{"id":35195,"date":"2024-12-01T13:00:25","date_gmt":"2024-12-01T13:00:25","guid":{"rendered":"https:\/\/toposuranos.com\/material\/?p=35195"},"modified":"2025-11-22T22:34:40","modified_gmt":"2025-11-22T22:34:40","slug":"regle-de-la-chaine-pour-la-derivee-de-la-composition-de-fonctions","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/fr\/regle-de-la-chaine-pour-la-derivee-de-la-composition-de-fonctions\/","title":{"rendered":"R\u00e8gle de la cha\u00eene pour la d\u00e9riv\u00e9e de la composition de fonctions"},"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>R\u00e8gle de la cha\u00eene pour la d\u00e9riv\u00e9e de la composition de fonctions<\/h1>\n<p><em>Avec ce que nous avons vu jusqu\u2019\u00e0 pr\u00e9sent, nous disposons d\u00e9j\u00e0 de tout le n\u00e9cessaire pour calculer presque n\u2019importe quelle d\u00e9riv\u00e9e. Toutefois, il convient de distinguer entre la possibilit\u00e9 de calculer une d\u00e9riv\u00e9e et l\u2019effort investi pour effectuer ces calculs, et c\u2019est ici que les th\u00e9or\u00e8mes tels que celui de la r\u00e8gle de la cha\u00eene pour le calcul d\u2019une variable entrent en jeu. La r\u00e8gle de la cha\u00eene nous permettra de calculer rapidement des d\u00e9riv\u00e9es qui, autrement, impliqueraient un travail assez fastidieux et complexe.<\/em><\/p>\n<p style=\"text-align:center;\" dir=\"ltr\">\n<b><u>INDICE DE CONTENUS<\/u><\/b><br \/>\n<b><a href=\"#1\">Le th\u00e9or\u00e8me de la r\u00e8gle de la cha\u00eene en une variable r\u00e9elle<\/a><\/b><br \/>\n<a href=\"#11\">D\u00e9monstration de la r\u00e8gle de la cha\u00eene<\/a><br \/>\n<a href=\"#12\">Exemples d\u2019utilisation de la r\u00e8gle de la cha\u00eene dans des fonctions d\u2019une variable<\/a><br \/>\n<a href=\"#13\">Pr\u00e9caution \u00e0 consid\u00e9rer face \u00e0 la r\u00e8gle de la cha\u00eene<\/a><br \/>\n<b><a href=\"#2\">R\u00e9sultats utiles obtenus \u00e0 partir de la r\u00e8gle de la cha\u00eene<\/a><\/b><br \/>\n<a href=\"#21\">Th\u00e9or\u00e8me de la fonction inverse<\/a><br \/>\n<a href=\"#211\">D\u00e9riv\u00e9e de la fonction exponentielle<\/a><br \/>\n<a href=\"#212\">D\u00e9riv\u00e9e des fonctions trigonom\u00e9triques inverses<\/a><br \/>\n<a href=\"#22\">D\u00e9rivation implicite<\/a><br \/>\n<a href=\"#221\">D\u00e9riv\u00e9es de puissances rationnelles<\/a><br \/>\n<a href=\"#221\">D\u00e9riv\u00e9es de puissances rationnelles<\/a><br \/>\n<b><a href=\"#3\">Guide d\u2019exercices<\/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>Le th\u00e9or\u00e8me de la r\u00e8gle de la cha\u00eene en une variable r\u00e9elle<\/h2>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=0y2SQpbRe3A&amp;t=165s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Soient <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">g<\/span> deux fonctions<\/span><\/a> susceptibles de composition<\/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>Si <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> est d\u00e9rivable en <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">g<\/span> est d\u00e9rivable en <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span>, alors la fonction compos\u00e9e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g\\circ f<\/span><\/span> est d\u00e9rivable pour tous les <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in A<\/span><\/span> et la formule suivante sera valable<\/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>D\u00e9monstration de la r\u00e8gle de la cha\u00eene<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=0y2SQpbRe3A&amp;t=242s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Consid\u00e9rons les fonctions<\/span><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">g<\/span> telles que d\u00e9finies pr\u00e9c\u00e9demment. Si nous calculons la d\u00e9riv\u00e9e de la composition, nous aurons alors<\/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>Ce qui \u00e9tait \u00e0 d\u00e9montrer.<\/p>\n<p><a name=\"12\"><\/a><\/p>\n<h3>Exemples d\u2019utilisation de la r\u00e8gle de la cha\u00eene dans des fonctions d\u2019une variable<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=0y2SQpbRe3A&amp;t=423s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Quelque chose qui para\u00eet \u00e9vident, du moins \u00e0 premi\u00e8re vue,<\/span><\/a> mais qui ne l\u2019est pas autant d\u2019un point de vista op\u00e9rationnel, est le fait que la r\u00e8gle de la cha\u00eene nous indique que, lorsqu\u2019on rencontre une composition de fonctions, on peut d\u00e9river \u00ab de l\u2019ext\u00e9rieur vers l\u2019int\u00e9rieur \u00bb. Pour l\u2019expliquer de mani\u00e8re simple, les exemples sont de loin le chemin le plus rapide.