{"id":35175,"date":"2024-12-01T13:00:45","date_gmt":"2024-12-01T13:00:45","guid":{"rendered":"https:\/\/toposuranos.com\/material\/?p=35175"},"modified":"2025-11-22T22:33:35","modified_gmt":"2025-11-22T22:33:35","slug":"regra-da-cadeia-para-a-derivada-da-composicao-de-funcoes","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/pt\/regra-da-cadeia-para-a-derivada-da-composicao-de-funcoes\/","title":{"rendered":"Regra da Cadeia para a derivada da composi\u00e7\u00e3o de fun\u00e7\u00f5es"},"content":{"rendered":"<style>\np, ul, ol{\ntext-align: justify;\n}\nh1{\ntext-align:center;\ntext-transform: uppercase;\n}\nh2{\ntext-align:center;\ntext-transform: uppercase;\nfont-size:24pt;\n}\nh3 { \n    text-align: center;\n    text-transform: uppercase;\n    font-size: 24px !important;\n}\n<\/style>\n<h1>Regra da Cadeia para a derivada da composi\u00e7\u00e3o de fun\u00e7\u00f5es<\/h1>\n<p><em>Com o que vimos at\u00e9 agora, j\u00e1 temos todo o b\u00e1sico para calcular quase qualquer derivada. No entanto, devemos distinguir entre a possibilidade de calcular uma derivada e o esfor\u00e7o que investimos em realizar tais c\u00e1lculos, e \u00e9 aqui que entram em jogo teoremas como o da regra da cadeia para o c\u00e1lculo em uma vari\u00e1vel. A regra da cadeia nos permitir\u00e1 calcular rapidamente derivadas que, de outra forma, implicariam um trabalho bastante tedioso e complicado.<\/em><\/p>\n<p style=\"text-align:center;\" dir=\"ltr\">\n<b><u>\u00cdNDICE DE CONTE\u00daDOS<\/u><\/b><br \/>\n<b><a href=\"#1\">O teorema da regra da cadeia em uma vari\u00e1vel real<\/a><\/b><br \/>\n<a href=\"#11\">Demonstra\u00e7\u00e3o da regra da cadeia<\/a><br \/>\n<a href=\"#12\">Exemplos de uso da regra da cadeia em fun\u00e7\u00f5es de uma vari\u00e1vel<\/a><br \/>\n<a href=\"#13\">Precau\u00e7\u00e3o a ter em conta ao aplicar a regra da cadeia<\/a><br \/>\n<b><a href=\"#2\">Resultados \u00fateis obtidos a partir da regra da cadeia<\/a><\/b><br \/>\n<a href=\"#21\">Teorema da fun\u00e7\u00e3o inversa<\/a><br \/>\n<a href=\"#211\">Derivada da fun\u00e7\u00e3o exponencial<\/a><br \/>\n<a href=\"#212\">Derivada das Trigonom\u00e9tricas Inversas<\/a><br \/>\n<a href=\"#22\">Deriva\u00e7\u00e3o Impl\u00edcita<\/a><br \/>\n<a href=\"#221\">Derivadas de pot\u00eancias racionais<\/a><br \/>\n<a href=\"#221\">Derivadas de pot\u00eancias racionais<\/a><br \/>\n<b><a href=\"#3\">Guia de Exerc\u00edcios<\/a><\/b>\n<\/p>\n<p><a name=\"1\"><\/a><br \/>\n<center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/0y2SQpbRe3A\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>O teorema da regra da cadeia em uma vari\u00e1vel real<\/h2>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=0y2SQpbRe3A&amp;t=165s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Sejam <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">g<\/span> duas fun\u00e7\u00f5es<\/span><\/a> suscet\u00edveis de composi\u00e7\u00e3o<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f: A\\subseteq \\mathbb{R} \\longmapsto B\\subseteq \\mathbb{R}<\/span>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g: B\\subseteq Dom(g) \\longmapsto D\\subseteq \\mathbb{R}<\/span>\n<p>Se <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> \u00e9 deriv\u00e1vel em <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">g<\/span> \u00e9 deriv\u00e1vel em <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span>, ent\u00e3o a fun\u00e7\u00e3o composta <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g\\circ f<\/span><\/span> \u00e9 deriv\u00e1vel para todos os <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in A<\/span><\/span> e valer\u00e1 a f\u00f3rmula<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dx}(g\\circ f)(x) = \\frac{d}{dx} g(f(x)) = \\frac{dg(f(x))}{df(x)} \\frac{df(x)}{dx}<\/span>\n<p><a name=\"11\"><\/a><\/p>\n<h3>Demonstra\u00e7\u00e3o da regra da cadeia<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=0y2SQpbRe3A&amp;t=242s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Consideremos as fun\u00e7\u00f5es<\/span><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">g<\/span> tal como definidas anteriormente. Se calcularmos a derivada da composi\u00e7\u00e3o, ent\u00e3o teremos<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rcl}\n\n\\dfrac{d}{dx} g(f(x))&amp; = &amp; \\displaystyle\\lim_{\\Delta x \\to 0} \\dfrac{g(f(x + \\Delta x)) - g(f(x))}{\\Delta x} \\\\ \\\\\n\n&amp;=&amp;\\displaystyle \\lim_{\\Delta x \\to 0} \\frac{g(f(x + \\Delta x)) - g(f(x))}{\\Delta x} \\cdot \\frac{f(x + \\Delta x) - f(x)}{f(x+\\Delta x) - f(x)} \\\\ \\\\\n\n&amp;=&amp; \\displaystyle \\lim_{\\Delta x \\to 0} \\frac{g(f(x + \\Delta x)) - g(f(x))}{f(x+\\Delta x) - f(x)} \\cdot \\frac{f(x + \\Delta x) - f(x)}{\\Delta x} \\\\ \\\\\n\n&amp;=&amp;\\displaystyle \\lim_{\\Delta x \\to 0} \\frac{g(f(x + \\Delta x)) - g(f(x))}{f(x+\\Delta x) - f(x)} \\cdot \\lim_{\\Delta x \\to 0} \\frac{f(x + \\Delta x) - f(x)}{\\Delta x}\\\\ \\\\\n\n&amp;=&amp; \\displaystyle \\lim_{f(x+\\Delta x) \\to f(x) } \\frac{g(f(x + \\Delta x)) - g(f(x))}{f(x+\\Delta x) - f(x)} \\cdot \\lim_{\\Delta x \\to 0} \\frac{f(x + \\Delta x) - f(x)}{\\Delta x}\\\\ \\\\\n\n&amp;=&amp; \\displaystyle \\frac{dg(f(x))}{df(x)} \\frac{df(x)}{dx}\n\n\\end{array}\n\n<\/span>\n<p>Que \u00e9 o que se queria demonstrar.