Misc 9 - Differentiate (sin x - cos x) sinx - cosx - Chapter 5 - Miscellaneous

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  1. Chapter 5 Class 12 Continuity and Differentiability
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Misc 9 Differentiate w.r.t. x the function, (sin⁑π‘₯βˆ’cos⁑π‘₯ )^((sin⁑〖π‘₯βˆ’cos⁑〖π‘₯)γ€— γ€— ), πœ‹/4 <π‘₯< 3πœ‹/4 Let y = (sin⁑π‘₯βˆ’cos⁑π‘₯ )^((sin⁑〖π‘₯βˆ’cos⁑〖π‘₯)γ€— γ€— ) Taking log on both sides log⁑𝑦 = log (sin⁑π‘₯βˆ’cos⁑π‘₯ )^((sin⁑〖π‘₯βˆ’cos⁑〖π‘₯)γ€— γ€— ) log⁑𝑦 = (sin⁑π‘₯βˆ’cos⁑π‘₯ ). γ€– log〗⁑〖 (sin⁑π‘₯βˆ’cos⁑π‘₯ )γ€— Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑(log⁑𝑦 )/𝑑π‘₯ = 𝑑((sin⁑π‘₯ βˆ’ cos⁑π‘₯ ). γ€– log〗⁑(sin⁑π‘₯ βˆ’ cos⁑π‘₯ ) )/𝑑π‘₯ 𝑑(log⁑𝑦 )/𝑑π‘₯ (𝑑𝑦/𝑑𝑦) = 𝑑((sin⁑π‘₯βˆ’cos⁑π‘₯ ). γ€– log〗⁑〖 (sin⁑π‘₯βˆ’cos⁑π‘₯ )γ€— )/𝑑π‘₯ 𝑑(log⁑𝑦 )/𝑑𝑦 (𝑑𝑦/𝑑π‘₯) = 𝑑((sin⁑π‘₯βˆ’cos⁑π‘₯ ). γ€– log〗⁑〖 (sin⁑π‘₯βˆ’cos⁑π‘₯ )γ€— )/𝑑π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = 𝑑((sin⁑π‘₯βˆ’cos⁑π‘₯ ). γ€– log〗⁑〖 (sin⁑π‘₯βˆ’cos⁑π‘₯ )γ€— " " )/𝑑π‘₯ " " Using product rule (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 where u = sin x βˆ’ cos x & v = log (sin x βˆ’ cos x) 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = 𝑑(sin⁑π‘₯βˆ’cos⁑π‘₯ )/𝑑π‘₯ . γ€– log 〗⁑(sin⁑π‘₯βˆ’cos⁑π‘₯ ) + 𝑑(γ€– log〗⁑〖 (sin⁑π‘₯βˆ’cos⁑π‘₯ )γ€— )/𝑑π‘₯ .(sin⁑π‘₯βˆ’cos⁑π‘₯ ) 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = (cos⁑π‘₯βˆ’(βˆ’sin⁑π‘₯ )). log⁑〖 (sin⁑π‘₯βˆ’cos⁑π‘₯ )γ€—+ 1/((sin⁑π‘₯βˆ’cos⁑π‘₯ ) ) . 𝑑(sin⁑π‘₯ βˆ’ cos⁑π‘₯ )/𝑑π‘₯ . (sin⁑π‘₯βˆ’cos⁑π‘₯ ) 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = (cos⁑π‘₯+sin⁑π‘₯ ) . log⁑〖 (sin⁑π‘₯βˆ’cos⁑π‘₯ )γ€—+ 1/((sin⁑π‘₯βˆ’cos⁑π‘₯ ) ) . (cos⁑π‘₯βˆ’(βˆ’sin⁑π‘₯ )) . (sin⁑π‘₯βˆ’cos⁑π‘₯ ) 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = (cos⁑π‘₯+sin⁑π‘₯ ) . log⁑〖 (sin⁑π‘₯βˆ’cos⁑π‘₯ )γ€—+ 1/((sin⁑π‘₯βˆ’cos⁑π‘₯ ) ) . (cos⁑π‘₯+sin⁑π‘₯ ) . (sin⁑π‘₯βˆ’cos⁑π‘₯ ) 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = (cos⁑π‘₯+sin⁑π‘₯ ) . log⁑〖 (sin⁑π‘₯βˆ’cos⁑π‘₯ )γ€— + (cos⁑π‘₯+sin⁑π‘₯ ) 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = (cos⁑π‘₯+sin⁑π‘₯ ) . (log⁑〖 (sin⁑π‘₯βˆ’cos⁑π‘₯ )+1γ€— ) 𝑑𝑦/𝑑π‘₯ = y(cos⁑π‘₯+sin⁑π‘₯ ) . (log⁑〖 (sin⁑π‘₯βˆ’cos⁑π‘₯ )+1γ€— ) π’…π’š/𝒅𝒙 = (π’”π’Šπ’β‘π’™βˆ’π’„π’π’”β‘π’™ )^((π’”π’Šπ’β‘γ€–π’™βˆ’π’„π’π’”β‘γ€–π’™)γ€— γ€— ) (𝒄𝒐𝒔⁑𝒙+π’”π’Šπ’β‘π’™ )(π’π’π’ˆβ‘γ€– (π’”π’Šπ’β‘π’™βˆ’π’„π’π’”β‘π’™ )+πŸγ€— )

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.