To find the derivative dydx\\frac{dy}{dx}dxdy\u200b at the point (2, 4) for the equation xy=yxxy = yxxy=yx, we will first differentiate the given equation implicitly.<\/p>\n\n\n\n
Step 1: Understand the equation<\/h3>\n\n\n\n
We are given that:xy=yxxy = yxxy=yx<\/p>\n\n\n\n
At first glance, this may appear trivial, but let’s delve deeper. The equation is symmetric, meaning both sides are similar, so it could represent a variety of solutions, not just a simple product of variables.<\/p>\n\n\n\n
Step 2: Differentiate implicitly<\/h3>\n\n\n\n
We will treat xxx and yyy as functions of xxx, meaning y=y(x)y = y(x)y=y(x) and apply the product rule of differentiation to both sides of the equation. Start by differentiating the left-hand side xyxyxy:ddx(xy)=ddx(y\u22c5x)\\frac{d}{dx}(xy) = \\frac{d}{dx}(y \\cdot x)dxd\u200b(xy)=dxd\u200b(y\u22c5x)<\/p>\n\n\n\n
Using the product rule, the derivative of xyxyxy is:xdydx+yx \\frac{dy}{dx} + yxdxdy\u200b+y<\/p>\n\n\n\n
Now, differentiate the right-hand side, yxyxyx:ddx(yx)=ddx(x\u22c5y)=dydx\u22c5x+y\\frac{d}{dx}(yx) = \\frac{d}{dx}(x \\cdot y) = \\frac{dy}{dx} \\cdot x + ydxd\u200b(yx)=dxd\u200b(x\u22c5y)=dxdy\u200b\u22c5x+y<\/p>\n\n\n\n
Step 3: Set up the equation<\/h3>\n\n\n\n
Since both sides are equal, we equate the derivatives:xdydx+y=dydx\u22c5x+yx \\frac{dy}{dx} + y = \\frac{dy}{dx} \\cdot x + yxdxdy\u200b+y=dxdy\u200b\u22c5x+y<\/p>\n\n\n\n
Step 4: Solve for dydx\\frac{dy}{dx}dxdy\u200b<\/h3>\n\n\n\n
Now, subtract yyy from both sides to eliminate it:xdydx=xdydxx \\frac{dy}{dx} = x \\frac{dy}{dx}xdxdy\u200b=xdxdy\u200b<\/p>\n\n\n\n
This simplifies to:0=00 = 00=0<\/p>\n\n\n\n
Which is always true. Thus, the equation does not provide a direct relationship for dydx\\frac{dy}{dx}dxdy\u200b, suggesting that the derivative dydx\\frac{dy}{dx}dxdy\u200b may not change depending on the values of xxx and yyy. In fact, the equation is an identity and holds for all values, meaning yyy does not depend on xxx in a typical way that we can express with a derivative.<\/p>\n\n\n\n
Step 5: Evaluate at (2, 4)<\/h3>\n\n\n\n
Since the equation xy=yxxy = yxxy=yx holds universally, the derivative dydx\\frac{dy}{dx}dxdy\u200b does not change at specific points like (2, 4). Therefore, dydx\\frac{dy}{dx}dxdy\u200b is undefined or indeterminate at that point.<\/p>\n\n\n\n
Conclusion<\/h3>\n\n\n\n
In conclusion, the equation xy=yxxy = yxxy=yx leads to an identity that holds for all xxx and yyy. The derivative dydx\\frac{dy}{dx}dxdy\u200b at any point, including (2, 4), is undefined, since no unique relationship between yyy and xxx can be derived from this equation. Thus, the answer is that dydx\\frac{dy}{dx}dxdy\u200b does not exist at this point.<\/p>\n\n\n\n If xy =yx, find dy\/dx at (2,4). The Correct Answer and Explanation is: To find the derivative dydx\\frac{dy}{dx}dxdy\u200b at the point (2, 4) for the equation xy=yxxy = yxxy=yx, we will first differentiate the given equation implicitly. Step 1: Understand the equation We are given that:xy=yxxy = yxxy=yx At first glance, this may appear trivial, […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-44318","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44318","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=44318"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44318\/revisions"}],"predecessor-version":[{"id":44325,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44318\/revisions\/44325"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=44318"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=44318"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=44318"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}![\"\"](\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1493.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"