{"id":183730,"date":"2025-01-17T03:00:02","date_gmt":"2025-01-17T03:00:02","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183730"},"modified":"2025-01-17T03:00:05","modified_gmt":"2025-01-17T03:00:05","slug":"11-fx-x-4-02-12-fx-tan-x-y-x-00-da-comparing-ay-and-dy-in-exercises-13-18","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/17\/11-fx-x-4-02-12-fx-tan-x-y-x-00-da-comparing-ay-and-dy-in-exercises-13-18\/","title":{"rendered":"11. f(x) = x + 4 (0,2) 12. f(x) = tan x y = x (0,0) DA Comparing Ay and dy In Exercises 13\u201318"},"content":{"rendered":"\n<ol start=\"11\" class=\"wp-block-list\">\n<li>f(x) = x + 4 (0,2) 12. f(x) = tan x y = x (0,0) DA Comparing Ay and dy In Exercises 13\u201318, use the information to find and compare Ay and dy. x-Value x = 1 x = -2 Function 13. y = 0.5&#215;3 14. y = 6 &#8211; 2&#215;2 15. y = x4 + 1 16, y = 2 &#8211; ** 17. y = x &#8211; 2r3 18. y = 7&#215;2 \u2013 5x Differential of x Ax = dx = 0.1 Ax = dx = 0.1 Ax = dx = 0.01 Ax = dx = 0.01 Ax = dx = 0.001 Ax = dx = 0.001 x = -4<\/li>\n<\/ol>\n\n\n\n<p>I need help with Question 16. Thank You.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Question 16:<\/h3>\n\n\n\n<p>The function given is y=2\u2212x2y = 2 &#8211; x^2, and you need to compare \u0394y\\Delta y and dydy at x=\u22124x = -4, where \u0394x=dx=0.01\\Delta x = dx = 0.01.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Step 1: Define the Changes<\/strong><\/h4>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Change in yy (\u0394y\\Delta y):<\/strong> \u0394y=y(x+\u0394x)\u2212y(x)\\Delta y = y(x + \\Delta x) &#8211; y(x).<\/li>\n\n\n\n<li><strong>Differential (dydy):<\/strong> The differential is an approximation of \u0394y\\Delta y and is calculated as: dy=dydx\u22c5dxdy = \\frac{dy}{dx} \\cdot dx where dydx\\frac{dy}{dx} is the derivative of yy with respect to xx.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Step 2: Derivative of y=2\u2212x2y = 2 &#8211; x^2<\/strong><\/h4>\n\n\n\n<p>The derivative is: dydx=\u22122x\\frac{dy}{dx} = -2x<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Step 3: Compute \u0394y\\Delta y<\/strong><\/h4>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Original yy:<br>At x=\u22124x = -4: y(\u22124)=2\u2212(\u22124)2=2\u221216=\u221214y(-4) = 2 &#8211; (-4)^2 = 2 &#8211; 16 = -14<\/li>\n\n\n\n<li>New xx:<br>x+\u0394x=\u22124+0.01=\u22123.99x + \\Delta x = -4 + 0.01 = -3.99.<\/li>\n\n\n\n<li>New yy: y(\u22123.99)=2\u2212(\u22123.99)2=2\u2212(15.9201)=\u221213.9201y(-3.99) = 2 &#8211; (-3.99)^2 = 2 &#8211; (15.9201) = -13.9201<\/li>\n\n\n\n<li>Change in yy: \u0394y=y(\u22123.99)\u2212y(\u22124)=\u221213.9201\u2212(\u221214)=0.0799\\Delta y = y(-3.99) &#8211; y(-4) = -13.9201 &#8211; (-14) = 0.0799<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Step 4: Compute dydy<\/strong><\/h4>\n\n\n\n<p>dy=dydx\u22c5dx=(\u22122x)\u22c5dxdy = \\frac{dy}{dx} \\cdot dx = (-2x) \\cdot dx<\/p>\n\n\n\n<p>At x=\u22124x = -4: dydx=\u22122(\u22124)=8\\frac{dy}{dx} = -2(-4) = 8 dy=8\u22c50.01=0.08dy = 8 \\cdot 0.01 = 0.08<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Step 5: Compare \u0394y\\Delta y and dydy<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u0394y=0.0799\\Delta y = 0.0799<\/li>\n\n\n\n<li>dy=0.08dy = 0.08<\/li>\n<\/ul>\n\n\n\n<p>The values are very close, showing that dydy provides a good linear approximation to \u0394y\\Delta y when \u0394x\\Delta x is small.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The differential dydy is a linear approximation of the actual change in yy, denoted \u0394y\\Delta y, based on the derivative of the function at a given point. Here, \u0394x=0.01\\Delta x = 0.01 is small, so dydy closely approximates \u0394y\\Delta y. This relationship is more accurate as \u0394x\\Delta x becomes smaller because the curve of the function is locally linear over tiny intervals.<\/p>\n\n\n\n<p>For y=2\u2212x2y = 2 &#8211; x^2, the derivative dydx=\u22122x\\frac{dy}{dx} = -2x reflects the instantaneous rate of change at any point xx. At x=\u22124x = -4, the rate of change is positive (dydx=8\\frac{dy}{dx} = 8), indicating that yy increases as xx increases near this point. Calculating both \u0394y\\Delta y and dydy confirms the accuracy of this approximation.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>I need help with Question 16. Thank You. The correct answer and explanation is: Question 16: The function given is y=2\u2212x2y = 2 &#8211; x^2, and you need to compare \u0394y\\Delta y and dydy at x=\u22124x = -4, where \u0394x=dx=0.01\\Delta x = dx = 0.01. Step 1: Define the Changes Step 2: Derivative of y=2\u2212x2y [&hellip;]<\/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-183730","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183730","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=183730"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183730\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183730"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183730"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183730"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}