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Transcript

Ex 12.2, 4 Find the derivative of the following functions from first principle. (iii) 1/๐‘ฅ^2 Let f (x) = 1/๐‘ฅ^2 We need to find derivative of f(x) i.e. fโ€™ (x) We know that fโ€™(x) = (๐‘™๐‘–๐‘š)โ”ฌ(โ„Žโ†’0) ๐‘“โกใ€–(๐‘ฅ + โ„Ž) โˆ’ ๐‘“(๐‘ฅ)ใ€—/โ„Ž Here, f (x) = 1/๐‘ฅ^2 So, f (x + h) = 1/ใ€–(๐‘ฅ + โ„Ž)ใ€—^2 Putting values fโ€™(x) = (๐‘™๐‘–๐‘š)โ”ฌ(โ„Žโ†’0)โกใ€–(1/ใ€–(๐‘ฅ + โ„Ž)ใ€—^2 โˆ’ 1/๐‘ฅ^2 )/โ„Žใ€— = (๐‘™๐‘–๐‘š)โ”ฌ(โ„Žโ†’0)โกใ€–(ใ€–ใ€–๐‘ฅ ใ€—^2 โˆ’ (๐‘ฅ + โ„Ž)ใ€—^2/(ใ€–(๐‘ฅ + โ„Ž)ใ€—^2 ๐‘ฅ^2 ))/โ„Žใ€— = (๐‘™๐‘–๐‘š)โ”ฌ(โ„Žโ†’0)โกใ€–( ๐‘ฅ2 โˆ’ (๐‘ฅ + โ„Ž)2)/(โ„Ž๐‘ฅ2 (๐‘ฅ + โ„Ž)2)ใ€— = (๐‘™๐‘–๐‘š)โ”ฌ(โ„Žโ†’0)โกใ€–((๐‘ฅ โˆ’ ( ๐‘ฅ + โ„Ž )) (๐‘ฅ + (๐‘ฅ + โ„Ž)))/(โ„Ž๐‘ฅ2 (๐‘ฅ + โ„Ž)2)ใ€— = (๐‘™๐‘–๐‘š)โ”ฌ(โ„Žโ†’0)โกใ€–( (๐‘ฅ โˆ’๐‘ฅ โˆ’ โ„Ž) (๐‘ฅ + ๐‘ฅ + โ„Ž))/(โ„Ž.๐‘ฅ2 (๐‘ฅ + โ„Ž)2)ใ€— (As a2 โ€“ b2 = (a โ€“ b) (a + b)) = (๐‘™๐‘–๐‘š)โ”ฌ(โ„Žโ†’0)โกใ€–((โˆ’โ„Ž)(2๐‘ฅ + โ„Ž))/(โ„Ž๐‘ฅ2 (๐‘ฅ + โ„Ž)2)ใ€— = (๐‘™๐‘–๐‘š)โ”ฌ(โ„Žโ†’0)โกใ€–((โˆ’1) 2๐‘ฅ + โ„Ž)/(๐‘ฅ2 (๐‘ฅ + โ„Ž)2)ใ€— Putting h = 0 = ((โˆ’1) 2๐‘ฅ + 0)/(๐‘ฅ2 (๐‘ฅ + 0)2) = ((โˆ’1) 2๐‘ฅ)/(๐‘ฅ2 (๐‘ฅ)2) = (โˆ’2๐‘ฅ)/๐‘ฅ4 = (โˆ’2)/๐‘ฅ^3 Thus, fโ€™(x) = (โˆ’๐Ÿ)/๐’™^๐Ÿ‘

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.