{"id":397,"date":"2023-07-03T02:11:56","date_gmt":"2023-07-03T02:11:56","guid":{"rendered":"https:\/\/mathority.org\/pt\/derivada-do-arco-secante-hiperbolico\/"},"modified":"2023-07-03T02:11:56","modified_gmt":"2023-07-03T02:11:56","slug":"derivada-do-arco-secante-hiperbolico","status":"publish","type":"post","link":"https:\/\/mathority.org\/pt\/derivada-do-arco-secante-hiperbolico\/","title":{"rendered":"Derivada do arco secante hiperb\u00f3lico"},"content":{"rendered":"<p>Aqui voc\u00ea descobrir\u00e1 como calcular a derivada do arco secante hiperb\u00f3lico de uma fun\u00e7\u00e3o. Al\u00e9m disso, voc\u00ea poder\u00e1 ver exemplos resolvidos da derivada do arco secante hiperb\u00f3lico. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"formula-de-la-derivada-de-la-arcosecante-hiperbolica\"><\/span> F\u00f3rmula derivada do arco secante hiperb\u00f3lico<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> <strong>A derivada do arco secante hiperb\u00f3lico de x \u00e9 igual a menos 1 dividido pelo produto de x vezes a raiz de um menos x ao quadrado.<\/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-30fd2d0fe7abd6d3774eaff22e8e762e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(x)=\\text{arcsech}(x) \\quad\\color{orange}\\bm{\\longrightarrow}\\quad\\color{black} f'(x)=\\cfrac{-1}{x\\sqrt{1-x^2}}\" title=\"Rendered by QuickLaTeX.com\" height=\"42\" width=\"435\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p> Portanto, a <strong>derivada do arco secante hiperb\u00f3lico de uma fun\u00e7\u00e3o<\/strong> \u00e9 menos a derivada dessa fun\u00e7\u00e3o dividida pelo produto da fun\u00e7\u00e3o vezes a raiz de um menos a fun\u00e7\u00e3o quadrada.<\/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-12cb116a10cdca4bf5b49f2d06d69a58_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(x)=\\text{arcsech}(u) \\quad\\color{orange}\\bm{\\longrightarrow}\\quad\\color{black} f'(x)=\\cfrac{-u'}{u\\sqrt{1-u^2}}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"435\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p> Resumindo, a f\u00f3rmula para a derivada da fun\u00e7\u00e3o arcosecante hiperb\u00f3lica \u00e9: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/derive-de-larcosecant-hyperbolique.webp\" alt=\"derivado do arco secante hiperb\u00f3lico\" class=\"wp-image-2786\" width=\"395\" height=\"281\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<p> Na verdade, ambas as express\u00f5es correspondem \u00e0 mesma f\u00f3rmula, mas a regra da cadeia \u00e9 aplicada \u00e0 segunda f\u00f3rmula. Na verdade, se voc\u00ea substituir u pela fun\u00e7\u00e3o identidade x, obter\u00e1 a primeira f\u00f3rmula, pois a derivada de x \u00e9 1. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"ejemplos-de-la-derivada-de-la-arcosecante-hiperbolica\"><\/span> Exemplos de derivada do arco secante hiperb\u00f3lico<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Depois de ver qual \u00e9 a f\u00f3rmula da derivada do arco secante hiperb\u00f3lico, resolveremos passo a passo dois exerc\u00edcios deste tipo de derivadas trigonom\u00e9tricas inversas. Assim, voc\u00ea pode ver exatamente como derivar o arco secante hiperb\u00f3lico de uma fun\u00e7\u00e3o.<\/p>\n<h3 class=\"wp-block-heading\"> Exemplo 1<\/h3>\n<p> Neste exemplo, determinaremos qual \u00e9 a derivada do arco secante hiperb\u00f3lico 2x.<\/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-a4e859277c6f50c7ea081153c8e79781_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(x)=\\text{arcsech}(2x)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"146\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> No argumento do arco secante hiperb\u00f3lico, temos uma fun\u00e7\u00e3o diferente de x, ent\u00e3o precisamos usar a f\u00f3rmula da regra da cadeia 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-12cb116a10cdca4bf5b49f2d06d69a58_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(x)=\\text{arcsech}(u) \\quad\\color{orange}\\bm{\\longrightarrow}\\quad\\color{black} f'(x)=\\cfrac{-u'}{u\\sqrt{1-u^2}}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"435\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p> A fun\u00e7\u00e3o 2x \u00e9 linear, ent\u00e3o