13.2.10 - first slide.jpg

Slide53.JPG
Slide54.JPG

  1. Chapter 13 Class 11 Limits and Derivatives
  2. Serial order wise

Transcript

Ex 13.2, 10 Find the derivative of cos x from first principle. Let f (x) = cos x We need to find 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) = cos x So, f (x + h) = cos (x + h) Putting values, f’ (x) = lim┬(hβ†’0)⁑〖(𝒄𝒐𝒔 (𝒙 + 𝒉) βˆ’γ€– 𝒄𝒐𝒔〗⁑𝒙)/hγ€— Using cos A – cos B = – 2 sin ((𝐴 + 𝐡)/2) sin ((𝐴 βˆ’ 𝐡)/2) = lim┬(hβ†’0)⁑〖(βˆ’πŸ π’”π’Šπ’((𝒙 + (𝒙 + 𝒉))/𝟐) . π’”π’Šπ’(((𝒙 + 𝒉) βˆ’ 𝒙)/𝟐))/hγ€— = lim┬(hβ†’0)⁑〖(βˆ’2 𝑠𝑖𝑛((2π‘₯ + β„Ž)/2) . 𝑠𝑖𝑛(β„Ž/2))/hγ€— = lim┬(hβ†’0)β‘γ€–βˆ’2 sin⁑((2π‘₯ + β„Ž)/2).γ€–sin γ€—β‘γ€–β„Ž/2γ€—/β„Žγ€— = lim┬(hβ†’0)β‘γ€–βˆ’sin⁑((2π‘₯ + β„Ž)/2).γ€–sin γ€—β‘γ€–β„Ž/2γ€—/(β„Ž/2)γ€— Using (π‘™π‘–π‘š)┬(π‘₯β†’0)⁑〖 𝑠𝑖𝑛⁑π‘₯/π‘₯γ€—=1 Replacing x by β„Ž/2 β‡’ (π‘™π‘–π‘š)┬(β„Žβ†’0) 𝑠𝑖𝑛⁑〖 β„Ž/2γ€—/(( β„Ž)/2) = 1 = lim┬(hβ†’0)β‘γ€–βˆ’sin⁑((2π‘₯ + β„Ž)/2).(π₯𝐒𝐦)┬(π‘β†’πŸŽ) 〖𝐬𝐒𝐧 〗⁑〖𝒉/πŸγ€—/(𝒉/𝟐)γ€— = lim┬(hβ†’0)β‘γ€–βˆ’sin⁑((2π‘₯ + β„Ž)/2).πŸγ€— = lim┬(hβ†’0)β‘γ€–βˆ’sin⁑((2π‘₯ + β„Ž)/2) γ€— Putting h = 0 = βˆ’sin⁑((2π‘₯ +0)/2) = βˆ’sin⁑(2π‘₯/2) = – sin x ∴ f’(x) = –sin x

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.