{"id":5131,"date":"2021-09-08T20:36:28","date_gmt":"2021-09-08T15:06:28","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5131"},"modified":"2021-11-22T21:41:06","modified_gmt":"2021-11-22T16:11:06","slug":"differentiation-of-implicit-function","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/differentiation-of-implicit-function\/","title":{"rendered":"Differentiation of Implicit Function"},"content":{"rendered":"
Here you will learn what is the differentiation of implicit function with examples.<\/p>\n
Let’s begin –<\/p>\n
If the variables x and y are connected by a relation of the form f(x,y) = 0 and it is not possible or convenient to express y as a function x in the form y\u00a0 = \\(\\phi (x)\\), then y is said to be an implicit function of x. To find \\(dy\\over dx\\) in such a case, we differentiate both sides of the given relation with respect to x, keeping in mind that the derivative of \\(\\phi (y)\\) with respect to x is \\(d\\phi\\over dy\\).\\(dy\\over dx\\).<\/p>\n
for example<\/strong>, \\(d\\over dx\\) (sin y) = cos y. \\(dy\\over dx\\) , \\(d\\over dx\\) (\\(y^2\\)) = 2y\\(dy\\over dx\\)<\/p>\n It should be noted that \\(d\\over dy\\) (sin y) = cos y but \\(d\\over dx\\) (sin y) = cos y. \\(dy\\over dx\\).<\/p>\n Similarly , \\(d\\over dy\\) \\((y^3)\\) = \\(3y^2\\) whereas \\(d\\over dx\\) \\((y^3)\\) = \\(3y^2\\) \\(dy\\over dx\\).<\/p>\n Example<\/strong><\/span> : If \\(ax^2 + 2hxy + by^2 + 2gx + 2fy + c\\) = 0, find \\(dy\\over dx\\).<\/p>\n Solution<\/span><\/strong> : We have,<\/p>\n \\(ax^2 + 2hxy + by^2 + 2gx + 2fy + c\\) = 0<\/p>\n Differentiating both sides of this with respect to x, we get<\/p>\n \\(d\\over dx\\) \\((ax^2)\\) + \\(d\\over dx\\)(2hxy) + \\(d\\over dx\\)\\((by^2)\\) + \\(d\\over dx\\)(2gx) + \\(d\\over dx\\)(2fy) + \\(d\\over dx\\)(c) = \\(d\\over dx\\)(0)<\/p>\n \\(\\implies\\) a\\(d\\over dx\\)\\((x^2)\\) + 2h\\(d\\over dx\\)(xy) + b\\(d\\over dx\\)\\((y^2)\\) + 2g\\(d\\over dx\\)(x) + 2f\\(d\\over dx\\)(y) + 0 = 0<\/p>\n \\(\\implies\\) 2ax + 2h\\((x{dy\\over dx} + y)\\) + b 2y\\(dy\\over dx\\) + 2g.1 + 2f.\\(dy\\over dx\\) = 0<\/p>\n = \\(dy\\over dx\\)(2hx + 2by + 2f) + 2ax + 2hy + 2g = 0<\/p>\n \\(dy\\over dx\\) = -\\(2(ax + hy + g)\\over 2(hx + by + f)\\)<\/p>\n \\(dy\\over dx\\) = -\\((ax + hy + g)\\over (hx + by + f)\\)<\/p>\n\n\n