<\/p>\n<ol>\n<li>Si l\u2019on nous demande de d\u00e9river <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x) = (2x^2+1)^{12}<\/span><\/span>, nous d\u00e9velopperions d\u2019abord les puissances puis appliquerions la d\u00e9riv\u00e9e de la puissance sur chacune des parties de ce grand polyn\u00f4me obtenu comme r\u00e9sultat. Un travail inutilement \u00e9puisant. Avec la r\u00e8gle de la cha\u00eene, le calcul de la d\u00e9riv\u00e9e peut \u00eatre r\u00e9alis\u00e9 en quelques lignes :<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>Essaie de calculer la d\u00e9riv\u00e9e de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g(x) = \\sin(\\cos(x))<\/span><\/span> uniquement avec les techniques basiques de d\u00e9rivation et tu conna\u00eetras la souffrance \u00e9ternelle. Fais-le en utilisant la r\u00e8gle de la cha\u00eene et le r\u00e9sultat appara\u00eetra sans larmes et en quelques \u00e9tapes :<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>Tu peux \u00e9galement calculer la d\u00e9riv\u00e9e de fonctions qui sont la composition de nombreuses fonctions. Si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\cos(\\cos(\\cos(x))),<\/span><\/span> la d\u00e9riv\u00e9e de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">df\/dx<\/span><\/span> devient :<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>Comme tu peux le voir, appliquer la r\u00e8gle de la cha\u00eene revient simplement \u00e0 d\u00e9river en cha\u00eene de l\u2019ext\u00e9rieur vers l\u2019int\u00e9rieur.<\/li>\n<\/ol>\n<p><a name=\"13\"><\/a><\/p>\n<h3>Pr\u00e9caution \u00e0 prendre en compte face \u00e0 la r\u00e8gle de la cha\u00eene<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=0y2SQpbRe3A&amp;t=607s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Dans la litt\u00e9rature, tous pr\u00e9sentent les grands avantages<\/span><\/a> de l\u2019utilisation de la r\u00e8gle de la cha\u00eene, mais tr\u00e8s peu insistent sur les pr\u00e9cautions \u00e0 prendre avant de l\u2019employer. Malgr\u00e9 la puissance de ce th\u00e9or\u00e8me, tu dois toujours pr\u00eater une grande attention aux domaines et aux images des fonctions avant d\u2019appliquer la r\u00e8gle de la cha\u00eene. Avant de travailler, tu dois t\u2019assurer que les domaines et les images des fonctions soient compatibles pour la composition ; autrement, tu risques de calculer des d\u00e9riv\u00e9es l\u00e0 o\u00f9 elles n\u2019existent pas. Si tu d\u00e9rives, par exemple, une fonction du type<\/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>et que tu fais confiance aveugl\u00e9ment \u00e0 la r\u00e8gle de la cha\u00eene, tu effectueras des calculs comme le suivant :<\/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>Il est clair que la fonction tangente est bien d\u00e9finie pour une valeur <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x=2\\pi\/3<\/span><\/span>, puisque sa valeur est <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\tan(2\\pi\/3) = -\\sqrt{3}<\/span><\/span>. Mais la fonction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\ln(\\cos(x))<\/span><\/span> n\u2019est pas bien d\u00e9finie \u00e0 cet endroit, car <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> et il n\u2019existe pas de logarithme de nombres n\u00e9gatifs ! Dans de tels cas, il est n\u00e9cessaire d\u2019indiquer, avant d\u2019appliquer la r\u00e8gle de la cha\u00eene, que les valeurs de <span class=\"katex-eq\" data-katex-display=\"false\">x<\/span> \u00e0 consid\u00e9rer sont telles qu\u2019elles maintiennent la fonction cosinus positive (afin d\u2019assurer la compatibilit\u00e9 sous composition), et seulement alors la r\u00e8gle de la cha\u00eene sera valable.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>R\u00e9sultats utiles obtenus \u00e0 partir de la r\u00e8gle de la cha\u00eene<\/h2>\n<p>La r\u00e8gle de la cha\u00eene n\u2019est pas seulement utile pour effectuer des calculs de d\u00e9riv\u00e9es qui seraient autrement p\u00e9nibles, elle sert aussi \u00e0 \u00e9tendre encore davantage les techniques de d\u00e9rivation \u00e0 de nombreuses autres fonctions. Nous examinerons ci-dessous ces techniques, leurs r\u00e9sultats et leurs d\u00e9monstrations.<\/p>\n<p><a name=\"21\"><\/a><br \/>\n<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><\/p>\n<p><a name=\"21\"><\/a><\/p>\n<h3>Th\u00e9or\u00e8me de la fonction inverse<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=5ddoUcIhgjU&amp;t=75s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Soit <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> une fonction bijective<\/span><\/a> et d\u00e9rivable sur un intervalle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I\\subseteq \\mathbb{R}<\/span><\/span>. En utilisant la r\u00e8gle de la cha\u00eene, il est possible de calculer la d\u00e9riv\u00e9e de la fonction identit\u00e9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(f^{-1}\\circ f)(x) = f^{-1}(f(x)) = x.