{"id":388,"date":"2023-07-03T10:10:06","date_gmt":"2023-07-03T10:10:06","guid":{"rendered":"https:\/\/mathority.org\/pt\/derivado-de-arcosina-hiperbolico\/"},"modified":"2023-07-03T10:10:06","modified_gmt":"2023-07-03T10:10:06","slug":"derivado-de-arcosina-hiperbolico","status":"publish","type":"post","link":"https:\/\/mathority.org\/pt\/derivado-de-arcosina-hiperbolico\/","title":{"rendered":"Derivada do arco cosseno hiperb\u00f3lico"},"content":{"rendered":"<p>Nesta p\u00e1gina voc\u00ea ver\u00e1 qual \u00e9 a derivada do arco cosseno hiperb\u00f3lico (f\u00f3rmula). Voc\u00ea tamb\u00e9m encontrar\u00e1 exerc\u00edcios resolvidos passo a passo para derivadas do arco cosseno hiperb\u00f3lico de uma fun\u00e7\u00e3o. E, por fim, voc\u00ea encontrar\u00e1 a demonstra\u00e7\u00e3o da f\u00f3rmula da derivada desse tipo de fun\u00e7\u00e3o trigonom\u00e9trica. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"formula-de-la-derivada-del-arcocoseno-hiperbolico\"><\/span> F\u00f3rmula para a derivada do arco cosseno hiperb\u00f3lico<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> <strong>A derivada do arco cosseno hiperb\u00f3lico de x \u00e9 um sobre a raiz quadrada de x ao quadrado menos 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> Portanto, a <strong>derivada do arco cosseno hiperb\u00f3lico de uma fun\u00e7\u00e3o<\/strong> \u00e9 igual ao quociente da derivada dessa fun\u00e7\u00e3o dividido pela raiz quadrada dessa fun\u00e7\u00e3o ao quadrado menos um.<\/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> A segunda f\u00f3rmula inclui a regra da cadeia e, portanto, pode ser usada para derivar qualquer arco cosseno hiperb\u00f3lico. Na verdade, se substituirmos x por u, obteremos a primeira f\u00f3rmula. Em vez disso, a primeira f\u00f3rmula s\u00f3 funciona para a derivada arcocoseno hiperb\u00f3lica de 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=\"derivada do arco cosseno hiperb\u00f3lico\" class=\"wp-image-2337\" width=\"403\" height=\"305\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<p> O arco cosseno hiperb\u00f3lico \u00e9 a fun\u00e7\u00e3o inversa do cosseno hiperb\u00f3lico e, portanto, as duas fun\u00e7\u00f5es est\u00e3o relacionadas. Voc\u00ea pode ver a f\u00f3rmula da derivada desta fun\u00e7\u00e3o trigonom\u00e9trica clicando aqui:<\/p>\n<p> <span style=\"color:#ff951b\">\u27a4<\/span> <strong>Veja:<\/strong> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/pt\/derivada-de-cosseno-hiperbolico\/\">f\u00f3rmula da derivada do cosseno hiperb\u00f3lico<\/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> Exemplos da derivada hiperb\u00f3lica do arco seno <span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\" id=\"block-46cfc7df-b680-41c2-ad53-bd8a19834b32\"> Exemplo 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\"> Para encontrar a derivada do arco cosseno hiperb\u00f3lico, precisamos usar sua f\u00f3rmula correspondente, que \u00e9:<\/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\"> Portanto, no numerador da fra\u00e7\u00e3o precisamos colocar a derivada de 5x, que \u00e9 5. E no denominador s\u00f3 precisamos colocar a raiz quadrada da fun\u00e7\u00e3o argumento ao quadrado menos 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-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\"> Exemplo 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\"> A fun\u00e7\u00e3o a ser derivada deste exerc\u00edcio \u00e9 um arco cosseno hiperb\u00f3lico, ent\u00e3o usamos a seguinte f\u00f3rmula para deriv\u00e1-la:<\/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\"> Assim, no numerador escrevemos a derivada do argumento da fun\u00e7\u00e3o e no denominador a raiz quadrada da fun\u00e7\u00e3o do argumento elevado a 2 menos 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> Prova da derivada do arco cosseno hiperb\u00f3lico<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Por fim, demonstraremos a f\u00f3rmula da derivada do arco cosseno hiperb\u00f3lico.<\/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> Primeiro, transformamos o cosseno do arco hiperb\u00f3lico em um cosseno hiperb\u00f3lico:<\/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> Deduzimos de ambos os lados da igualdade:<\/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> N\u00f3s limpamos voc\u00ea:<\/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> Agora usamos a identidade trigonom\u00e9trica que relaciona o seno hiperb\u00f3lico e o cosseno hiperb\u00f3lico para modificar o denominador:<\/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> Por\u00e9m, primeiro, deduzimos que x \u00e9 equivalente ao cosseno hiperb\u00f3lico de y, ent\u00e3o a equa\u00e7\u00e3o permanece: <\/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> itens similares<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/pt\/derivada-da-cossecante-hiperbolica\/\">F\u00f3rmula para a derivada da cossecante hiperb\u00f3lica<\/a><\/span><\/li>\n<li> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/pt\/derivado-de-arcossecante\/\">F\u00f3rmula derivada de arcossecante<\/a><\/span><\/li>\n<li> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/pt\/deriva-da-cossecante\/\">F\u00f3rmula derivada cossecante<\/a><\/span><\/li>\n<li> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/pt\/derivado-de-larcosina-1\/\">F\u00f3rmula derivada de arcosina<\/a><\/span><\/li>\n<li> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/pt\/deriva-do-cosseno\/\">F\u00f3rmula derivada de cosseno<\/a><\/span><\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Nesta p\u00e1gina voc\u00ea ver\u00e1 qual \u00e9 a derivada do arco cosseno hiperb\u00f3lico (f\u00f3rmula). Voc\u00ea tamb\u00e9m encontrar\u00e1 exerc\u00edcios resolvidos passo a passo para derivadas do arco cosseno hiperb\u00f3lico de uma fun\u00e7\u00e3o. E, por fim, voc\u00ea encontrar\u00e1 a demonstra\u00e7\u00e3o da f\u00f3rmula da derivada desse tipo de fun\u00e7\u00e3o trigonom\u00e9trica. F\u00f3rmula para a derivada do arco cosseno hiperb\u00f3lico A &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathority.org\/pt\/derivado-de-arcosina-hiperbolico\/\"> <span class=\"screen-reader-text\">Derivada do arco cosseno hiperb\u00f3lico<\/span> Leia mais &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":[11],"tags":[],"class_list":["post-388","post","type-post","status-publish","format-standard","hentry","category-derivados"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.2 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Derivada do cosseno do arco hiperb\u00f3lico - 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\/pt\/derivado-de-arcosina-hiperbolico\/\" \/>\n<meta property=\"og:locale\" content=\"pt_BR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Derivada do cosseno do arco hiperb\u00f3lico - Mathority\" \/>\n<meta property=\"og:description\" content=\"Nesta p\u00e1gina voc\u00ea ver\u00e1 qual \u00e9 a derivada do arco cosseno hiperb\u00f3lico (f\u00f3rmula). Voc\u00ea tamb\u00e9m encontrar\u00e1 exerc\u00edcios resolvidos passo a passo para derivadas do arco cosseno hiperb\u00f3lico de uma fun\u00e7\u00e3o. E, por fim, voc\u00ea encontrar\u00e1 a demonstra\u00e7\u00e3o da f\u00f3rmula da derivada desse tipo de fun\u00e7\u00e3o trigonom\u00e9trica. 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