{"id":35213,"date":"2024-12-01T13:00:57","date_gmt":"2024-12-01T13:00:57","guid":{"rendered":"https:\/\/toposuranos.com\/material\/?p=35213"},"modified":"2025-11-22T22:34:54","modified_gmt":"2025-11-22T22:34:54","slug":"regula-catenae-pro-derivata-compositionis-functionum","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/la\/regula-catenae-pro-derivata-compositionis-functionum\/","title":{"rendered":"Regula Catenae pro derivata compositionis functionum"},"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>Regula catenae pro derivata compositionis functionum<\/h1>\n<p><em>Ex iis quae hactenus vidimus, iam omnia fundamentalia habemus ad fere quamlibet derivatam computandam. Attamen distinguere debemus inter facultatem derivatam computandi et laborem quem in tales computationes impendimus; hic autem in scaenam prodeunt theoremat a, ut est regula catenae pro calculo unius variabilis. Regula catenae nobis permittet celeriter derivatas computare quae aliter opus satis taediosum atque implicatum requirerent.<\/em><\/p>\n<p style=\"text-align:center;\" dir=\"ltr\">\n<b><u>INDEX CONTENTORUM<\/u><\/b><br \/>\n<b><a href=\"#1\">Theorema regulae catenae in una variabili reali<\/a><\/b><br \/>\n<a href=\"#11\">Demonstratio regulae catenae<\/a><br \/>\n<a href=\"#12\">Exempla usus regulae catenae in functionibus unius variabilis<\/a><br \/>\n<a href=\"#13\">Cautio circa regulam catenae observanda<\/a><br \/>\n<b><a href=\"#2\">Resultata utilia ex regula catenae obtenta<\/a><\/b><br \/>\n<a href=\"#21\">Theorema functionis inversae<\/a><br \/>\n<a href=\"#211\">Derivata functionis exponentialis<\/a><br \/>\n<a href=\"#212\">Derivata functionum trigonometricarum inversarum<\/a><br \/>\n<a href=\"#22\">Derivatio implicita<\/a><br \/>\n<a href=\"#221\">Derivatae potestatum rationalium<\/a><br \/>\n<a href=\"#221\">Derivatae potestatum rationalium<\/a><br \/>\n<b><a href=\"#3\">Exercitationum index<\/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>Theorema regulae catenae in una variabili reali<\/h2>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=0y2SQpbRe3A&amp;t=165s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Sint <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">g<\/span> duae functiones<\/span><\/a> componibiles<\/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 derivabilis in <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">g<\/span> est derivabilis in <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span>, tum functio composita <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g\\circ f<\/span><\/span> est derivabilis pro omnibus <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in A<\/span><\/span> et valebit formula<\/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>Demonstratio regulae catenae<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=0y2SQpbRe3A&amp;t=242s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Consideremus functiones<\/span><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">g<\/span> ut supra definitae sunt. Si derivatam compositionis computamus, tunc obtinebitur<\/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>Quod erat demonstrandum.<\/p>\n<p><a name=\"12\"><\/a><\/p>\n<h3>Exempla usus regulae catenae in functionibus unius variabilis<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=0y2SQpbRe3A&amp;t=423s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Quod quidem videtur evidens, saltem prima facie,<\/span><\/a> sed non ita perspicuum est ex parte operationis, est hoc: regula catenae nos docet, cum compositionem functionum invenimus, nos posse derivare \u201cab exteriori ad interius\u201d. Ut hoc modo explanetur faciliori intellectu, exempla longe via celerrima sunt.<\/p>\n<ol>\n<li>Si postulatur ut derivemus <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x) = (2x^2+1)^{12}<\/span><\/span>, primum evolveremus potestates atque deinde derivationem potentiae in singulas partis illius magni polynomii, quod ex hoc processu emergere posset, applicaremus. Labor prorsus superfluus et gravis. Regula catenae adhibita, calculus derivatae paucis lineis perfici potest:<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>Istud conare: computare derivatam <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g(x) = \\sin(\\cos(x))<\/span><\/span> solis technicis fundamentalibus derivationis; id te ad aeternas miserias duceret. Id vero fac per regulam catenae, et eventus sine lacrimis et paucis gradibus apparebit:<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>Potes etiam derivatam computare functionum quae ex plurimis functionibus compositae sunt. Si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\cos(\\cos(\\cos(x))),<\/span><\/span> derivata <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">df\/dx<\/span><\/span> talis erit:<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>Ut videre licet, regulam catenae applicare idem est ac derivare concatenatim ab exterioribus ad interiora.<\/li>\n<\/ol>\n<p><a name=\"13\"><\/a><\/p>\n<h3>Monendum quod attinet ad regulam catenae<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=0y2SQpbRe3A&amp;t=607s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">In litteris omnes ostendunt maximos usus<\/span><\/a> regulae catenae, sed pauci sunt qui de cautionibus, antequam ea utatur, satis emphatice loquantur. Quamvis hoc theorema sit potens, semper diligenter attendere debes ad dominia et imagines functionum antequam regulam catenae applices. Antequam laboras, tibi confirmandum est dominia atque imagines functionum esse inter se compatibilia ad compositionem; sin minus, periculum subis derivatas computandi ubi non existunt. Si, exempli gratia, derivaveris functionem huius formae<\/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>si regulae catenae caeca fide adhaereas, computationes huiusmodi facies:<\/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>Plane functio tangens bene definita est pro valore <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x=2\\pi\/3<\/span><\/span>, cum eius valor sit <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\tan(2\\pi\/3) = -\\sqrt{3}<\/span><\/span>. Attamen functio <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\ln(\\cos(x))<\/span><\/span> ibi bene definita non est, quia <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> nec est logaritmus numerorum negativorum! In talibus casibus necesse est antequam regulam catenae applices declarare valores <span class=\"katex-eq\" data-katex-display=\"false\">x<\/span> eligendos esse tales qui functionem cosinus positivam servant (quo fit ut compositio sit legitima), atque solum tunc regula catenae valebit.