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Ex 13.2, 4 (iii) - Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)

Last updated at March 30, 2023 by

Ex 13.2, 4 - Chapter 13 Class 11 Limits and Derivatives - Part 7

Ex 13.2, 4 - Chapter 13 Class 11 Limits and Derivatives - Part 8
Ex 13.2, 4 - Chapter 13 Class 11 Limits and Derivatives - Part 9

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Ex 13.2, 4 Find the derivative of the following functions from first principle. (iii) 1/π‘₯^2 Let f (x) = 1/π‘₯^2 We need to find derivative of f(x) i.e. f’ (x) We know that f’(x) = (π‘™π‘–π‘š)┬(β„Žβ†’0) 𝑓⁑〖(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯)γ€—/β„Ž Here, f (x) = 1/π‘₯^2 So, f (x + h) = 1/γ€–(π‘₯ + β„Ž)γ€—^2 Putting values f’(x) = (π‘™π‘–π‘š)┬(β„Žβ†’0)⁑〖(1/γ€–(π‘₯ + β„Ž)γ€—^2 βˆ’ 1/π‘₯^2 )/β„Žγ€— = (π‘™π‘–π‘š)┬(β„Žβ†’0)⁑〖(γ€–γ€–π‘₯ γ€—^2 βˆ’ (π‘₯ + β„Ž)γ€—^2/(γ€–(π‘₯ + β„Ž)γ€—^2 π‘₯^2 ))/β„Žγ€— = (π‘™π‘–π‘š)┬(β„Žβ†’0)⁑〖( π‘₯2 βˆ’ (π‘₯ + β„Ž)2)/(β„Žπ‘₯2 (π‘₯ + β„Ž)2)γ€— = (π‘™π‘–π‘š)┬(β„Žβ†’0)⁑〖((π‘₯ βˆ’ ( π‘₯ + β„Ž )) (π‘₯ + (π‘₯ + β„Ž)))/(β„Žπ‘₯2 (π‘₯ + β„Ž)2)γ€— = (π‘™π‘–π‘š)┬(β„Žβ†’0)⁑〖( (π‘₯ βˆ’π‘₯ βˆ’ β„Ž) (π‘₯ + π‘₯ + β„Ž))/(β„Ž.π‘₯2 (π‘₯ + β„Ž)2)γ€— (As a2 – b2 = (a – b) (a + b)) = (π‘™π‘–π‘š)┬(β„Žβ†’0)⁑〖((βˆ’β„Ž)(2π‘₯ + β„Ž))/(β„Žπ‘₯2 (π‘₯ + β„Ž)2)γ€— = (π‘™π‘–π‘š)┬(β„Žβ†’0)⁑〖((βˆ’1) 2π‘₯ + β„Ž)/(π‘₯2 (π‘₯ + β„Ž)2)γ€— Putting h = 0 = ((βˆ’1) 2π‘₯ + 0)/(π‘₯2 (π‘₯ + 0)2) = ((βˆ’1) 2π‘₯)/(π‘₯2 (π‘₯)2) = (βˆ’2π‘₯)/π‘₯4 = (βˆ’2)/π‘₯^3 Thus, f’(x) = (βˆ’πŸ)/𝒙^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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.