{"id":5135,"date":"2021-09-08T20:41:13","date_gmt":"2021-09-08T15:11:13","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5135"},"modified":"2021-11-22T20:59:39","modified_gmt":"2021-11-22T15:29:39","slug":"differentiation-of-infinite-series","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/differentiation-of-infinite-series\/","title":{"rendered":"Differentiation of Infinite Series Class 12"},"content":{"rendered":"
Here you will learn what is differentiation of infinite series class 12 with examples.<\/p>\n
Let’s begin –<\/p>\n
Sometimes the value of y is given as an infinite series and we are asked to find \\(dy\\over dx\\). In such cases we use the fact that if a term is deleted from a infinite series, it remains unaffected. The method of finding \\(dy\\over dx\\) is explained in the following examples.<\/p>\n
Example 1<\/span><\/strong> : If y = \\(x^{x^{x^{…\\infty}}}\\), find \\(dy\\over dx\\).<\/p>\n Solution<\/span><\/strong> : Since by deleting a single term from an infinite series, it remains same. Therefore, the given function may be written as<\/p>\n y = \\(x^y\\)<\/p>\n Taking log on both sides,<\/p>\n \\(\\implies\\) log y = y logx<\/p>\n Differentiating both sides with respect to x,<\/p>\n \\(1\\over y\\)\\(dy\\over dx\\) = \\(dy\\over dx\\) log x + y \\(d\\over dx\\) (log x)<\/p>\n \\(1\\over y\\)\\(dy\\over dx\\) = \\(dy\\over dx\\) log x + \\(y\\over x\\)<\/p>\n \\(dy\\over dx\\){\\({{1\\over y} – log x}\\)} = \\(y\\over x\\)<\/p>\n \\(\\implies\\) \\(dy\\over dx\\)\\((1 – y log x)\\over y\\) = \\(y\\over x\\)<\/p>\n \\(\\implies\\) \\(dy\\over dx\\) = \\(y^2\\over {x(1 – ylog x)}\\)<\/p>\n Example 2<\/span><\/strong> : If y = \\(\\sqrt{sinx + \\sqrt{sinx + \\sqrt{sinx + ……. to \\infty}}}\\), find \\(dy\\over dx\\).<\/p>\n Solution<\/span><\/strong> : Since by deleting a single term from an infinite series, it remains same. Therefore, the given function may be written as<\/p>\n y = \\(\\sqrt{sin x + y}\\)<\/p>\n Squaring on both sides,<\/p>\n \\(\\implies\\) \\(y^2\\) = sin x + y<\/p>\n Differentiating both sides with respect to x,<\/p>\n 2y \\(dy\\over dx\\) =cosx + \\(dy\\over dx\\)<\/p>\n \\(\\implies\\) \\(dy\\over dx\\)\\((2y – 1)\\) = cos x<\/p>\n \\(\\implies\\) \\(dy\\over dx\\) = \\(cos x\\over {2y – 1}\\)<\/p>\n\n\n