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Ex 13.2, 5 - For f(x) = x100/100 + x99/100 + ... + x2/2 + x + 1 - Derivatives by formula - x^n formula

Ex 13.2, 5 - Chapter 13 Class 11 Limits and Derivatives - Part 2
Ex 13.2, 5 - Chapter 13 Class 11 Limits and Derivatives - Part 3

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Ex 13.2, 5 For the function f(x) = x100﷮100﷯ + x99﷮100﷯ +….+ x2﷮2﷯ + x + 1. Prove that f’(1) = 100 f’(0) We have f (x) = 𝑥﷮100﷯﷮100﷯ + 𝑥﷮99﷯﷮99﷯ + …… + 𝑥﷮2﷯﷮2﷯ + x + 1 f’ (x) = 1﷮100﷯ x100 + 1﷮99﷯ x99 + …… + 1﷮2﷯ x2 + x1 + 1﷯﷮′﷯ f’ (x) = 1﷮100﷯ × 100x100 – 1 + 1﷮99﷯ × 99x99 – 1 + … + 1﷮2﷯ × 2x2 – 1 + 1.x1-1 + 0 = 100﷮100﷯ x99 + 99﷮99﷯ x98 + …+ 2﷮2﷯ x1 + x0 + 0 = x99 + x98 + …..+ x + 1 + 0 = x99 + x98 + … + x + 1 Hence , f’ (x) = x99 + x98 + … + x + 1 We need to prove f’(1) = 100 f’(0) Hence R.H.S = L.H.S Hence proved

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.