Misc 1 - Chapter 13 Class 11 Limits and Derivatives - Part 6

Advertisement

Misc 1 - Chapter 13 Class 11 Limits and Derivatives - Part 7

Advertisement

Misc 1 - Chapter 13 Class 11 Limits and Derivatives - Part 8

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

Transcript

Misc 1 Find the derivative of the following functions from first principle: (iii) sin (x + 1) Let f (x) = sin (x + 1) We need to find Derivative of f(x) We know that f’(x) = lim┬(hβ†’0) 𝑓⁑〖(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯)γ€—/β„Ž Here, f (x) = sin (x + 1) So, f (x + h) = sin ((x + h) + 1) Putting values f’(x) = lim┬(hβ†’0)⁑〖sin⁑〖(( π‘₯ + β„Ž ) + 1 ) βˆ’γ€– sin〗⁑〖 (π‘₯ + 1)γ€— γ€—/hγ€— = lim┬(hβ†’0)⁑〖(𝑠𝑖𝑛( π‘₯ + 1 + β„Ž) βˆ’γ€– sin〗⁑〖 ( π‘₯ + 1 )γ€—)/hγ€— Using sin A – sin B = 2 cos ((𝐴 + 𝐡)/2) sin ((𝐴 βˆ’ 𝐡)/2) = lim┬(hβ†’0)⁑〖(2 cos⁑(( ( π‘₯ + 1 + β„Ž ) + ( π‘₯ + 1))/2).sin⁑((( π‘₯ + 1 + β„Ž ) βˆ’ ( π‘₯ + 1 ))/2))/hγ€— = lim┬(hβ†’0)⁑〖(2 cos⁑〖(( 2(π‘₯ + 1 )+ β„Ž))/2γ€— . γ€–sin γ€—β‘γ€–β„Ž/2γ€—)/hγ€— = lim┬(hβ†’0)⁑〖(cos⁑〖(( 2(π‘₯ + 1 )+ β„Ž))/2γ€— . γ€–sin γ€—β‘γ€–β„Ž/2γ€—)/(β„Ž/2)γ€— = lim┬(hβ†’0)⁑〖〖cos 〗⁑〖(( 2(π‘₯ + 1 )+ β„Ž))/2γ€— . sin⁑〖 β„Ž/2γ€—/(( β„Ž)/2)γ€— = lim┬(hβ†’0)⁑〖cos 〗⁑〖(( 2(π‘₯ + 1 )+ β„Ž))/2γ€— .(π₯𝐒𝐦)┬(π‘β†’πŸŽ) π’”π’Šπ’β‘γ€– 𝒉/πŸγ€—/(( 𝒉)/𝟐) = lim┬(hβ†’0)⁑〖cos 〗⁑〖(( 2(π‘₯ + 1 ) + β„Ž))/2γ€— .𝟏 = lim┬(hβ†’0)⁑〖cos 〗⁑〖(( 2(π‘₯ + 1 )+ β„Ž))/2γ€— Putting h = 0 =γ€– cos 〗⁑〖(( 2(π‘₯ + 1 ) + 0))/2γ€— = cos ( x + 1) Hence f’ (x) = cos (x + 1) 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

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.