{"id":5983,"date":"2021-10-04T21:05:19","date_gmt":"2021-10-04T15:35:19","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5983"},"modified":"2021-10-04T21:06:05","modified_gmt":"2021-10-04T15:36:05","slug":"differentiability-examples","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/differentiability-examples\/","title":{"rendered":"Differentiability Examples"},"content":{"rendered":"

Here you will learn some differentiability examples for better understanding of differentiability concepts<\/a><\/span>.<\/p>\n\n\n

\n
\n\n \n

Example 1 : <\/span> If f(x + y) = f(x) + f(y) – 2xy – 1 for all x and y. If f'(0) exists and f'(0) = -sin\\(\\alpha\\), then find f{f'(0)}.<\/p>\n

Solution : <\/span>f'(x) = \\(\\displaystyle{\\lim_{h \\to 0}}\\) \\(f(x+h) – f(x)\\over h\\)

\n = \\(\\displaystyle{\\lim_{h \\to 0}}\\) \\({f(x)+f(h)-2xh-1} – f(x)\\over h\\)

\n = \\(\\displaystyle{\\lim_{h \\to 0}}\\) -2x + \\(\\displaystyle{\\lim_{h \\to 0}}\\) \\(f(h)-1\\over h\\) = -2x + \n \\(\\displaystyle{\\lim_{h \\to 0}}\\) \\(f(h) – f(0)\\over h\\)

\n [Putting x = 0 = y in the given relation we find f(0) = f(0) + f(0) + 0 – 1 \\(\\implies\\) f(0) = 1]

\n \\(\\therefore\\)   f'(x) = -2x + f'(0) = -2x – sin\\(\\alpha\\)

\n \\(\\implies\\)   f(x) = -\\(x^2\\) – (sin\\(\\alpha\\)).x + c

\n       f(0) = – 0 – 0 + c \\(\\implies\\) c = 1

\n \\(\\therefore\\)   f(x) = -\\(x^2\\) – (sin\\(\\alpha\\)).x + 1

\n so, f{f'(0)} = f(-sin\\(\\alpha\\)) = -\\(sin^2\\alpha\\) + \\(sin^2\\alpha\\) + 1 = 1<\/p>\n


\n\n \n

Example 2 : <\/span>Discuss the continuity and differentiability of the function y = f(x) defined parametrically; x = 2t – |t-1| and y = 2\\(t^2\\) + t|t|.<\/p>\n\t\t

Solution : <\/span>Here x = 2t – |t-1| and y = 2\\(t^2\\) + t|t|.

\n\t\t Now when t < 0;

\n\t\t x = 2t – {-(t-1)} = 3t – 1 and y = 2\\(t^2\\) – \\(t^2\\) = \\(t^2\\) \\(\\implies\\) = y = \\({1\\over 9}{(x+1)}^2\\)

\n\t\t = when 0 \\(\\le\\) t < 1

\n\t\t x = 2t – {-(t-1)} = 3t – 1 and y = 2\\(t^2\\) – \\(t^2\\) = 3\\(t^2\\) \\(\\implies\\) = y = \\({1\\over 3}{(x+1)}^2\\)

\n\t\t when t \\(\\ge\\) 1;

\n\t\t x = 2t – (t-1) = t + 1 and y = 2\\(t^2\\) + \\(t^2\\) = 3\\(t^2\\) \\(\\implies\\) = y = 3\\({(x+1)}^2\\)

\n\t\t Thus, y = f(x) = \\({1\\over 9}{(x+1)}^2\\), x < -1

\n\t\t y = f(x) = \\({1\\over 3}{(x+1)}^2\\), -1\\(\\le\\)x < 2

\n\t\t y = f(x) = 3\\({(x+1)}^2\\), x\\(\\ge\\) 2

\n\t\t We have to check differentiability at x = -1 and 2.

\n\t\t Differentiabilty at x = -1;\n\t\t LHD = f'(\\(-1)^-\\) = \\(\\displaystyle{\\lim_{h \\to 0}}\\)\\(f(-1-h) – f(-1)\\over -h\\) = \n\t\t \\(\\displaystyle{\\lim_{h \\to 0}}\\) \\({1\\over 9}(-1-h+1)^2 – 0\\over -h\\) = 0

\n\t\t RHD = f'(\\(-1)^+\\) = \\(\\displaystyle{\\lim_{h \\to 0}}\\)\\(f(-1+h) – f(-1)\\over h\\) = \n\t\t \\(\\displaystyle{\\lim_{h \\to 0}}\\) \\({1\\over 3}(-1+h+1)^2 – 0\\over h\\) = 0

\n\t\t Hence f(x) is differentiable at x = -1

\n\t\t \\(\\implies\\)     continuous at x = -1.

\n\t\t To check differentiability at x = 2;

\n\t\t LHD = f'(\\(2)^-\\) = \\(\\displaystyle{\\lim_{h \\to 0}}\\)\\({1\\over 3}(2-h+1)^2 – 3\\over -h\\) = 2

\n\t\t RHD = f'(\\(2)^+\\) = \\(\\displaystyle{\\lim_{h \\to 0}}\\)\\(3(2+h-1)^2 – 3\\over h\\) = 6

\n\t\t Hence f(x) is not differentiable at x = 2.

\n\t\t But continuous at x = 2, because LHD and RHD both are finite.

\n\t\t f(x) is continuous for all x and differentiable for all x, except x = 2.\n
\n<\/p>

Practice these given differentiability examples to test your knowledge on concepts of differentiability.<\/p>\n \n <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":"

Here you will learn some differentiability examples for better understanding of differentiability concepts. Example 1 : If f(x + y) = f(x) + f(y) – 2xy – 1 for all x and y. If f'(0) exists and f'(0) = -sin\\(\\alpha\\), then find f{f'(0)}. Solution : f'(x) = \\(\\displaystyle{\\lim_{h \\to 0}}\\) \\(f(x+h) – f(x)\\over h\\) = …<\/p>\n

Differentiability Examples<\/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":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[21],"tags":[],"yoast_head":"\nDifferentiability Examples - Mathemerize<\/title>\n<meta name=\"description\" content=\"Solve these differentiability examples to practice and test your knowledge of differentiability concepts.\" \/>\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\/differentiability-examples\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Differentiability Examples - 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