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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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