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Ex 13.2, 10 - Chapter 13 Class 11 Limits and Derivatives - Part 2
Ex 13.2, 10 - Chapter 13 Class 11 Limits and Derivatives - Part 3

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.