<\/p>\n<p><a name=\"12\"><\/a><\/p>\n<h3>Exemplos de uso da regra da cadeia em fun\u00e7\u00f5es de uma vari\u00e1vel<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=0y2SQpbRe3A&amp;t=423s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Algo que fica claro, ao menos \u00e0 primeira vista,<\/span><\/a> por\u00e9m n\u00e3o tanto desde uma perspectiva operacional, \u00e9 o fato de que a regra da cadeia nos diz que, quando encontramos uma composi\u00e7\u00e3o de fun\u00e7\u00f5es, podemos derivar \u201cde fora para dentro\u201d. Para explicar isso de uma forma f\u00e1cil de entender, os exemplos s\u00e3o de longe o caminho mais r\u00e1pido.<\/p>\n<ol>\n<li>Se nos pedem para derivar <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x) = (2x^2+1)^{12}<\/span><\/span>, primeiro desenvolver\u00edamos as pot\u00eancias e depois aplicar\u00edamos a derivada da pot\u00eancia sobre cada uma das partes desse grande polin\u00f4mio que ter\u00edamos obtido como resultado. Um trabalho desnecessariamente exaustivo. Com a regra da cadeia, o c\u00e1lculo da derivada pode ser feito em poucas linhas:<br \/>\n<\/p>\n<p style=\"align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dx} (2x^2+1)^{12} = 12(2x^2+1)^{11}(4x)= 48x(2x^2+1)^{11}<\/span>\n<\/li>\n<li>Tenta calcular a derivada de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g(x) = \\sin(\\cos(x))<\/span><\/span> apenas com as t\u00e9cnicas b\u00e1sicas de deriva\u00e7\u00e3o e enfrenta o sofrimento eterno. Faz isso usando a regra da cadeia e o resultado aparecer\u00e1 sem l\u00e1grimas e em poucos passos:<br \/>\n<\/p>\n<p style=\"align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dx} \\sin(\\cos(x))= -\\cos(cos(x))\\sin(x) <\/span>\n<\/li>\n<li>Tamb\u00e9m podes calcular a derivada de fun\u00e7\u00f5es que s\u00e3o a composi\u00e7\u00e3o de muitas fun\u00e7\u00f5es. Se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\cos(\\cos(\\cos(x))),<\/span><\/span> a derivada de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">df\/dx<\/span><\/span> fica como:<br \/>\n<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rcl}\n\n\\displaystyle \\frac{d}{dx} \\cos(\\cos(\\cos(x))) &amp;=&amp; -\\sin(\\cos(\\cos(x)))\\cdot(-\\sin(\\cos(x))\\cdot(-\\sin(x)) \\\\ \\\\\n\n&amp;=&amp; -\\sin(\\cos(\\cos(x)))\\cdot\\sin(\\cos(x))\\cdot\\sin(x)\n\n\\end{array}\n\n<\/span>\n<p>Como podes ver, aplicar a regra da cadeia \u00e9 simplesmente derivar de maneira encadeada desde fora at\u00e9 dentro.<\/li>\n<\/ol>\n<p><a name=\"13\"><\/a><\/p>\n<h3>Precau\u00e7\u00e3o a ter em conta diante da regra da cadeia<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=0y2SQpbRe3A&amp;t=607s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Na literatura, todos mostram os grandes benef\u00edcios<\/span><\/a> de usar a regra da cadeia, mas muito poucos s\u00e3o enf\u00e1ticos nas precau\u00e7\u00f5es que devem ser tomadas antes de utiliz\u00e1-la. Apesar da pot\u00eancia desse teorema, deves sempre prestar muita aten\u00e7\u00e3o aos dom\u00ednios e contradom\u00ednios das fun\u00e7\u00f5es antes de aplicar a regra da cadeia. Antes de trabalhar, deves assegurar que os dom\u00ednios e contradom\u00ednios das fun\u00e7\u00f5es sejam compat\u00edveis para a composi\u00e7\u00e3o; caso contr\u00e1rio, corres o risco de calcular derivadas onde elas n\u00e3o existem. Se, por exemplo, derivaras uma fun\u00e7\u00e3o do tipo<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\ln(\\cos(x))<\/span>\n<p>se confias cegamente na regra da cadeia, far\u00e1s c\u00e1lculos como o seguinte:<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dx}\\ln(\\cos(x)) = -\\frac{1}{\\cos(x)}\\sin(x) = -\\tan(x)<\/span>\n<p>Claramente, a fun\u00e7\u00e3o tangente est\u00e1 bem definida para um valor de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x=2\\pi\/3<\/span><\/span>, pois seu valor \u00e9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\tan(2\\pi\/3) = -\\sqrt{3}<\/span><\/span>. Por\u00e9m, a fun\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\ln(\\cos(x))<\/span><\/span> n\u00e3o est\u00e1 bem definida a\u00ed porque <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(2\\pi\/3) = \\ln(\\cos(2\\pi\/3)) = \\ln(-1\/2),<\/span><\/span> e n\u00e3o existe logaritmo de n\u00fameros negativos! Em casos como este, \u00e9 necess\u00e1rio indicar, antes de aplicar a regra da cadeia, que os valores de <span class=\"katex-eq\" data-katex-display=\"false\">x<\/span> a serem considerados s\u00e3o tais que mant\u00eam positiva a fun\u00e7\u00e3o cosseno (de modo a assegurar a compatibilidade sob composi\u00e7\u00e3o) e s\u00f3 ent\u00e3o a regra da cadeia ser\u00e1 v\u00e1lida.