}(y)\\cdot y'\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"115\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Wij zuiveren u:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7848ebafab6456313be4d737917b898a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y'=\\cfrac{1}{\\text{senh}(y)}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"97\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p> We gebruiken nu de trigonometrische identiteit die de hyperbolische sinus en de hyperbolische cosinus met elkaar in verband brengt om de noemer te wijzigen:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c45c9cefda55a71bba2eb8cfe1c3a861_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{cosh}^2(y)-\\text{senh}^2(y)=1 \\ \\longrightarrow \\ \\text{senh}(y)=\\sqrt{\\text{cosh}^2(y)-1}\" title=\"Rendered by QuickLaTeX.com\" height=\"32\" width=\"428\" style=\"vertical-align: -9px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5b7a01a3b50f15a8b5ac0df596cfb1e0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y'=\\cfrac{1}{\\sqrt{\\text{cosh}^2(y)-1}}\" title=\"Rendered by QuickLaTeX.com\" height=\"56\" width=\"152\" style=\"vertical-align: -30px;\"><\/p>\n<\/p>\n<p> Maar eerst hebben we afgeleid dat x equivalent is aan de hyperbolische cosinus van y, dus de vergelijking blijft: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-01f9e07a0c0f68a771edda2395685299_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y'=\\cfrac{1}{\\sqrt{x^2-1}}\" title=\"Rendered by QuickLaTeX.com\" height=\"42\" width=\"103\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"articulos-relacionados\"><\/span> Gelijkwaardige producten<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/nl\/afgeleide-van-de-hyperbolische-cosecans\/\">Formule voor de afgeleide van de hyperbolische cosecans<\/a><\/span><\/li>\n<li> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/nl\/arccosecante-afgeleide\/\">Arccosecant-afgeleide formule<\/a><\/span><\/li>\n<li> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/nl\/is-afgeleid-van-de-cosecans\/\">Cosecant-afgeleide formule<\/a><\/span><\/li>\n<li> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/nl\/larccosinederivaat\/\">Arccosine-afgeleide formule<\/a><\/span><\/li>\n<li> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/nl\/komt-voort-uit-de-cosinus\/\">Cosinus afgeleide formule<\/a><\/span><\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Op deze pagina ziet u wat de afgeleide is van de hyperbolische boogcosinus (formule). Je vindt ook oefeningen die stap voor stap worden opgelost voor afgeleiden van de hyperbolische boogcosinus van een functie. En ten slotte vind je de demonstratie van de formule voor de afgeleide van dit type trigonometrische functie. Formule voor de afgeleide &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathority.org\/nl\/hyperbolisch-arccosinederivaat\/\"> <span class=\"screen-reader-text\">Afgeleide van hyperbolische boogcosinus<\/span> Lees meer &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"","footnotes":""},"categories":[38],"tags":[],"class_list":["post-433","post","type-post","status-publish","format-standard","hentry","category-derivaten"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.2 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Afgeleide van hyperbolische boogcosinus - Mathority<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathority.org\/nl\/hyperbolisch-arccosinederivaat\/\" \/>\n<meta property=\"og:locale\" content=\"nl_NL\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Afgeleide van hyperbolische boogcosinus - Mathority\" \/>\n<meta property=\"og:description\" content=\"Op deze pagina ziet u wat de afgeleide is van de hyperbolische boogcosinus (formule). 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Formule voor de afgeleide &hellip; Afgeleide van hyperbolische boogcosinus Lees meer &raquo;\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mathority.org\/nl\/hyperbolisch-arccosinederivaat\/\" \/>\n<meta property=\"article:published_time\" content=\"2023-07-03T10:10:06+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-cf6cb0ef7aae071322695ae7c8455d1a_l3.png\" \/>\n<meta name=\"author\" content=\"Redactioneel Team\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Geschreven door\" \/>\n\t<meta name=\"twitter:data1\" content=\"Redactioneel Team\" \/>\n\t<meta name=\"twitter:label2\" content=\"Geschatte leestijd\" \/>\n\t<meta name=\"twitter:data2\" content=\"2 minuten\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"WebPage\",\"@id\":\"https:\/\/mathority.org\/nl\/hyperbolisch-arccosinederivaat\/\",\"url\":\"https:\/\/mathority.org\/nl\/hyperbolisch-arccosinederivaat\/\",\"name\":\"Afgeleide van hyperbolische boogcosinus - Mathority\",\"isPartOf\":{\"@id\":\"https:\/\/mathority.org\/nl\/#website\"},\"datePublished\":\"2023-07-03T10:10:06+00:00\",\"dateModified\":\"2023-07-03T10:10:06+00:00\",\"author\":{\"@id\":\"https:\/\/mathority.org\/nl\/#\/schema\/person\/19b550cef1a9fbd238be112b7b7bbf64\"},\"breadcrumb\":{\"@id\":\"https:\/\/mathority.org\/nl\/hyperbolisch-arccosinederivaat\/#breadcrumb\"},\"inLanguage\":\"nl-NL\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/mathority.org\/nl\/hyperbolisch-arccosinederivaat\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/mathority.org\/nl\/hyperbolisch-arccosinederivaat\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/mathority.org\/nl\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Afgeleide van hyperbolische boogcosinus\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/mathority.org\/nl\/#website\",\"url\":\"https:\/\/mathority.org\/nl\/\",\"name\":\"\",\"description\":\"Waar nieuwsgierigheid en berekening elkaar ontmoeten!\",\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/mathority.org\/nl\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"nl-NL\"},{\"@type\":\"Person\",\"@id\":\"https:\/\/mathority.org\/nl\/#\/schema\/person\/19b550cef1a9fbd238be112b7b7bbf64\",\"name\":\"Redactioneel Team\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"nl-NL\",\"@id\":\"https:\/\/mathority.org\/nl\/#\/schema\/person\/image\/\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/8a35e4c8616d1c34c03ca02862b580f4372c5650665668489db53a09579bbc4f?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/8a35e4c8616d1c34c03ca02862b580f4372c5650665668489db53a09579bbc4f?s=96&d=mm&r=g\",\"caption\":\"Redactioneel Team\"},\"sameAs\":[\"http:\/\/mathority.org\/nl\"]}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"Afgeleide van hyperbolische boogcosinus - Mathority","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/mathority.org\/nl\/hyperbolisch-arccosinederivaat\/","og_locale":"nl_NL","og_type":"article","og_title":"Afgeleide van hyperbolische boogcosinus - Mathority","og_description":"Op deze pagina ziet u wat de afgeleide is van de hyperbolische boogcosinus (formule). Je vindt ook oefeningen die stap voor stap worden opgelost voor afgeleiden van de hyperbolische boogcosinus van een functie. En ten slotte vind je de demonstratie van de formule voor de afgeleide van dit type trigonometrische functie. 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