Find derivative by first principle f(x) = sin x + cos x [Video]

Example 20 (i) - Chapter 13 Class 11 Limits and Derivatives - Part 2
Example 20 (i) - Chapter 13 Class 11 Limits and Derivatives - Part 3

 

 

 

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Example 20 Find the derivative of f(x) from the first principle, where f(x) is (i) sin x + cos x Given f (x) = sin x + cos x 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 + cos x f (x + h) = sin (x + h) + cos (x + h) Putting values f’(x) = lim┬(hβ†’0)⁑〖(sin⁑〖(π‘₯ + β„Ž)γ€— + cos⁑(π‘₯ + β„Ž) βˆ’ (sin⁑π‘₯ + cos⁑〖π‘₯)γ€—)/β„Žγ€— Using sin (A + B) = sin A cos B + cos A sin B & cos (A + B) = cos A cos B – sin A sin B = lim┬(hβ†’0)⁑〖sin⁑〖π‘₯ cosβ‘γ€–β„Ž +γ€– cos〗⁑〖π‘₯ sinβ‘γ€–β„Ž + cos⁑〖π‘₯ cosβ‘γ€–β„Ž βˆ’ sin⁑〖π‘₯ γ€– sinγ€—β‘γ€–β„Ž βˆ’γ€– sin〗⁑〖π‘₯ βˆ’γ€– cos〗⁑π‘₯ γ€— γ€— γ€— γ€— γ€— γ€— γ€— γ€— γ€—/hγ€— = lim┬(hβ†’0)⁑〖cos⁑〖π‘₯ sinβ‘γ€–β„Ž βˆ’γ€– sin〗⁑〖π‘₯ sinβ‘γ€–β„Ž + sin⁑〖π‘₯ cosβ‘γ€–β„Ž βˆ’ sin⁑〖π‘₯ +γ€– cos〗⁑〖π‘₯ cosβ‘γ€–β„Ž βˆ’γ€– cos〗⁑π‘₯ γ€— γ€— γ€— γ€— γ€— γ€— γ€— γ€— γ€—/hγ€— = lim┬(hβ†’0)⁑〖sinβ‘γ€–β„Ž γ€–(cos〗⁑〖π‘₯ βˆ’ sin⁑〖π‘₯) + sin⁑〖π‘₯ (cosβ‘γ€–β„Ž βˆ’ 1) + cos⁑π‘₯ (cosβ‘γ€–β„Ž βˆ’ 1)γ€— γ€— γ€— γ€— γ€— γ€—/hγ€— = lim┬(hβ†’0)⁑(sinβ‘γ€–β„Ž γ€–(cos〗⁑〖π‘₯ βˆ’ sin⁑〖π‘₯)γ€— γ€— γ€—/h+sin⁑〖π‘₯ (cosβ‘γ€–β„Ž βˆ’ 1)γ€— γ€—/h+cos⁑〖π‘₯ (cosβ‘γ€–β„Ž βˆ’ 1)γ€— γ€—/β„Ž)" " = lim┬(hβ†’0)⁑〖sinβ‘γ€–β„Ž γ€–(cos〗⁑〖π‘₯ βˆ’γ€– sin〗⁑〖π‘₯)γ€— γ€— γ€—/h+lim┬(hβ†’0) sin⁑〖π‘₯ (cosβ‘γ€–β„Ž βˆ’ 1)γ€— γ€—/h+lim┬(hβ†’0) cos⁑〖π‘₯ (cosβ‘γ€–β„Ž βˆ’ 1)γ€— γ€—/β„Žγ€— = lim┬(hβ†’0)⁑〖"(cos x – sin x)" sinβ‘β„Ž/β„Ž+lim┬(hβ†’0) "(– sin x)" ((1 βˆ’ cosβ‘γ€–β„Ž)γ€—)/β„Žγ€—+lim┬(hβ†’0) "(– cos x)" ((1 βˆ’ cosβ‘γ€–β„Ž)γ€—)/β„Ž = "(cos x – sin x)" (π₯𝐒𝐦)┬(π‘β†’πŸŽ)⁑〖𝐬𝐒𝐧⁑𝒉/π’‰βˆ’sin⁑〖π‘₯ (π₯𝐒𝐦)┬(π‘β†’πŸŽ) γ€— ((𝟏 βˆ’ 𝒄𝒐𝒔⁑〖𝒉)γ€—)/π’‰βˆ’cos⁑π‘₯ (π₯𝐒𝐦)┬(π‘β†’πŸŽ) ((𝟏 βˆ’ 𝒄𝒐𝒔⁑〖𝒉)γ€—)/𝒉〗Using (π‘™π‘–π‘š)┬(β„Žβ†’0) π‘ π‘–π‘›β‘β„Ž/β„Ž = 1 & (π‘™π‘–π‘š)┬(β„Žβ†’0) γ€–(1 βˆ’ π‘π‘œπ‘ γ€—β‘γ€–β„Ž)γ€—/β„Ž = 0 Using (π‘™π‘–π‘š)┬(β„Žβ†’0) π‘ π‘–π‘›β‘β„Ž/β„Ž = 1 & (π‘™π‘–π‘š)┬(β„Žβ†’0) γ€–(1 βˆ’ π‘π‘œπ‘ γ€—β‘γ€–β„Ž)γ€—/β„Ž = 0

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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.