We are asked to differentiate the implicit equation:cos\u2061(xy)=x+3y\\cos(xy) = x + 3ycos(xy)=x+3y<\/p>\n\n\n\n
To solve this, we will use implicit differentiation<\/strong>. The idea is to differentiate both sides of the equation with respect to xxx, treating yyy as an implicit function of xxx (i.e., y=y(x)y = y(x)y=y(x)).<\/p>\n\n\n\n The left-hand side involves the composite function cos\u2061(xy)\\cos(xy)cos(xy). We’ll need to apply the chain rule<\/strong> here. The derivative of cos\u2061(u)\\cos(u)cos(u) is \u2212sin\u2061(u)-\\sin(u)\u2212sin(u), where u=xyu = xyu=xy, so we need to differentiate xyxyxy with respect to xxx.<\/p>\n\n\n\n Using the product rule<\/strong> to differentiate xyxyxy, we get:ddx(xy)=xdydx+y\\frac{d}{dx}(xy) = x \\frac{dy}{dx} + ydxd\u200b(xy)=xdxdy\u200b+y<\/p>\n\n\n\n Thus, the derivative of cos\u2061(xy)\\cos(xy)cos(xy) with respect to xxx is:ddx[cos\u2061(xy)]=\u2212sin\u2061(xy)\u22c5(xdydx+y)\\frac{d}{dx} \\left[ \\cos(xy) \\right] = -\\sin(xy) \\cdot \\left( x \\frac{dy}{dx} + y \\right)dxd\u200b[cos(xy)]=\u2212sin(xy)\u22c5(xdxdy\u200b+y)<\/p>\n\n\n\n The right-hand side of the equation is x+3yx + 3yx+3y. We differentiate each term:ddx(x)=1andddx(3y)=3dydx\\frac{d}{dx}(x) = 1 \\quad \\text{and} \\quad \\frac{d}{dx}(3y) = 3 \\frac{dy}{dx}dxd\u200b(x)=1anddxd\u200b(3y)=3dxdy\u200b<\/p>\n\n\n\n So, the derivative of the right-hand side is:1+3dydx1 + 3 \\frac{dy}{dx}1+3dxdy\u200b<\/p>\n\n\n\n Now, we can equate the derivatives of both sides:\u2212sin\u2061(xy)\u22c5(xdydx+y)=1+3dydx-\\sin(xy) \\cdot \\left( x \\frac{dy}{dx} + y \\right) = 1 + 3 \\frac{dy}{dx}\u2212sin(xy)\u22c5(xdxdy\u200b+y)=1+3dxdy\u200b<\/p>\n\n\n\n To isolate dydx\\frac{dy}{dx}dxdy\u200b, distribute \u2212sin\u2061(xy)-\\sin(xy)\u2212sin(xy) on the left side:\u2212sin\u2061(xy)\u22c5xdydx\u2212sin\u2061(xy)\u22c5y=1+3dydx-\\sin(xy) \\cdot x \\frac{dy}{dx} – \\sin(xy) \\cdot y = 1 + 3 \\frac{dy}{dx}\u2212sin(xy)\u22c5xdxdy\u200b\u2212sin(xy)\u22c5y=1+3dxdy\u200b<\/p>\n\n\n\n Now, collect the terms involving dydx\\frac{dy}{dx}dxdy\u200b on one side:\u2212sin\u2061(xy)\u22c5xdydx\u22123dydx=1+sin\u2061(xy)\u22c5y-\\sin(xy) \\cdot x \\frac{dy}{dx} – 3 \\frac{dy}{dx} = 1 + \\sin(xy) \\cdot y\u2212sin(xy)\u22c5xdxdy\u200b\u22123dxdy\u200b=1+sin(xy)\u22c5y<\/p>\n\n\n\n Factor out dydx\\frac{dy}{dx}dxdy\u200b:(\u2212sin\u2061(xy)\u22c5x\u22123)dydx=1+sin\u2061(xy)\u22c5y\\left( -\\sin(xy) \\cdot x – 3 \\right) \\frac{dy}{dx} = 1 + \\sin(xy) \\cdot y(\u2212sin(xy)\u22c5x\u22123)dxdy\u200b=1+sin(xy)\u22c5y<\/p>\n\n\n\n Finally, solve for dydx\\frac{dy}{dx}dxdy\u200b:dydx=1+sin\u2061(xy)\u22c5y\u2212sin\u2061(xy)\u22c5x\u22123\\frac{dy}{dx} = \\frac{1 + \\sin(xy) \\cdot y}{-\\sin(xy) \\cdot x – 3}dxdy\u200b=\u2212sin(xy)\u22c5x\u221231+sin(xy)\u22c5y\u200b<\/p>\n\n\n\n The derivative of yyy with respect to xxx is:dydx=1+sin\u2061(xy)\u22c5y\u2212sin\u2061(xy)\u22c5x\u22123\\frac{dy}{dx} = \\frac{1 + \\sin(xy) \\cdot y}{-\\sin(xy) \\cdot x – 3}dxdy\u200b=\u2212sin(xy)\u22c5x\u221231+sin(xy)\u22c5y\u200b<\/p>\n\n\n\n This gives the rate of change of yyy with respect to xxx in terms of both xxx and yyy.<\/p>\n\n\n\n Differentiate: cosxy=x+3y Differentiate: cos xy = x + 3y The Correct Answer and Explanation is: We are asked to differentiate the implicit equation:cos\u2061(xy)=x+3y\\cos(xy) = x + 3ycos(xy)=x+3y To solve this, we will use implicit differentiation. The idea is to differentiate both sides of the equation with respect to xxx, treating yyy as an implicit function […]<\/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-268258","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/268258","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=268258"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/268258\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=268258"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=268258"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=268258"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Differentiate the left side<\/h3>\n\n\n\n
Step 2: Differentiate the right side<\/h3>\n\n\n\n
Step 3: Combine the derivatives<\/h3>\n\n\n\n
Step 4: Solve for dydx\\frac{dy}{dx}dxdy\u200b<\/h3>\n\n\n\n
Conclusion<\/h3>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-1775.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"