{"id":268254,"date":"2025-07-24T02:18:55","date_gmt":"2025-07-24T02:18:55","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=268254"},"modified":"2025-07-24T02:18:57","modified_gmt":"2025-07-24T02:18:57","slug":"find-dy-dx-by-implicit-differentiation-cosxy-sinx-y","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/24\/find-dy-dx-by-implicit-differentiation-cosxy-sinx-y\/","title":{"rendered":"Find dy\/dx by implicit differentiation. cos(xy) = sin(x + y)"},"content":{"rendered":"\n
Find dy\/dx by implicit differentiation. cos(xy) = sin(x + y)<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/em><\/p>\n\n\n\n
To find dydx\\frac{dy}{dx}dxdy\u200b by implicit differentiation of the equation cos\u2061(xy)=sin\u2061(x+y)\\cos(xy) = \\sin(x + y)cos(xy)=sin(x+y), follow these steps:<\/p>\n\n\n\n
1. Differentiate both sides of the equation with respect to xxx.<\/h3>\n\n\n\n
We will apply the chain rule and product rule as necessary since both sides of the equation involve functions of xxx and yyy.<\/p>\n\n\n\n
Differentiate cos\u2061(xy)\\cos(xy)cos(xy) with respect to xxx. Since the argument xyxyxy is a product of xxx and yyy, use the chain rule: ddx[cos\u2061(xy)]=\u2212sin\u2061(xy)\u22c5ddx(xy)\\frac{d}{dx} \\left[ \\cos(xy) \\right] = -\\sin(xy) \\cdot \\frac{d}{dx}(xy)dxd\u200b[cos(xy)]=\u2212sin(xy)\u22c5dxd\u200b(xy)<\/li>\n\n\n\n
To differentiate xyxyxy, apply the product rule: ddx(xy)=ddx(x)\u22c5y+x\u22c5ddx(y)=y+xdydx\\frac{d}{dx}(xy) = \\frac{d}{dx}(x) \\cdot y + x \\cdot \\frac{d}{dx}(y) = y + x \\frac{dy}{dx}dxd\u200b(xy)=dxd\u200b(x)\u22c5y+x\u22c5dxd\u200b(y)=y+xdxdy\u200b<\/li>\n\n\n\n
Thus, the derivative of the left-hand side is: \u2212sin\u2061(xy)\u22c5(y+xdydx)-\\sin(xy) \\cdot (y + x \\frac{dy}{dx})\u2212sin(xy)\u22c5(y+xdxdy\u200b)<\/li>\n<\/ul>\n\n\n\n
Differentiate sin\u2061(x+y)\\sin(x + y)sin(x+y) with respect to xxx using the chain rule: ddx[sin\u2061(x+y)]=cos\u2061(x+y)\u22c5ddx(x+y)\\frac{d}{dx} \\left[ \\sin(x + y) \\right] = \\cos(x + y) \\cdot \\frac{d}{dx}(x + y)dxd\u200b[sin(x+y)]=cos(x+y)\u22c5dxd\u200b(x+y)<\/li>\n\n\n\n
The derivative of x+yx + yx+y is: ddx(x+y)=1+dydx\\frac{d}{dx}(x + y) = 1 + \\frac{dy}{dx}dxd\u200b(x+y)=1+dxdy\u200b<\/li>\n\n\n\n
So, the derivative of the right-hand side is: cos\u2061(x+y)\u22c5(1+dydx)\\cos(x + y) \\cdot (1 + \\frac{dy}{dx})cos(x+y)\u22c5(1+dxdy\u200b)<\/li>\n<\/ul>\n\n\n\n
2. Set up the equation for implicit differentiation:<\/h3>\n\n\n\n
Now, equate the derivatives of the left and right sides:\u2212sin\u2061(xy)\u22c5(y+xdydx)=cos\u2061(x+y)\u22c5(1+dydx)-\\sin(xy) \\cdot (y + x \\frac{dy}{dx}) = \\cos(x + y) \\cdot (1 + \\frac{dy}{dx})\u2212sin(xy)\u22c5(y+xdxdy\u200b)=cos(x+y)\u22c5(1+dxdy\u200b)<\/p>\n\n\n\n
3. Solve for dydx\\frac{dy}{dx}dxdy\u200b:<\/h3>\n\n\n\n
To isolate dydx\\frac{dy}{dx}dxdy\u200b, expand both sides:\u2212sin\u2061(xy)\u22c5y\u2212sin\u2061(xy)\u22c5xdydx=cos\u2061(x+y)\u22c51+cos\u2061(x+y)\u22c5dydx-\\sin(xy) \\cdot y – \\sin(xy) \\cdot x \\frac{dy}{dx} = \\cos(x + y) \\cdot 1 + \\cos(x + y) \\cdot \\frac{dy}{dx}\u2212sin(xy)\u22c5y\u2212sin(xy)\u22c5xdxdy\u200b=cos(x+y)\u22c51+cos(x+y)\u22c5dxdy\u200b<\/p>\n\n\n\n
Group the terms involving dydx\\frac{dy}{dx}dxdy\u200b on one side:\u2212sin\u2061(xy)\u22c5xdydx\u2212cos\u2061(x+y)\u22c5dydx=cos\u2061(x+y)+sin\u2061(xy)\u22c5y-\\sin(xy) \\cdot x \\frac{dy}{dx} – \\cos(x + y) \\cdot \\frac{dy}{dx} = \\cos(x + y) + \\sin(xy) \\cdot y\u2212sin(xy)\u22c5xdxdy\u200b\u2212cos(x+y)\u22c5dxdy\u200b=cos(x+y)+sin(xy)\u22c5y<\/p>\n\n\n\n
Factor out dydx\\frac{dy}{dx}dxdy\u200b:(\u2212sin\u2061(xy)\u22c5x\u2212cos\u2061(x+y))dydx=cos\u2061(x+y)+sin\u2061(xy)\u22c5y\\left( -\\sin(xy) \\cdot x – \\cos(x + y) \\right) \\frac{dy}{dx} = \\cos(x + y) + \\sin(xy) \\cdot y(\u2212sin(xy)\u22c5x\u2212cos(x+y))dxdy\u200b=cos(x+y)+sin(xy)\u22c5y<\/p>\n\n\n\n
Solve for dydx\\frac{dy}{dx}dxdy\u200b:dydx=cos\u2061(x+y)+sin\u2061(xy)\u22c5y\u2212sin\u2061(xy)\u22c5x\u2212cos\u2061(x+y)\\frac{dy}{dx} = \\frac{\\cos(x + y) + \\sin(xy) \\cdot y}{-\\sin(xy) \\cdot x – \\cos(x + y)}dxdy\u200b=\u2212sin(xy)\u22c5x\u2212cos(x+y)cos(x+y)+sin(xy)\u22c5y\u200b<\/p>\n\n\n\n
This is the derivative of yyy with respect to xxx using implicit differentiation.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
Find dy\/dx by implicit differentiation. cos(xy) = sin(x + y) The Correct Answer and Explanation is: To find dydx\\frac{dy}{dx}dxdy\u200b by implicit differentiation of the equation cos\u2061(xy)=sin\u2061(x+y)\\cos(xy) = \\sin(x + y)cos(xy)=sin(x+y), follow these steps: 1. Differentiate both sides of the equation with respect to xxx. We will apply the chain rule and product rule as necessary […]<\/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-268254","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/268254","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=268254"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/268254\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=268254"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=268254"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=268254"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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