Misc 1 (ii) - Chapter 12 Class 11 Limits and Derivatives

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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

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 and Computer Science at Teachoo