sua derivada \u00e9 2. Portanto, para encontrar a derivada, simplesmente substitu\u00edmos 2x por u e 2 por u&#8217; na f\u00f3rmula:<\/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-85f083f6c0009277265cca483ec04ac9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(x)=\\text{arcsech}(2x) \\quad\\color{orange}\\bm{\\longrightarrow}\\quad\\color{black} f'(x)=\\cfrac{-2}{2x\\sqrt{1-(2x)^2}}=\\cfrac{-2}{2x\\sqrt{1-4x^2}}\" title=\"Rendered by QuickLaTeX.com\" height=\"92\" width=\"582\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<h3 class=\"wp-block-heading\"> Exemplo 2<\/h3>\n<p> Neste segundo exerc\u00edcio, derivaremos o arco secante hiperb\u00f3lico de uma fun\u00e7\u00e3o polinomial:<\/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-0fc71b56d622872d79c469d74504f0fe_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(x)=\\text{arcsech}(x^3-4x)\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"185\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> A fun\u00e7\u00e3o deste exerc\u00edcio \u00e9 composta, pois o arco secante hiperb\u00f3lico tem outra fun\u00e7\u00e3o em seu argumento. Portanto, precisamos usar a f\u00f3rmula da derivada do arco secante hiperb\u00f3lico com a regra da cadeia para fazer sua deriva\u00e7\u00e3o:<\/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-12cb116a10cdca4bf5b49f2d06d69a58_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(x)=\\text{arcsech}(u) \\quad\\color{orange}\\bm{\\longrightarrow}\\quad\\color{black} f'(x)=\\cfrac{-u'}{u\\sqrt{1-u^2}}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"435\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p> Portanto, no numerador da fra\u00e7\u00e3o colocamos a derivada da fun\u00e7\u00e3o polinomial do argumento, e no denominador trocamos o u pela fun\u00e7\u00e3o polinomial: <\/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-1f6389de5c7761fb5d35a9861156eec1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{aligned}f(x)=\\text{arcsech}(x^3-4x) \\quad\\color{orange}\\bm{\\longrightarrow}\\quad\\color{black}f'(x)&amp;=\\cfrac{-(3x^2-4)}{(x^3-4x)\\sqrt{1-(x^3-4x)^2}}\\\\[1.5ex] &amp;=\\cfrac{-3x^2+4}{(x^3-4x)\\sqrt{1-(x^3-4x)^2}}\\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"117\" width=\"610\" style=\"vertical-align: 0px;\"><\/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-secante-hiperbolica\/\">Derivada secante hiperb\u00f3lica<\/a><\/span><\/li>\n<li> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/pt\/derivado-hiperbolico-da-larcosina\/\">Derivado de arco seno hiperb\u00f3lico<\/a><\/span><\/li>\n<li> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/pt\/derivada-seno-hiperbolica\/\">Derivada do seno hiperb\u00f3lico<\/a><\/span><\/li>\n<li> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/pt\/deriva-de-arco-secante\/\">Derivado de arco secante<\/a><\/span><\/li>\n<li> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/pt\/derivada-da-secante\/\">derivada de secante<\/a><\/span><\/li>\n<li><span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/pt\/derivado-de-larcosina\/\">derivada de arco seno<\/a><\/span><\/li>\n<li> <span style=\"text-decoration: underline;\"><a href=\"https:\/\/mathority.org\/pt\/derivada-sinusal\/\">derivado do seno<\/a><\/span><\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Aqui voc\u00ea descobrir\u00e1 como calcular a derivada do arco secante hiperb\u00f3lico de uma fun\u00e7\u00e3o. Al\u00e9m disso, voc\u00ea poder\u00e1 ver exemplos resolvidos da derivada do arco secante hiperb\u00f3lico. F\u00f3rmula derivada do arco secante hiperb\u00f3lico A derivada do arco secante hiperb\u00f3lico de x \u00e9 igual a menos 1 dividido pelo produto de x vezes a raiz de &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathority.org\/pt\/derivada-do-arco-secante-hiperbolico\/\"> <span class=\"screen-reader-text\">Derivada do arco secante 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-397","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 arco secante 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\/derivada-do-arco-secante-hiperbolico\/\" \/>\n<meta property=\"og:locale\" content=\"pt_BR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Derivada do arco