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

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

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

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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

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Davneet Singh

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.