<\/span><\/span> Les calculs donnent le r\u00e9sultat suivant :<\/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>\u00c0 partir de l\u00e0, on peut isoler <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">df^{-1}(f(x))\/df(x)<\/span><\/span> et obtenir le r\u00e9sultat suivant :<\/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>C\u2019est ce qu\u2019on appelle le th\u00e9or\u00e8me de la fonction inverse pour le calcul des d\u00e9riv\u00e9es. Dans la litt\u00e9rature, il est courant de voir ce th\u00e9or\u00e8me \u00e9crit sous la forme<\/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>Les deux formes d\u2019expression du th\u00e9or\u00e8me de la fonction inverse sont \u00e9quivalentes et s\u2019obtiennent en \u00e9crivant <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=f(x)<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x=f^{-1}(y).<\/span><\/span><\/p>\n<p>Jusqu\u2019ici, nous avons vu tout ce qu\u2019il est possible de dire sur le contenu du th\u00e9or\u00e8me de la fonction inverse ; nous verrons maintenant comment l\u2019utiliser pour calculer quelques d\u00e9riv\u00e9es qui, autrement, seraient assez difficiles.<\/p>\n<p><a name=\"211\"><\/a><\/p>\n<h4>D\u00e9riv\u00e9e de la fonction exponentielle<\/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\">Lorsque nous avons \u00e9tudi\u00e9 les techniques<\/a><\/span> \u00e9l\u00e9mentaires de d\u00e9rivation, nous avons vu 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>Avec ce r\u00e9sultat et le th\u00e9or\u00e8me de la fonction inverse, il est facile de montrer 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>D\u00c9MONSTRATION :<\/strong><\/p>\n<p>Il est clair que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\ln(x)<\/span><\/span> est \u00e9quivalent \u00e0 dire que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x=e^y.<\/span><\/span> Ensuite, en appliquant le th\u00e9or\u00e8me de la fonction inverse, on obtient :<\/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>C\u2019est-\u00e0-dire :<\/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>Si, dans cette derni\u00e8re expression, nous rempla\u00e7ons les \u00ab y \u00bb par \u00ab x \u00bb, nous obtenons ce qu\u2019il fallait d\u00e9montrer :<\/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>D\u00e9riv\u00e9e des fonctions trigonom\u00e9triques inverses<\/h4>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=5ddoUcIhgjU\" target:=\"\" span=\"\" style=\"color: #ff0000;\" 0=\"\" a=\"\">Le th\u00e9or\u00e8me de la fonction inverse<\/a> nous permettra \u00e9galement d\u2019obtenir les d\u00e9riv\u00e9es de toutes les fonctions trigonom\u00e9triques inverses. Celles-ci sont :<\/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>D\u00c9MONSTRATION<\/strong><\/p>\n<h5>Arc sinus<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f916425ed\"  tabindex=\"0\" title=\"Mostrar Demostraci\u00f3n\"    >Mostrar Demostraci\u00f3n<\/span><div id=\"target-id69e3f916425ed\" class=\"collapseomatic_content \">\n<p>La fonction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin(x)<\/span><\/span> est bijective \u00e0 condition de restreindre son domaine \u00e0 un ensemble de la forme <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> o\u00f9 <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> est un entier quelconque. Sans perte de g\u00e9n\u00e9ralit\u00e9, il est possible de se limiter au cas principal, o\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, de sorte que la fonction sinus bijective sera de la forme<\/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>et dans ces conditions, on a<\/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>Si l\u2019on applique le th\u00e9or\u00e8me de la fonction inverse, on obtient :<\/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>Rappelons maintenant l\u2019identit\u00e9 trigonom\u00e9trique<\/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>d\u2019o\u00f9 il s\u2019ensuit que, si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in [-\\pi\/2, \\pi\/2]<\/span><\/span>, alors<\/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>En rempla\u00e7ant cela dans la d\u00e9riv\u00e9e de l\u2019arc sinus, on obtient :<\/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>Et comme <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>Finalement, en rempla\u00e7ant les \u00ab y \u00bb par \u00ab x \u00bb dans cette derni\u00e8re expression, nous obtenons ce qu\u2019il fallait d\u00e9montrer :<\/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>Arc