{"id":8508,"date":"2021-11-22T21:17:50","date_gmt":"2021-11-22T15:47:50","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8508"},"modified":"2021-12-05T17:47:27","modified_gmt":"2021-12-05T12:17:27","slug":"if-y-sqrtsinx-sqrtsinx-sqrtsinx-to-infty-find-dyover-dx","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/if-y-sqrtsinx-sqrtsinx-sqrtsinx-to-infty-find-dyover-dx\/","title":{"rendered":"If y = \\(\\sqrt{sinx + \\sqrt{sinx + \\sqrt{sinx + ……. to \\infty}}}\\), find \\(dy\\over dx\\)."},"content":{"rendered":"

Solution :<\/h2>\n

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

By using differentiation of infinite series<\/a>,<\/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","protected":false},"excerpt":{"rendered":"

Solution : Since by deleting a single term from an infinite series, it remains same. Therefore, the given function may be written as y = \\(\\sqrt{sin x + y}\\) Squaring on both sides, \\(\\implies\\)  \\(y^2\\)  = sin x + y By using differentiation of infinite series, Differentiating both sides with respect to x, 2y \\(dy\\over …<\/p>\n

If y = \\(\\sqrt{sinx + \\sqrt{sinx + \\sqrt{sinx + ……. to \\infty}}}\\), find \\(dy\\over dx\\).<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","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":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[65,43],"tags":[],"yoast_head":"\nIf y = \\(\\sqrt{sinx + \\sqrt{sinx + \\sqrt{sinx + ....... to \\infty}}}\\), find \\(dy\\over dx\\).<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/if-y-sqrtsinx-sqrtsinx-sqrtsinx-to-infty-find-dyover-dx\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"If y = \\(\\sqrt{sinx + \\sqrt{sinx + \\sqrt{sinx + ....... to \\infty}}}\\), find \\(dy\\over dx\\).\" \/>\n<meta property=\"og:description\" content=\"Solution : Since by deleting a single term from an infinite series, it remains same. Therefore, the given function may be written as y = (sqrt{sin x + y}) Squaring on both sides, (implies)  (y^2)  = sin x + y By using differentiation of infinite series, Differentiating both sides with respect to x, 2y (dyover … If y = (sqrt{sinx + sqrt{sinx + sqrt{sinx + ……. to infty}}}), find (dyover dx). 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