secante hiperb\u00f3lico - Mathority\" \/>\n<meta property=\"og:description\" content=\"Aqui voc\u00ea descobrir\u00e1 como calcular a derivada do arco secante hiperb\u00f3lico de uma fun\u00e7\u00e3o. Al\u00e9m disso, voc\u00ea poder\u00e1 ver exemplos resolvidos da derivada do arco secante hiperb\u00f3lico. F\u00f3rmula derivada do arco secante hiperb\u00f3lico A derivada do arco secante hiperb\u00f3lico de x \u00e9 igual a menos 1 dividido pelo produto de x vezes a raiz de &hellip; Derivada do arco secante hiperb\u00f3lico Leia mais &raquo;\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mathority.org\/pt\/derivada-do-arco-secante-hiperbolico\/\" \/>\n<meta property=\"article:published_time\" content=\"2023-07-03T02:11:56+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-30fd2d0fe7abd6d3774eaff22e8e762e_l3.png\" \/>\n<meta name=\"author\" content=\"Equipe Mathoridade\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Escrito por\" \/>\n\t<meta name=\"twitter:data1\" content=\"Equipe Mathoridade\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. tempo de leitura\" \/>\n\t<meta name=\"twitter:data2\" content=\"2 minutos\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/mathority.org\/pt\/derivada-do-arco-secante-hiperbolico\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/mathority.org\/pt\/derivada-do-arco-secante-hiperbolico\/\"},\"author\":{\"name\":\"Equipe Mathoridade\",\"@id\":\"https:\/\/mathority.org\/pt\/#\/schema\/person\/26defeb7b79f5baaedafa33a1ac6ac00\"},\"headline\":\"Derivada do arco secante hiperb\u00f3lico\",\"datePublished\":\"2023-07-03T02:11:56+00:00\",\"dateModified\":\"2023-07-03T02:11:56+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/mathority.org\/pt\/derivada-do-arco-secante-hiperbolico\/\"},\"wordCount\":392,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/mathority.org\/pt\/#organization\"},\"articleSection\":[\"Derivados\"],\"inLanguage\":\"pt-BR\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/mathority.org\/pt\/derivada-do-arco-secante-hiperbolico\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/mathority.org\/pt\/derivada-do-arco-secante-hiperbolico\/\",\"url\":\"https:\/\/mathority.org\/pt\/derivada-do-arco-secante-hiperbolico\/\",\"name\":\"Derivada do arco secante hiperb\u00f3lico - 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Al\u00e9m disso, voc\u00ea poder\u00e1 ver exemplos resolvidos da derivada do arco secante hiperb\u00f3lico. F\u00f3rmula derivada do arco secante hiperb\u00f3lico A derivada do arco secante hiperb\u00f3lico de x \u00e9 igual a menos 1 dividido pelo produto de x vezes a raiz de &hellip; Derivada do arco secante hiperb\u00f3lico Leia mais &raquo;","og_url":"https:\/\/mathority.org\/pt\/derivada-do-arco-secante-hiperbolico\/","article_published_time":"2023-07-03T02:11:56+00:00","og_image":[{"url":"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-30fd2d0fe7abd6d3774eaff22e8e762e_l3.png"}],"author":"Equipe Mathoridade","twitter_card":"summary_large_image","twitter_misc":{"Escrito por":"Equipe Mathoridade","Est. tempo de leitura":"2 minutos"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/mathority.org\/pt\/derivada-do-arco-secante-hiperbolico\/#article","isPartOf":{"@id":"https:\/\/mathority.org\/pt\/derivada-do-arco-secante-hiperbolico\/"},"author":{"name":"Equipe Mathoridade","@id":"https:\/\/mathority.org\/pt\/#\/schema\/person\/26defeb7b79f5baaedafa33a1ac6ac00"},"headline":"Derivada do arco secante hiperb\u00f3lico","datePublished":"2023-07-03T02:11:56+00:00","dateModified":"2023-07-03T02:11:56+00:00","mainEntityOfPage":{"@id":"https:\/\/mathority.org\/pt\/derivada-do-arco-secante-hiperbolico\/"},"wordCount":392,"commentCount":0,"publisher":{"@id":"https:\/\/mathority.org\/pt\/#organization"},"articleSection":["Derivados"],"inLanguage":"pt-BR","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/mathority.org\/pt\/derivada-do-arco-secante-hiperbolico\/#respond"]}]},{"@type":"WebPage","@id":"https:\/\/mathority.org\/pt\/derivada-do-arco-secante-hiperbolico\/","url":"https:\/\/mathority.org\/pt\/derivada-do-arco-secante-hiperbolico\/","name":"Derivada do arco secante hiperb\u00f3lico - 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