Check sibling questions

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

Introducing your new favourite teacher - Teachoo Black, at only β‚Ή83 per month


Transcript

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‘

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.