cosinus<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f916428b7\"  tabindex=\"0\" title=\"Mostrar Demostraci\u00f3n\"    >Mostrar Demostraci\u00f3n<\/span><div id=\"target-id69e3f916428b7\" class=\"collapseomatic_content \">\n<p>La fonction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\cos(x)<\/span><\/span> est bijective \u00e0 condition de restreindre son domaine \u00e0 un ensemble de la forme <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left[0+k\\pi , \\pi+ k\\pi \\right],<\/span><\/span>, o\u00f9 <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> est un entier quelconque. Sans perte de g\u00e9n\u00e9ralit\u00e9, il est possible de se limiter au cas principal, o\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, de sorte que la fonction cosinus bijective sera de la forme<\/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>et dans ces conditions on a<\/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>Si nous appliquons le th\u00e9or\u00e8me de la fonction inverse, nous obtenons :<\/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>Rappelons maintenant l\u2019identit\u00e9 trigonom\u00e9trique<\/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>d\u2019o\u00f9 il s\u2019ensuit que, si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in [0, \\pi]<\/span><\/span>, alors<\/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>En rempla\u00e7ant cela dans la d\u00e9riv\u00e9e de l\u2019arc cosinus, on obtient :<\/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>Et comme <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>Finalement, en rempla\u00e7ant les \u00ab y \u00bb par \u00ab x \u00bb dans cette derni\u00e8re expression, nous obtenons ce qu\u2019il fallait d\u00e9montrer :<\/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>Arc tangente<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f91642a22\"  tabindex=\"0\" title=\"Mostrar Demostraci\u00f3n\"    >Mostrar Demostraci\u00f3n<\/span><div id=\"target-id69e3f91642a22\" class=\"collapseomatic_content \">\n<p>La fonction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\tan(x)<\/span><\/span> est bijective \u00e0 condition de restreindre son domaine \u00e0 un ensemble de la forme <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>, o\u00f9 <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> est un entier quelconque. Sans perte de g\u00e9n\u00e9ralit\u00e9, on peut se limiter au cas principal, o\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0,<\/span><\/span> de sorte que la fonction tangente bijective sera de la forme<\/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>et dans ces conditions on a<\/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>Si nous appliquons le th\u00e9or\u00e8me de la fonction inverse, nous obtenons :<\/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>Rappelons maintenant l\u2019identit\u00e9 trigonom\u00e9trique<\/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>d\u2019o\u00f9 il s\u2019ensuit 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>En rempla\u00e7ant cela dans la d\u00e9riv\u00e9e de l\u2019arc tangente, on obtient :<\/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>Et comme <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>Finalement, en rempla\u00e7ant les \u00ab y \u00bb par \u00ab x \u00bb dans cette derni\u00e8re expression, nous obtenons ce qu\u2019il fallait d\u00e9montrer :<\/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>Arc cotangente<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f91642b80\"  tabindex=\"0\" title=\"Mostrar Demostraci\u00f3n\"    >Mostrar Demostraci\u00f3n<\/span><div id=\"target-id69e3f91642b80\" class=\"collapseomatic_content \">\n<p>La fonction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">cot(x)<\/span><\/span> est bijective \u00e0 condition de restreindre son domaine \u00e0 un ensemble de la forme <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left[0+k\\pi , \\pi+ k\\pi \\right],<\/span><\/span>, o\u00f9 <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> est un entier quelconque. Sans perte de g\u00e9n\u00e9ralit\u00e9, il est possible de se limiter au cas principal, o\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, de sorte que la fonction cotangente bijective sera de la forme<\/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>et dans ces conditions on a<\/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>Si nous appliquons le th\u00e9or\u00e8me de la fonction inverse, nous obtenons :<\/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>Rappelons maintenant l\u2019identit\u00e9 trigonom\u00e9trique<\/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>d\u2019o\u00f9 il s\u2019ensuit 