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Resultados \u00fateis obtidos a partir da regra da cadeia<\/h2>\n<p>A regra da cadeia n\u00e3o \u00e9 \u00fatil apenas para realizar c\u00e1lculos de derivadas que de outra forma seriam insuport\u00e1veis; ela tamb\u00e9m \u00e9 \u00fatil para expandir ainda mais as t\u00e9cnicas de deriva\u00e7\u00e3o para muitas outras fun\u00e7\u00f5es. A seguir, revisaremos essas t\u00e9cnicas, seus resultados e demonstra\u00e7\u00f5es.<\/p>\n<p><a name=\"21\"><\/a><\/p>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/5ddoUcIhgjU\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><a name=\"21\"><\/a><\/p>\n<h3>Teorema da fun\u00e7\u00e3o inversa<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=5ddoUcIhgjU&amp;t=75s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Seja <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> uma fun\u00e7\u00e3o bijetiva<\/span><\/a> e deriv\u00e1vel em algum intervalo <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I\\subseteq \\mathbb{R}<\/span><\/span>. Utilizando a regra da cadeia, \u00e9 poss\u00edvel calcular a derivada da fun\u00e7\u00e3o identidade <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(f^{-1}\\circ f)(x) = f^{-1}(f(x)) = x.<\/span><\/span> Os c\u00e1lculos d\u00e3o o seguinte resultado:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">1 = \\displaystyle \\frac{d}{dx} x = \\frac{d}{dx} f^{-1}(f(x)) = \\frac{df^{-1}(f(x))}{df(x)}\\frac{df(x)}{dx}<\/span>\n<p>A partir disso, pode-se isolar <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">df^{-1}(f(x))\/df(x)<\/span><\/span> e obt\u00e9m-se o resultado:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\frac{df^{-1}(f(x))}{df(x)}= \\frac{1}{\\frac{df(x)}{dx}}}<\/span>\n<p>Isto \u00e9 o que se conhece como o teorema da fun\u00e7\u00e3o inversa para o c\u00e1lculo das derivadas. Na literatura, \u00e9 comum encontrar esse teorema escrito na forma<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\frac{dx}{dy}= \\frac{1}{\\frac{dy}{dx}}}<\/span>\n<p>Ambas as formas de expressar o teorema da fun\u00e7\u00e3o inversa s\u00e3o equivalentes e resultam de escrever <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=f(x)<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x=f^{-1}(y).<\/span><\/span><\/p>\n<p>At\u00e9 aqui vimos tudo o que se pode dizer sobre o conte\u00fado do teorema da fun\u00e7\u00e3o inversa; agora veremos como podemos utiliz\u00e1-lo para calcular algumas derivadas que, de outra forma, seriam bastante dif\u00edceis.<\/p>\n<p><a name=\"211\"><\/a><\/p>\n<h4>Derivada da fun\u00e7\u00e3o exponencial<\/h4>\n<p><span style=\"color: #ff0000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=5ddoUcIhgjU&amp;t=215s\" target=\"_blank\" style=\"color: #ff0000;\" rel=\"noopener\">Quando estudamos as t\u00e9cnicas<\/a><\/span> b\u00e1sicas de deriva\u00e7\u00e3o, vimos que<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dx}\\ln(x) = \\frac{1}{x}<\/span>\n<p>Com este resultado e o teorema da fun\u00e7\u00e3o inversa, \u00e9 f\u00e1cil provar que<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dx}e^x = e^x<\/span>\n<p style=\"text-align: justify;color: #000080;\"><strong>DEMONSTRA\u00c7\u00c3O:<\/strong><\/p>\n<p>\u00c9 claro que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\ln(x)<\/span><\/span> \u00e9 equivalente a dizer que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x=e^y.<\/span><\/span> Em seguida, aplicando o teorema da fun\u00e7\u00e3o inversa, tem-se:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}e^y = \\frac{dx}{dy} = \\frac{1}{\\frac{dy}{dx}} = \\frac{1}{\\frac{d}{dx}\\ln(x)} = x = e^y<\/span>\n<p>\u00c9 dizer:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}e^y = e^y<\/span>\n<p>Se nesta \u00faltima express\u00e3o substituirmos os \u201cy\u201d por \u201cx\u201d, obtemos o que se queria demonstrar:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dx}e^x = e^x.