If z = (x + 6y)e^(x+y), x = u, y = ln(v), find (\u201a\u00e0\u00c7z\/\u201a\u00e0\u00c7u) and (\u201a\u00e0\u00c7z\/\u201a\u00e0\u00c7v)?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Para encontrar las derivadas parciales (\\frac{\\partial z}{\\partial u}) y (\\frac{\\partial z}{\\partial v}) de la funci\u00f3n (z = (x + 6y)e^{(x+y)}) donde (x = u) y (y = \\ln(v)), primero debemos expresar (z) en t\u00e9rminos de las nuevas variables (u) y (v).<\/p>\n\n\n\n Finalmente, las derivadas parciales son:<\/p>\n\n\n\n [ [ If z = (x + 6y)e^(x+y), x = u, y = ln(v), find (\u201a\u00e0\u00c7z\/\u201a\u00e0\u00c7u) and (\u201a\u00e0\u00c7z\/\u201a\u00e0\u00c7v)? The Correct Answer and Explanation is : Para encontrar las derivadas parciales (\\frac{\\partial z}{\\partial u}) y (\\frac{\\partial z}{\\partial v}) de la funci\u00f3n (z = (x + 6y)e^{(x+y)}) donde (x = u) y (y = \\ln(v)), primero debemos expresar (z) […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-152278","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/152278","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=152278"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/152278\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=152278"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=152278"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=152278"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
Dado que (x = u) y (y = \\ln(v)), podemos sustituir estas variables en la expresi\u00f3n de (z): [
z = (u + 6\\ln(v)) e^{(u + \\ln(v))}
] Simplificando la parte exponencial: [
e^{(u + \\ln(v))} = e^u \\cdot e^{\\ln(v)} = e^u \\cdot v
] Entonces, la expresi\u00f3n de (z) se convierte en: [
z = (u + 6\\ln(v)) (e^u \\cdot v) = v (u + 6\\ln(v)) e^u
]<\/li>\n\n\n\n
Para calcular (\\frac{\\partial z}{\\partial u}), tratamos (v) como una constante: [
z = v (u + 6\\ln(v)) e^u
] Aplicamos la regla del producto y la regla de la cadena: [
\\frac{\\partial z}{\\partial u} = v e^u + v (u + 6\\ln(v)) e^u = v e^u (1 + (u + 6\\ln(v)))
] Simplificando, obtenemos: [
\\frac{\\partial z}{\\partial u} = v e^u (u + 6\\ln(v) + 1)
]<\/li>\n\n\n\n
Ahora, para (\\frac{\\partial z}{\\partial v}), consideramos a (u) constante: [
z = v (u + 6\\ln(v)) e^u
] Usando la regla del producto y derivando (6\\ln(v)): [
\\frac{\\partial z}{\\partial v} = (u + 6\\ln(v)) e^u + v \\cdot \\frac{6}{v} e^u
] Simplificando: [
\\frac{\\partial z}{\\partial v} = (u + 6\\ln(v)) e^u + 6 e^u = (u + 6\\ln(v) + 6)e^u
]<\/li>\n<\/ol>\n\n\n\n
\\frac{\\partial z}{\\partial u} = v e^u (u + 6\\ln(v) + 1)
]<\/p>\n\n\n\n
\\frac{\\partial z}{\\partial v} = (u + 6\\ln(v) + 6)e^u
]<\/p>\n","protected":false},"excerpt":{"rendered":"