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>En rempla\u00e7ant cela dans la d\u00e9riv\u00e9e de l\u2019arc cotangente, on obtient :<\/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>Et comme <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>Finalement, en rempla\u00e7ant les \u00ab y \u00bb par \u00ab x \u00bb dans cette derni\u00e8re expression, nous obtenons ce qu\u2019il fallait d\u00e9montrer :<\/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>Arc s\u00e9cante<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f91642cd8\"  tabindex=\"0\" title=\"Mostrar Demostraci\u00f3n\"    >Mostrar Demostraci\u00f3n<\/span><div id=\"target-id69e3f91642cd8\" class=\"collapseomatic_content \">\n<p>La fonction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sec(x)<\/span><\/span> est bijective \u00e0 condition de restreindre son domaine \u00e0 un ensemble de la forme <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> o\u00f9 <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> est un entier quelconque. Sans perte de g\u00e9n\u00e9ralit\u00e9, il est possible de se limiter au cas principal, o\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, de sorte que la fonction s\u00e9cante bijective sera de la forme<\/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>et dans ces conditions on a<\/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>Si nous appliquons le th\u00e9or\u00e8me de la fonction inverse, nous obtenons :<\/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>Rappelons maintenant l\u2019identit\u00e9 trigonom\u00e9trique<\/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>d\u2019o\u00f9 il s\u2019ensuit 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>En rempla\u00e7ant cela dans la d\u00e9riv\u00e9e de l\u2019arc s\u00e9cante, on obtient :<\/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>Et comme <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>Finalement, en rempla\u00e7ant les \u00ab y \u00bb par \u00ab x \u00bb dans cette derni\u00e8re expression, nous obtenons ce qu\u2019il fallait d\u00e9montrer :<\/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>Arc cos\u00e9cante<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f91642e2e\"  tabindex=\"0\" title=\"Mostrar Demostraci\u00f3n\"    >Mostrar Demostraci\u00f3n<\/span><div id=\"target-id69e3f91642e2e\" class=\"collapseomatic_content \">\n<p>La fonction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\csc(x)<\/span><\/span> est bijective \u00e0 condition de restreindre son domaine \u00e0 un ensemble de la forme <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>, o\u00f9 <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> est un entier quelconque. Sans perte de g\u00e9n\u00e9ralit\u00e9, il est possible de se limiter au cas principal, o\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, de sorte que la fonction cos\u00e9cante bijective sera de la forme<\/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>et dans ces conditions, on a<\/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>Si nous appliquons le th\u00e9or\u00e8me de la fonction inverse, nous obtenons :<\/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>Rappelons maintenant l\u2019identit\u00e9 trigonom\u00e9trique<\/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>d\u2019o\u00f9 il s\u2019ensuit 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>En rempla\u00e7ant cela dans la d\u00e9riv\u00e9e de l\u2019arc cos\u00e9cante, on obtient :<\/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>Et comme <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>Finalement, en rempla\u00e7ant les \u00ab y \u00bb par \u00ab x \u00bb dans cette derni\u00e8re expression, nous obtenons ce qu\u2019il fallait d\u00e9montrer :<\/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>Diff\u00e9rentiation implicite<\/h3>\n<p>Toutes les d\u00e9riv\u00e9es que nous avons calcul\u00e9es jusqu\u2019\u00e0 pr\u00e9sent ont \u00e9t\u00e9 r\u00e9alis\u00e9es sur des fonctions d\u00e9finies de mani\u00e8re explicite : <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=f(x)<\/span><\/span>. Cependant, il existe des situations o\u00f9, \u00e0 partir de la relation entre variables, soit il n\u2019est pas simple d\u2019obtenir l\u2019expression explicite de la fonction, soit une telle t\u00e2che n\u2019est tout simplement pas r\u00e9alisable. Pour ce type de cas, la technique de la diff\u00e9rentiation implicite est utile, et ses fondements se trouvent, encore une fois, dans la r\u00e8gle de la cha\u00eene.