<\/span>\n<p><a name=\"212\"><\/a><\/p>\n<h4>Derivada das Trigonom\u00e9tricas Inversas<\/h4>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=5ddoUcIhgjU\" target:=\"\" span=\"\" style=\"color: #ff0000;\" 0=\"\" a=\"\">O teorema da fun\u00e7\u00e3o inversa<\/a> tamb\u00e9m nos permitir\u00e1 obter as derivadas de todas as inversas trigonom\u00e9tricas. Estas s\u00e3o:<\/p>\n<p style=\"text-align:center;\" dir=\"ltr;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{ccccccc}\n\n\\dfrac{d}{dx}\\text{Arcsin}(x) &amp;=&amp; \\dfrac{1}{\\sqrt{1-x^2}} &amp;\\phantom{asd}&amp;\\dfrac{d}{dx}\\text{Arccos}(x) &amp;=&amp; \\dfrac{-1}{\\sqrt{1-x^2}} \\\\ \\\\\n\n\\dfrac{d}{dx}\\text{Arctan}(x) &amp;=&amp; \\dfrac{1}{1+x^2} &amp;\\phantom{asd}&amp;\\dfrac{d}{dx}\\text{Arccot}(x) &amp;=&amp; \\dfrac{-1}{1-x^2} \\\\ \\\\\n\n\\dfrac{d}{dx}\\text{Arcsec}(x) &amp;=&amp; \\dfrac{1}{x\\sqrt{x^2-1}} &amp;\\phantom{asd}&amp;\\dfrac{d}{dx}\\text{Arccsc}(x) &amp;=&amp; \\dfrac{-1}{x\\sqrt{x^2-1}}\n\n\\end{array}<\/span>\n<p style=\"text-align: justify; color: #000080;\"><strong>DEMONSTRA\u00c7\u00c3O<\/strong><\/p>\n<h5>Arco seno<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f8fe50991\"  tabindex=\"0\" title=\"Mostrar Demostraci\u00f3n\"    >Mostrar Demostraci\u00f3n<\/span><div id=\"target-id69e3f8fe50991\" class=\"collapseomatic_content \">\n<p>A fun\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin(x)<\/span><\/span> \u00e9 bijetiva sempre que restrinjamos seu dom\u00ednio a um conjunto da forma <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[\\frac{-\\pi}{2}+k\\pi , \\frac{\\pi}{2}+ k\\pi \\right],<\/span><\/span> sendo <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> um inteiro qualquer. Sem perda de generalidade, \u00e9 poss\u00edvel limitar-se ao caso principal, onde <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, de modo que a fun\u00e7\u00e3o seno bijetiva ser\u00e1 da forma<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\\sin : \\left[-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right] \\longrightarrow [-1,1]<\/span>\n<p>e sob estas condi\u00e7\u00f5es cumpre-se que<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\sin(x) \\longleftrightarrow x=arcsin(y).<\/span>\n<p>Se aplicarmos o teorema da fun\u00e7\u00e3o inversa, tem-se:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arcsin(y) = \\frac{1}{\\frac{d}{dx}\\sin(x)} = \\frac{1}{\\cos(x)}<\/span>\n<p>Agora, recordemos a identidade trigonom\u00e9trica<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin^2(x) + \\cos^2(x) = 1<\/span>\n<p>da qual se infere que, se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in [-\\pi\/2, \\pi\/2]<\/span><\/span>, ent\u00e3o se cumpre<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\cos(x) = \\sqrt{1 - \\sin^2(x)}<\/span>\n<p>Depois, se substituirmos isso na derivada do arco seno, chegaremos \u00e0 express\u00e3o<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arcsin(y) = \\frac{1}{\\cos(x)} = \\frac{1}{ \\sqrt{1 - \\sin^2(x)}}<\/span>\n<p>E como <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\sin(x)<\/span><\/span><\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arcsin(y) = \\frac{1}{ \\sqrt{1 - y^2}}<\/span>\n<p>Finalmente, substituindo os \u201cy\u201d por \u201cx\u201d nesta \u00faltima express\u00e3o, chegamos ao que se queria demonstrar:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\frac{d}{dx}arcsin(x) = \\frac{1}{ \\sqrt{1 - x^2}}}<\/span>\n<\/div>\n<h5>Arco cosseno<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f8fe50c6e\"  tabindex=\"0\" title=\"Mostrar Demostraci\u00f3n\"    >Mostrar Demostraci\u00f3n<\/span><div id=\"target-id69e3f8fe50c6e\" class=\"collapseomatic_content \">\n<p>A fun\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\cos(x)<\/span><\/span> \u00e9 bijetiva sempre que restrinjamos seu dom\u00ednio a um conjunto da forma <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left[0+k\\pi , \\pi+ k\\pi \\right],<\/span><\/span> sendo <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> um inteiro qualquer. Sem perda de generalidade, \u00e9 poss\u00edvel limitar-se ao caso principal, onde <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, de modo que a fun\u00e7\u00e3o cosseno bijetiva ser\u00e1 da forma<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\cos : \\left[0, \\pi\\right] \\longrightarrow [-1,1]<\/span>\n<p>e sob estas condi\u00e7\u00f5es cumpre-se que<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\cos(x) \\longleftrightarrow x=arccos(y).