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Misc 1 (ii) - Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)

Last updated at Sept. 6, 2021 by

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Misc 1 Find the derivative of the following functions from first principle: (ii) (βˆ’π‘₯)^(βˆ’1) Let f (x) = (βˆ’π‘₯)^(βˆ’1) We need to find Derivative of f(x) i.e. f’ (x) We know that f’(x) = lim┬(hβ†’0) f⁑〖(x + h) βˆ’ f(x)γ€—/h Here, f (x) = (βˆ’π‘₯)^(βˆ’1) So, f (x + h) = (βˆ’(π‘₯+β„Ž))^(βˆ’1) Putting values f’(x) = lim┬(hβ†’0)⁑〖(γ€–(βˆ’(π‘₯ + β„Ž)) γ€—^(βˆ’1) γ€–βˆ’ (βˆ’ π‘₯)γ€—^(βˆ’1))/β„Žγ€— = lim┬(hβ†’0)⁑〖((1/(βˆ’(π‘₯ + β„Ž))) βˆ’(1/(βˆ’π‘₯)))/β„Žγ€— = lim┬(hβ†’0)⁑〖((βˆ’1)/(π‘₯ + β„Ž) + 1/π‘₯)/β„Žγ€— = lim┬(hβ†’0)⁑〖((βˆ’ π‘₯ + π‘₯ + β„Ž)/(π‘₯ (π‘₯ + β„Ž) ))/β„Žγ€— = lim┬(hβ†’0)⁑〖(βˆ’ π‘₯ + π‘₯ + β„Ž)/(β„Žπ‘₯ (π‘₯ + β„Ž))γ€— = lim┬(hβ†’0)β‘γ€–β„Ž/(β„Žπ‘₯ (π‘₯ + β„Ž))γ€— = lim┬(hβ†’0)⁑〖1/(π‘₯ (π‘₯ + β„Ž))γ€— Putting h = 0 = 1/(π‘₯ (π‘₯ + 0)) = 𝟏/π’™πŸ

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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.