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Resultata utilia ex regula catenae obtenta<\/h2>\n<p>Regula catenae non solum utilis est ad derivatas computandas quae aliter taediosissimae essent, sed etiam ad amplificandas rationes derivationis in multas alias functiones. Infra has rationes, earum eventus et demonstrationes recognoscemus.<\/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>Theorema functionis inversae<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=5ddoUcIhgjU&amp;t=75s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Sit <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> functio biiectiva<\/span><\/a> et derivabilis in aliquo intervallo <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I\\subseteq \\mathbb{R}<\/span><\/span>. Utendo regula catenae possibile est derivatam functionis identitatis computare <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(f^{-1}\\circ f)(x) = f^{-1}(f(x)) = x.<\/span><\/span> Computationes dant sequentem eventum:<\/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>Ex hoc elici potest <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">df^{-1}(f(x))\/df(x)<\/span><\/span> et obtinetur:<\/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>Hoc est quod vocatur theorema functionis inversae ad calculum derivatarum. In litteris frequens est hoc theorema ita scribere:<\/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>Utraque forma theorematis functionis inversae aequivalet et obtinetur ex scriptura <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>Hactenus vidimus quid theorematis functionis inversae substantia sit; nunc videre possumus quomodo eo uti possimus ad derivatas computandas quae aliter satis difficiles essent.<\/p>\n<p><a name=\"211\"><\/a><\/p>\n<h4>Derivata functionis exponentialis<\/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\">Cum rationes derivandi<\/a><\/span> fundamentales studeremus, vidimus:<\/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>Hoc cum theorema functionis inversae facile demonstrat:<\/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>DEMONSTRATIO:<\/strong><\/p>\n<p>Manifestum est <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\ln(x)<\/span><\/span> idem esse quod dicere <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x=e^y.<\/span><\/span> Quare, applicato theorema functionis inversae, habemus:<\/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>Id est:<\/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 in hac ultima expressione litteras \u201cy\u201d pro \u201cx\u201d substituimus, obtinemus quod demonstrandum erat:<\/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>Derivata Trigonometricae Inversae<\/h4>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=5ddoUcIhgjU\" target:=\"\" span=\"\" style=\"color: #ff0000;\" 0=\"\" a=\"\">Theorema functionis inversae<\/a> etiam nobis concedet obtinere derivatas omnium functionum trigonometricarum inversarum. Hae sunt:<\/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>DEMONSTRATIO<\/strong><\/p>\n<h5>Arcsinus<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f8fe49132\"  tabindex=\"0\" title=\"Mostrar Demostraci\u00f3n\"    >Mostrar Demostraci\u00f3n<\/span><div id=\"target-id69e3f8fe49132\" class=\"collapseomatic_content \">\n<p>Functio <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin(x)<\/span><\/span> est biiectiva, dummodo eius domain restringatur ad inter\u00advallum formae <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> ubi <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> est numerus integer quivis. Sine detrimento generalitatis limitari licet ad casum principalem, ubi <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, ita functio sinus biiectiva erit huiusmodi:<\/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>His condicionibus satisfactis obtinemus relationem<\/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 theorema functionis inversae applicamus, habemus:<\/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>Nunc memorare oportet identitatem trigonometricam<\/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>unde sequitur, si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in [-\\pi\/2, \\pi\/2]<\/span><\/span>, valere<\/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>Quam relationem si in derivatam arcsin substituimus, consequimur:<\/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 quoniam <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>Denique, substitutis \u201cy\u201d pro \u201cx\u201d in hac ultima expressione, pervenimus ad id quod demonstrandum erat:<\/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>Arccosinus<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f8fe492e5\"  tabindex=\"0\" title=\"Mostrar Demostraci\u00f3n\"    >Mostrar Demostraci\u00f3n<\/span><div id=\"target-id69e3f8fe492e5\" class=\"collapseomatic_content \">\n<p>Functio <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\cos(x)<\/span><\/span> est biiectiva dummodo eius domain ad intervallum huius formae restringatur: <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left[0+k\\pi , \\pi+ k\\pi \\right],<\/span><\/span> ubi <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> est integer quivis. Sine detrimento generalitatis ad casum principalem, ubi <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, nos limitare possumus, unde functio cosinus biiectiva erit:<\/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>His positis habetur:<\/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>Applicato theorema functionis inversae, consequitur:<\/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>Nunc recordemur identitatem trigonometricam:<\/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>unde colligitur, si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in [0, \\pi]<\/span><\/span>, valere:<\/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>Quam relationem si in derivatam arccosini substituimus, pervenitur ad formulam:<\/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 quoniam <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>Denique, substitutis \u201cy\u201d pro \u201cx\u201d in hac expressione, obtinemus quod demonstrandum erat:<\/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>Arctangens<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f8fe49442\"  tabindex=\"0\" title=\"Mostrar Demostraci\u00f3n\"    >Mostrar Demostraci\u00f3n<\/span><div id=\"target-id69e3f8fe49442\" class=\"collapseomatic_content \">\n<p>Functio <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\tan(x)<\/span><\/span> est biiectiva si eius domain restringatur ad intervallum formae <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> ubi <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> est integer quivis. Sine iactura generalitatis licet nos limitare ad casum principalem, ubi <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0,<\/span><\/span> ita functio tangens biiectiva erit:<\/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>His positis habetur:<\/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>Applicato theorema functionis inversae, consequitur:<\/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>Recordemur nunc identitatem trigonometricam:<\/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>unde obtinetur:<\/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>Quam relationem si in derivatam arctangentis substituimus, pervenitur ad formulam:<\/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 quoniam <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>Denique, substitutis \u201cy\u201d pro \u201cx\u201d in hac expressione, pervenimus ad quod demonstrandum erat:<\/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>Arccotangens<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f8fe4958c\"  tabindex=\"0\" title=\"Mostrar Demostraci\u00f3n\"    >Mostrar Demostraci\u00f3n<\/span><div id=\"target-id69e3f8fe4958c\" class=\"collapseomatic_content \">\n<p>Functio <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">cot(x)<\/span><\/span> est biiectiva dummodo eius domain restringatur ad intervallum huiusmodi: <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left[0+k\\pi , \\pi+ k\\pi \\right],<\/span><\/span> ubi <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> est integer quivis. Sine detrimento generalitatis licet nos limitare ad casum principalem, ubi <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, ita functio cotangens biiectiva erit:<\/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>His positis habetur:<\/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>Applicato theorema functionis inversae obtinemus:<\/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>Recordemur identitatem trigonometricam:<\/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>unde sequitur:<\/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>Substituendo hanc relationem in derivatam arcctg habetur:<\/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 cum <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>Denique, substitutis \u201cy\u201d pro \u201cx\u201d in hac expressione, pervenimus ad quod demonstrandum erat:<\/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>Arcsecans<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f8fe496da\"  tabindex=\"0\" title=\"Mostrar Demostraci\u00f3n\"    >Mostrar Demostraci\u00f3n<\/span><div id=\"target-id69e3f8fe496da\" class=\"collapseomatic_content \">\n<p>Functio <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sec(x)<\/span><\/span> est biiectiva dummodo eius domain restringatur ad intervallum <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> ubi <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> est integer quivis. Sine detrimento generalitatis licet nos limitare ad casum principalem, ubi <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, ita functio secans biiectiva erit:<\/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>His positis habetur:<\/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>Applicato theorema functionis inversae habemus:<\/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>Recordemur identitatem trigonometricam:<\/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>unde sequitur:<\/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>Substituendo hanc relationem in derivatam arcsecantis, obtinemus:<\/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 quoniam <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>Denique, substitutis \u201cy\u201d pro \u201cx\u201d in hac expressione, pervenimus ad quod demonstrandum erat:<\/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>Arccosecans<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f8fe49817\"  tabindex=\"0\" title=\"Mostrar Demostraci\u00f3n\"    >Mostrar Demostraci\u00f3n<\/span><div id=\"target-id69e3f8fe49817\" class=\"collapseomatic_content \">\n<p>Functio <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\csc(x)<\/span><\/span> est biiectiva dummodo eius domain restringatur ad intervallum huius formae: <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>, ubi <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> est integer quivis. Sine detrimento generalitatis licet nos limitare ad casum principalem, ubi <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, ita functio cosecans biiectiva erit:<\/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>His positis habetur:<\/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>Applicato theorema functionis inversae, obtinemus:<\/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>Nunc recordemur identitatem trigonometricam:<\/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>unde sequitur:<\/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>Substituendo hanc relationem in derivatam arccosecantis, pervenitur ad formulam:<\/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 quoniam <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>Denique, substitutis \u201cy\u201d pro \u201cx\u201d in hac expressione, obtinemus quod demonstrandum erat:<\/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>Derivatio Implicita<\/h3>\n<p>Omnes derivationes quas hactenus computavimus factae sunt super functionibus quae explicite definitae sunt: <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=f(x)<\/span><\/span>. Attamen adsunt casus in quibus, ex relatione inter variabiles data, vel non facile est expressionem explicitam obtinere, vel illud plane fieri non potest. In eiusmodi casibus adhibetur ratio derivationis implicitae, cuius fundamenta, iterum, in regula catenae inveniuntur.<\/p>\n<p>Ad hanc rationem intellegendam plus valent exempla quam demonstrationes; consideremus ergo relationem inter variabiles <span class=\"katex-eq\" data-katex-display=\"false\">x<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span> datam aequatione:<\/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 hanc relationem graphice repraesentamus, intellegemus eam non esse graphicon ullius functionis. Est graphicon curvae quae vocatur \u201cfolium Descartesii\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>Nunc, si velimus computare, exempli gratia, derivatam <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span> respectu <span class=\"katex-eq\" data-katex-display=\"false\">x<\/span>, graves difficultates haberemus in inveniendis explicite expressionem <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)<\/span><\/span> quae aequationi <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=f(x)<\/span><\/span> satisfaciat ut deinde derivemus. Quod tamen facimus est hoc: illum gradum omittimus et implicite assumimus <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span> esse functionem <span class=\"katex-eq\" data-katex-display=\"false\">x<\/span>, id est: <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=y(x)<\/span><\/span>. Hoc posito, relatio folii Descartesiani fit:<\/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 consequenter possumus totum derivare utens regula catenae. Si ita facimus, perveniemus ad hunc eventum:<\/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>Ex hoc possumus computare, si punctum curva notum est, declivitatem rectae tangentis per illud punctum transeuntis. Exempli gratia, ex graphico intellegere possumus punctum <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(2,4)<\/span><\/span> in curva iacere; idque revera confirmatur quia <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> Hoc scientes possumus celeriter dicere declivitatem rectae tangentis per illud punctum esse:<\/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>Derivata potestatum rationalium<\/h4>\n<p>Derivando implicite possibile est ampliare ambitum unius ex technicis fundamentalibus derivationis. Haec est derivata functionum generis <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=x^n<\/span><\/span>, cum <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n\\in\\mathbb{Z}<\/span><\/span>. Nunc possumus ex integralibus ad rationales transire atque sine difficultate demonstrare:<\/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>Ubi <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>Ad hoc demonstrandum dicimus: sit <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=x^{p\/q}<\/span><\/span> et logarithmum naturalem utrique lateri applicamus, unde obtinemus:<\/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>Iam, derivando implicite hanc expressionem, habebimus:<\/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>Index Exercitationum:<\/h2>\n<h4>Regula Catenae Unius Variabilis<\/h4>\n<ol>\n<li>Computa derivatas sequentium functionum:<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>Computa derivatam sequentium functionum:<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>Regula catenae pro derivata compositionis functionum Ex iis quae hactenus vidimus, iam omnia fundamentalia habemus ad fere quamlibet derivatam computandam. Attamen distinguere debemus inter facultatem derivatam computandi et laborem quem in tales computationes impendimus; hic autem in scaenam prodeunt theoremat a, ut est regula catenae pro calculo unius variabilis. Regula catenae nobis permittet celeriter derivatas [&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":2,"footnotes":""},"categories":[1328,1298],"tags":[],"class_list":["post-35213","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-calculus-differentialis","category-mathematica"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Regula Catenae pro derivata compositionis functionum - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Regula catenae in calculo: explicatio simplex, demonstrationes, exempla et exercitationes. 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