<\/span>\n<p>Se aplicarmos o teorema da fun\u00e7\u00e3o inversa, tem-se:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arccos(y) = \\frac{1}{\\frac{d}{dx}\\cos(x)} = \\frac{-1}{\\sin(x)}<\/span>\n<p>Agora, recordemos a identidade trigonom\u00e9trica<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin^2(x) + \\cos^2(x) = 1<\/span>\n<p>da qual se infere que, se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in [0, \\pi]<\/span><\/span>, ent\u00e3o se cumpre<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\sin(x) = \\sqrt{1 - \\cos^2(x)}<\/span>\n<p>Depois, se substituirmos isso na derivada do arco cosseno, chegaremos \u00e0 express\u00e3o<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arccos(y) = \\frac{-1}{\\sin(x)} = \\frac{-1}{ \\sqrt{1 - \\cos^2(x)}}<\/span>\n<p>E como <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\cos(x)<\/span><\/span><\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arccos(y) = \\frac{-1}{ \\sqrt{1 - y^2}}<\/span>\n<p>Finalmente, substituindo os \u201cy\u201d por \u201cx\u201d nesta \u00faltima express\u00e3o, chegamos ao que se queria demonstrar:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\frac{d}{dx}arccos(x) = \\frac{-1}{ \\sqrt{1 - x^2}}}<\/span>\n<\/div>\n<h5>Arco tangente<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f8fe50e14\"  tabindex=\"0\" title=\"Mostrar Demostraci\u00f3n\"    >Mostrar Demostraci\u00f3n<\/span><div id=\"target-id69e3f8fe50e14\" class=\"collapseomatic_content \">\n<p>A fun\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\tan(x)<\/span><\/span> \u00e9 bijetiva sempre que restrinjamos seu dom\u00ednio a um conjunto da forma <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[-\\frac{\\pi}{2}+k\\pi , \\frac{\\pi}{2}+ k\\pi \\right],<\/span><\/span> sendo <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> um inteiro qualquer. Sem perda de generalidade, \u00e9 poss\u00edvel limitar-se ao caso principal, onde <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0,<\/span><\/span> de modo que a fun\u00e7\u00e3o tangente bijetiva ser\u00e1 da forma<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\tan : \\left[-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right] \\longrightarrow \\mathbb{R}<\/span>\n<p>e sob estas condi\u00e7\u00f5es cumpre-se que<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\tan(x) \\longleftrightarrow x=arctan(y).<\/span>\n<p>Se aplicarmos o teorema da fun\u00e7\u00e3o inversa, tem-se:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arctan(y) = \\frac{1}{\\frac{d}{dx}\\tan(x)} = \\frac{1}{\\sec^2(x)}<\/span>\n<p>Agora, recordemos a identidade trigonom\u00e9trica<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin^2(x) + \\cos^2(x) = 1<\/span>\n<p>da qual se infere que<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\sec^2(x) =1+\\tan^2(x)<\/span>\n<p>Depois, se substituirmos isso na derivada do arco tangente, chegaremos \u00e0 express\u00e3o<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arctan(y) = \\frac{1}{\\sec^2(x)} = \\frac{1}{ 1+\\tan^2(x)}<\/span>\n<p>E como <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\tan(x)<\/span><\/span><\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arctan(y) = \\frac{1}{1 + y^2}<\/span>\n<p>Finalmente, substituindo os \u201cy\u201d por \u201cx\u201d nesta \u00faltima express\u00e3o, chegamos ao que se queria demonstrar:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\frac{d}{dx}arctan(x) = \\frac{1}{1+ x^2}}<\/span>\n<\/div>\n<h5>Arco cotangente<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f8fe50f89\"  tabindex=\"0\" title=\"Mostrar Demostraci\u00f3n\"    >Mostrar Demostraci\u00f3n<\/span><div id=\"target-id69e3f8fe50f89\" class=\"collapseomatic_content \">\n<p>A fun\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">cot(x)<\/span><\/span> \u00e9 bijetiva sempre que restrinjamos seu dom\u00ednio a um conjunto da forma <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left[0+k\\pi , \\pi+ k\\pi \\right],<\/span><\/span> sendo <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> um inteiro qualquer. Sem perda de generalidade, \u00e9 poss\u00edvel limitar-se ao caso principal, onde <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, de modo que a fun\u00e7\u00e3o cotangente bijetiva ser\u00e1 da forma<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">ctg : \\left[0, \\pi\\right] \\longrightarrow \\mathbb{R}<\/span>\n<p>e sob estas condi\u00e7\u00f5es cumpre-se que<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=ctg(x) \\longleftrightarrow x=arcctg(y).<\/span>\n<p>Se aplicarmos o teorema da fun\u00e7\u00e3o inversa, tem-se:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arcctg(y) = \\frac{1}{\\frac{d}{dx}ctg(x)} = \\frac{-1}{\\csc^2(x)}<\/span>\n<p>Agora, recordemos a identidade trigonom\u00e9trica<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin^2(x) + \\cos^2(x) = 1<\/span>\n<p>da qual se infere que<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\csc^2(x) =1+ctg^2(x)<\/span>\n<p>Depois, se substituirmos isso na derivada do arco cotangente, chegaremos \u00e0 express\u00e3o<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arcctg(y) = \\frac{-1}{\\csc^2(x)} = \\frac{-1}{ 1+ctg^2(x)}<\/span>\n<p>E como <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=ctg(x)<\/span><\/span><\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arcctg(y) = \\frac{-1}{1 + y^2}<\/span>\n<p>Finalmente, substituindo os \u201cy\u201d por \u201cx\u201d nesta \u00faltima express\u00e3o, chegamos ao que se queria demonstrar:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\frac{d}{dx}arcctg(x) = \\frac{-1}{1+ x^2}}<\/span>\n<\/div>\n<h5>Arco secante<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f8fe51104\"  tabindex=\"0\" title=\"Mostrar Demostraci\u00f3n\"    >Mostrar Demostraci\u00f3n<\/span><div id=\"target-id69e3f8fe51104\" class=\"collapseomatic_content \">\n<p>A fun\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sec(x)<\/span><\/span> \u00e9 bijetiva sempre que restrinjamos seu dom\u00ednio a um conjunto da forma <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[0+k\\pi , \\pi+ k\\pi \\right]\\setminus\\left\\{\\frac{\\pi}{2} + k\\pi\\right\\},<\/span><\/span> sendo <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> um inteiro qualquer. Sem perda de generalidade, \u00e9 poss\u00edvel limitar-se ao caso principal, onde <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, de modo que a fun\u00e7\u00e3o secante bijetiva ser\u00e1 da forma<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sec : \\left[0, \\pi\\right]\\setminus\\{\\pi\/2\\} \\longrightarrow \\mathbb{R}\\setminus]-1,1[<\/span>\n<p>e sob estas condi\u00e7\u00f5es cumpre-se que<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\sec(x) \\longleftrightarrow x={arcsec}(y).<\/span>\n<p>Se aplicarmos o teorema da fun\u00e7\u00e3o inversa, tem-se:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}{arcsec}(y) = \\frac{1}{\\frac{d}{dx}\\sec(x)} = \\frac{1}{\\sec(x)\\tan(x)}<\/span>\n<p>Agora, recordemos a identidade trigonom\u00e9trica<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin^2(x) + \\cos^2(x) = 1<\/span>\n<p>da qual se infere que<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\tan^2(x) =\\sec^2(x)-1<\/span>\n<p>Depois, se substituirmos isso na derivada do arco secante, chegaremos \u00e0 express\u00e3o<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}{arcsec}(y) = \\frac{1}{\\sec(x)\\tan(x)} = \\frac{1}{sec(x)\\sqrt{\\sec^2(x)-1}}<\/span>\n<p>E como <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\sec(x)<\/span><\/span><\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}{arcsec}(y) = \\frac{1}{y\\sqrt{y^2-1}}<\/span>\n<p>Finalmente, substituindo os \u201cy\u201d por \u201cx\u201d nesta \u00faltima express\u00e3o, chegamos ao que se queria demonstrar:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\frac{d}{dx}{arcsec}(x) = \\frac{1}{x\\sqrt{x^2-1}}}<\/span>\n<\/div>\n<h5>Arco cosecante<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f8fe51259\"  tabindex=\"0\" title=\"Mostrar Demostraci\u00f3n\"    >Mostrar Demostraci\u00f3n<\/span><div id=\"target-id69e3f8fe51259\" class=\"collapseomatic_content \">\n<p>A fun\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\csc(x)<\/span><\/span> \u00e9 bijetiva sempre que restrinjamos seu dom\u00ednio a um conjunto da forma <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[-\\frac{\\pi}{2}+k\\pi , \\frac{\\pi}{2} + k\\pi \\right]\\setminus\\left\\{0+k\\pi\\right\\}<\/span><\/span>, sendo <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> um inteiro qualquer. Sem perda de generalidade, \u00e9 poss\u00edvel limitar-se ao caso principal, onde <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, de modo que a fun\u00e7\u00e3o cossecante bijetiva ser\u00e1 da forma<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\csc : \\left[-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right]\\setminus\\{0\\} \\longrightarrow \\mathbb{R}\\setminus]-1,1[<\/span>\n<p>e sob estas condi\u00e7\u00f5es cumpre-se que<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\csc(x) \\longleftrightarrow x={arccsc}(y).<\/span>\n<p>Se aplicarmos o teorema da fun\u00e7\u00e3o inversa, tem-se:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}{arccsc}(y) = \\frac{1}{\\frac{d}{dx}\\csc(x)} = \\frac{-1}{\\csc(x)ctg(x)}<\/span>\n<p>Agora, recordemos a identidade trigonom\u00e9trica<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin^2(x) + \\cos^2(x) = 1<\/span>\n<p>da qual se infere que<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> ctg^2(x) =\\csc^2(x)-1<\/span>\n<p>Depois, se substituirmos isso na derivada do arco cossecante, chegaremos \u00e0 express\u00e3o<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}{arcsec}(y) = \\frac{-1}{\\csc(x)ctg(x)} = \\frac{-1}{csc(x)\\sqrt{\\csc^2(x)-1}}<\/span>\n<p>E como <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\csc(x)<\/span><\/span><\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}{arccsc}(y) = \\frac{-1}{y\\sqrt{y^2-1}}<\/span>\n<p>Finalmente, substituindo