<\/p>\n<p>Pour comprendre cette technique, les exemples valent mieux que les d\u00e9monstrations ; consid\u00e9rons donc la relation entre les variables <span class=\"katex-eq\" data-katex-display=\"false\">x<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span> donn\u00e9e par l\u2019\u00e9quation<\/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>Si nous repr\u00e9sentons graphiquement cette relation, nous nous rendrons compte qu\u2019il ne s\u2019agit pas du graphe d\u2019une fonction. Il s\u2019agit du graphe d\u2019une courbe appel\u00e9e \u00ab feuille de Descartes \u00bb.<\/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>Maintenant, si nous voulions calculer, par exemple, la d\u00e9riv\u00e9e de <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span> par rapport \u00e0 <span class=\"katex-eq\" data-katex-display=\"false\">x<\/span>, nous aurions alors de grandes difficult\u00e9s \u00e0 trouver explicitement l\u2019expression <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)<\/span><\/span> qui satisfait l\u2019\u00e9quation <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=f(x)<\/span><\/span> pour ensuite d\u00e9river. Ce que nous faisons, toutefois, est de sauter cette \u00e9tape et d\u2019assumer implicitement que <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span> est une fonction de <span class=\"katex-eq\" data-katex-display=\"false\">x<\/span>, c\u2019est-\u00e0-dire : <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=y(x)<\/span><\/span>. En proc\u00e9dant ainsi, la relation de la feuille de Descartes devient :<\/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>Et nous pouvons, en cons\u00e9quence, tout d\u00e9river en utilisant la r\u00e8gle de la cha\u00eene. Si nous le faisons, nous arriverons au r\u00e9sultat suivant :<\/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>\u00c0 partir de cela, si nous connaissons un point de la courbe, nous pouvons calculer la pente de la droite tangente qui passe par ce point. Par exemple, \u00e0 partir du graphique, nous pouvons supposer que le point <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(2,4)<\/span><\/span> appartient \u00e0 la courbe ; et en effet, cela se v\u00e9rifie puisque <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> Sachant cela, nous pouvons dire imm\u00e9diatement que la pente de la tangente en ce point sera :<\/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>D\u00e9riv\u00e9es de puissances rationnelles<\/h4>\n<p>En d\u00e9rivant implicitement, il est possible d\u2019\u00e9largir la port\u00e9e d\u2019une des techniques fondamentales de d\u00e9rivation. Il s\u2019agit de la d\u00e9riv\u00e9e des fonctions du type <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=x^n<\/span><\/span>, avec <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n\\in\\mathbb{Z}<\/span><\/span>. Nous pouvons d\u00e9sormais passer d\u2019entiers \u00e0 rationnels et d\u00e9montrer sans difficult\u00e9 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>o\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">p,q\\in\\mathbb{Z}<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q\\neq 0<\/span><\/span>.<\/p>\n<p>Pour d\u00e9montrer cela, disons : soit <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=x^{p\/q}<\/span><\/span> et appliquons le logarithme naturel pour obtenir :<\/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>Ensuite, en d\u00e9rivant implicitement cette expression, nous avons :<\/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}}<\/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>Guide d\u2019exercices :<\/h2>\n<h4>R\u00e8gle de la cha\u00eene \u2013 Une variable<\/h4>\n<ol>\n<li>Calculez les d\u00e9riv\u00e9es du groupe de fonctions suivant :<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>Calculez la d\u00e9riv\u00e9e du groupe de fonctions suivant :<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>R\u00e8gle de la cha\u00eene pour la d\u00e9riv\u00e9e de la composition de fonctions Avec ce que nous avons vu jusqu\u2019\u00e0 pr\u00e9sent, nous disposons d\u00e9j\u00e0 de tout le n\u00e9cessaire pour calculer presque n\u2019importe quelle d\u00e9riv\u00e9e. Toutefois, il convient de distinguer entre la possibilit\u00e9 de calculer une d\u00e9riv\u00e9e et l\u2019effort investi pour effectuer ces calculs, et c\u2019est ici [&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":[866,569],"tags":[],"class_list":["post-35195","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-calcul-differentiel","category-mathematiques"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>R\u00e8gle de la cha\u00eene pour la d\u00e9riv\u00e9e de la composition de fonctions - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"R\u00e8gle de la cha\u00eene en calcul diff\u00e9rentiel : explication simple, d\u00e9monstrations, exemples et exercices. 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