os \u201cy\u201d por \u201cx\u201d nesta \u00faltima express\u00e3o, chegamos ao que se queria demonstrar:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\frac{d}{dx}{arccsc}(x) = \\frac{-1}{x\\sqrt{x^2-1}}}<\/span>\n<\/div>\n<p><a name=\"22\"><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/hOAydWcd6zw\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/center><\/a><\/p>\n<h3>Deriva\u00e7\u00e3o Impl\u00edcita<\/h3>\n<p>Todas as derivadas que calculamos at\u00e9 agora foram realizadas sobre fun\u00e7\u00f5es definidas de forma expl\u00edcita: <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=f(x)<\/span><\/span>. No entanto, h\u00e1 situa\u00e7\u00f5es em que, a partir da rela\u00e7\u00e3o entre vari\u00e1veis, ou n\u00e3o \u00e9 simples obter a express\u00e3o expl\u00edcita da fun\u00e7\u00e3o, ou simplesmente tal tarefa n\u00e3o \u00e9 realiz\u00e1vel. Para esse tipo de caso serve a t\u00e9cnica da deriva\u00e7\u00e3o impl\u00edcita, e seus fundamentos se encontram, mais uma vez, na regra da cadeia.<\/p>\n<p>Para entender essa t\u00e9cnica, valem mais os exemplos do que as demonstra\u00e7\u00f5es; portanto, consideremos a rela\u00e7\u00e3o entre as vari\u00e1veis <span class=\"katex-eq\" data-katex-display=\"false\">x<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span> dada pela equa\u00e7\u00e3o<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x^3 +y^3- 9xy=0<\/span>\n<p>Se representarmos graficamente essa rela\u00e7\u00e3o, perceberemos que n\u00e3o \u00e9 o gr\u00e1fico de nenhuma fun\u00e7\u00e3o. \u00c9 o gr\u00e1fico de uma curva chamada \u201cfolha de Descartes\u201d.<\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-l30tAMcTkk0\/YLCIuWcDueI\/AAAAAAAAFIY\/K7uSR44DepgIjBlSVV7mCQO-Z0iy_RnRQCLcBGAsYHQ\/s0\/hojaDeDescartes.PNG\" alt=\"hoja de descartes\" class=\"alignnone size-full lazyload\" width=\"690\" height=\"515\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-l30tAMcTkk0\/YLCIuWcDueI\/AAAAAAAAFIY\/K7uSR44DepgIjBlSVV7mCQO-Z0iy_RnRQCLcBGAsYHQ\/s0\/hojaDeDescartes.PNG\" alt=\"hoja de descartes\" class=\"alignnone size-full lazyload\" width=\"690\" height=\"515\" \/><\/noscript><\/center><\/p>\n<p>Agora, se quis\u00e9ssemos calcular, por exemplo, a derivada de <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span> com respeito a <span class=\"katex-eq\" data-katex-display=\"false\">x<\/span>, ent\u00e3o ter\u00edamos s\u00e9rias dificuldades para encontrar explicitamente a express\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)<\/span><\/span> que satisfaz a equa\u00e7\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=f(x)<\/span><\/span> para depois derivar. O que fazemos, por\u00e9m, \u00e9 saltar esse passo e assumir implicitamente que <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span> \u00e9 fun\u00e7\u00e3o de <span class=\"katex-eq\" data-katex-display=\"false\">x<\/span>, ou seja: <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=y(x)<\/span><\/span>. Fazendo isso, a rela\u00e7\u00e3o da folha de Descartes transforma-se em:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x^3 +y^3(x)- 9xy(x)=0<\/span>\n<p>E podemos, consequentemente, derivar tudo utilizando a regra da cadeia. Se o fizermos, chegaremos ao seguinte resultado:<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rcl}\n\n\\displaystyle 3x^{2} + 3\\,y(x)^{2}\\,\\frac{dy}{dx} - \\left(9\\,y(x) + 9x\\,\\frac{dy}{dx}\\right) &amp;=&amp; 0 \\\\ \\\\\n\n\\displaystyle 3x^{2} + 3\\,y(x)^{2}\\,\\frac{dy}{dx} - 9\\,y(x) - 9x\\,\\frac{dy}{dx} &amp;=&amp; 0 \\\\ \\\\\n\n\\displaystyle \\frac{dy}{dx}\\,\\big(3\\,y(x)^{2} - 9x\\big) &amp;=&amp; 9\\,y(x) - 3x^{2} \\\\ \\\\\n\n\\displaystyle \\frac{dy}{dx} &amp;=&amp; \\dfrac{9\\,y(x) - 3x^{2}}{3\\,y(x)^{2} - 9x} \\\\ \\\\\n\n\\displaystyle \\color{blue}{\\frac{dy}{dx}} &amp;\\color{blue}{=}&amp; \\color{blue}{\\dfrac{3\\,y(x) - x^{2}}{y(x)^{2} - 3x}}\n\n\\end{array}\n\n<\/span>\n<p>A partir disso, podemos calcular, se conhecemos um ponto da curva, a inclina\u00e7\u00e3o da reta tangente que passa por esse ponto. Por exemplo, a partir do gr\u00e1fico podemos intuir que o ponto <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(2,4)<\/span><\/span> est\u00e1 sobre a curva; e, de fato, isso se comprova porque <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">2^3 + 4^3 - 9\\cdot 2\\cdot 4 = 8+64 - 72 = 0. <\/span><\/span> Sabendo isso, podemos dizer rapidamente que a inclina\u00e7\u00e3o da reta tangente que passa por esse ponto ser\u00e1:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\left.\\frac{dy}{dx}\\right|_{(2,4)}= \\frac{3\\cdot 4 - 2^2}{4^2 - 3\\cdot 2}= \\frac{8}{10}= \\frac{4}{5}}<\/span>\n<p><a name=\"221\"><\/a><\/p>\n<h4>Derivadas de pot\u00eancias racionais<\/h4>\n<p>Derivando implicitamente, \u00e9 poss\u00edvel ampliar o alcance de uma das t\u00e9cnicas b\u00e1sicas de deriva\u00e7\u00e3o. Esta \u00e9 a derivada de fun\u00e7\u00f5es do tipo <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=x^n<\/span><\/span>, com <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n\\in\\mathbb{Z}<\/span><\/span>. Agora podemos passar de considerar inteiros a racionais e demonstrar sem dificuldade que<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dx}x^{p\/q}= \\frac{p}{q}x^{(p\/q) -1}<\/span>\n<p>Onde <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">p,q\\in\\mathbb{Z}<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q\\neq 0<\/span><\/span>.<\/p>\n<p>Para demonstrar isso, dizemos: seja <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=x^{p\/q}<\/span><\/span> e aplicamos logaritmo natural para obter:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln(y) = \\displaystyle \\frac{p}{q}\\ln(x)<\/span>\n<p>Agora, derivando implicitamente esta express\u00e3o teremos:<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\displaystyle \\frac{1}{y}\\frac{dy}{dx} = \\frac{p}{q}\\frac{1}{x}<\/span>\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\frac{dy}{dx} = \\frac{p}{q}\\frac{1}{x}y(x)= \\frac{p}{q}\\frac{1}{x}x^{p\/q} = \\frac{p}{q}x^{(p\/q) - 1}}\n\n<\/span>\n<p><a name=\"3\"><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/KwJ5Bb5Ch_o\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/center><\/a><\/p>\n<h2>Guia de Exerc\u00edcios:<\/h2>\n<h4>Regra da Cadeia Uma Vari\u00e1vel<\/h4>\n<ol>\n<li>Calcule as derivadas do seguinte grupo de fun\u00e7\u00f5es:<br \/>\n<table>\n<tbody>\n<tr>\n<td width=\"20px\">a.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=(x^2-3)^{12}<\/span><\/span><\/td>\n<td width=\"20px\">b.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\displaystyle \\left(\\frac{4x^3 - x\\cos(2x) - 1}{\\sin(2x) + 2} \\right)^5<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td width=\"20px\">c.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\cos(1-x^2)<\/span><\/span><\/td>\n<td width=\"20px\">d.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\tan(x\\cos(3-x^2))<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td width=\"20px\">e.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\displaystyle \\frac{1}{(\\sec(2x)-1)^{3\/2}}<\/span><\/span><\/td>\n<td width=\"20px\">f.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\displaystyle \\frac{\\tan(2x)}{1-\\cot(2x)}<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td width=\"20px\">g.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\displaystyle \\ln\\left(\\frac{\\tan(x)}{x^2+1}\\right)<\/span><\/span><\/td>\n<td width=\"20px\">h.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=3^{\\csc(4x)}<\/span><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>Calcule a derivada do seguinte grupo de fun\u00e7\u00f5es:<br \/>\n<table>\n<tbody>\n<tr>\n<td width=\"20px\">a.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\displaystyle \\frac{1}{\\sqrt{x}arctan\\left(x^3\\right)}<\/span><\/span><\/td>\n<td width=\"20px\">b.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\displaystyle \\frac{{arcsec}(x^2-x+2)}{\\sqrt{x^2+1}}<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td width=\"20px\">c.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=x^x<\/span><\/span><\/td>\n<td width=\"20px\">d.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)={arccsc}\\left(x^{\\ln(x)}\\right)<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td width=\"20px\">e.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\ln\\left(arctan(e^x)\\right)<\/span><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>Regra da Cadeia para a derivada da composi\u00e7\u00e3o de fun\u00e7\u00f5es Com o que vimos at\u00e9 agora, j\u00e1 temos todo o b\u00e1sico para calcular quase qualquer derivada. No entanto, devemos distinguir entre a possibilidade de calcular uma derivada e o esfor\u00e7o que investimos em realizar tais c\u00e1lculos, e \u00e9 aqui que entram em jogo teoremas como [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":35164,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":0,"footnotes":""},"categories":[856,571],"tags":[],"class_list":["post-35175","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-calculo-diferencial-pt","category-matematica-pt"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Regra da Cadeia para a derivada da composi\u00e7\u00e3o de fun\u00e7\u00f5es - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Regra da cadeia em c\u00e1lculo: explica\u00e7\u00e3o simples, demonstra\u00e7\u00f5es, exemplos e exerc\u00edcios. 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