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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Ex 5.5, 10 Differentiate the functions in, π‘₯^(π‘₯ π‘π‘œπ‘ β‘π‘₯ ) + (π‘₯2+ 1)/(π‘₯2βˆ’ 1) Let y = π‘₯^(π‘₯ π‘π‘œπ‘ β‘π‘₯ ) + (π‘₯2+ 1)/(π‘₯2βˆ’ 1) Let 𝑒 =π‘₯^(π‘₯ π‘π‘œπ‘ β‘π‘₯ ) & 𝑣 =(π‘₯2+ 1)/(π‘₯2βˆ’ 1) ∴ 𝑦 = 𝑒+𝑣 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑦/𝑑π‘₯ = (𝑑 (𝑒 + 𝑣))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ Calculating 𝒅𝒖/𝒅𝒙 𝑒 =π‘₯^(π‘₯ π‘π‘œπ‘ β‘π‘₯ ) Taking log both sides log⁑𝑒=log⁑〖π‘₯^(π‘₯ π‘π‘œπ‘ β‘π‘₯ ) γ€— log⁑𝑒=π‘₯ π‘π‘œπ‘ β‘π‘₯. log⁑〖 π‘₯γ€— Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. (𝑑(log⁑𝑒))/𝑑π‘₯ = 𝑑(π‘₯ π‘π‘œπ‘ β‘π‘₯ log⁑π‘₯ )/𝑑π‘₯ (𝑑(log⁑𝑒))/𝑑π‘₯ . 𝑑𝑒/𝑑𝑒 = 𝑑(π‘₯ π‘π‘œπ‘ β‘π‘₯ log⁑π‘₯ )/𝑑π‘₯ (𝑑(log⁑𝑒))/𝑑𝑒 . 𝑑𝑒/𝑑π‘₯ = 𝑑(π‘₯ π‘π‘œπ‘ β‘π‘₯ log⁑π‘₯ )/𝑑π‘₯ (As π‘™π‘œπ‘”β‘(π‘Ž^𝑏) = 𝑏 . π‘™π‘œπ‘”β‘π‘Ž) 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = 𝑑(π‘₯ π‘π‘œπ‘ β‘π‘₯ log⁑π‘₯ )/𝑑π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = 𝑑(π‘₯ cos⁑π‘₯ )/𝑑π‘₯ log⁑π‘₯ + π‘₯ π‘π‘œπ‘ β‘π‘₯ (𝑑(log⁑π‘₯))/𝑑π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = 𝑑(π‘₯ cos⁑π‘₯ )/𝑑π‘₯ log⁑π‘₯ + π‘₯ π‘π‘œπ‘ β‘π‘₯ Γ—1/π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = [𝑑(π‘₯)/𝑑π‘₯ cos⁑π‘₯+π‘₯ 𝑑(cos⁑π‘₯ )/𝑑π‘₯]log⁑π‘₯ + π‘π‘œπ‘ β‘π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = [cos⁑π‘₯βˆ’π‘₯ 𝑠𝑖𝑛 π‘₯]log⁑π‘₯ + π‘π‘œπ‘ β‘π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = cos⁑π‘₯ . log⁑π‘₯ βˆ’ π‘₯ sin π‘₯ log⁑π‘₯+cos⁑π‘₯ Using product Rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 Where u = x cos x, v = log x 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = cos⁑π‘₯ . log⁑π‘₯+cos⁑π‘₯βˆ’ π‘₯ sin π‘₯ log⁑π‘₯ 1/𝑒 𝑑𝑒/𝑑π‘₯ = cos⁑π‘₯ (log⁑π‘₯+1)βˆ’π‘₯ sin⁑〖π‘₯ log⁑π‘₯ γ€— 𝑑𝑒/𝑑π‘₯ = u (cos⁑π‘₯ (log⁑π‘₯+1)βˆ’π‘₯ sin⁑〖π‘₯ log⁑π‘₯ γ€— ) 𝑑𝑒/𝑑π‘₯ = π‘₯^(π‘₯ π‘π‘œπ‘ β‘π‘₯ ) (cos⁑〖 (log⁑〖π‘₯+1γ€— )βˆ’π‘₯ sin⁑〖π‘₯ log⁑π‘₯ γ€— γ€— ) Calculating 𝒅𝒗/𝒅𝒙 𝑣= (π‘₯2 + 1)/(π‘₯2 βˆ’ 1) Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑(𝑣)/𝑑π‘₯ = 𝑑((π‘₯2 + 1)/(π‘₯2 βˆ’ 1))/𝑑π‘₯ 𝑑(𝑣)/𝑑π‘₯ . 𝑑𝑣/𝑑𝑣 = 𝑑((π‘₯2 + 1)/(π‘₯2 βˆ’ 1))/𝑑π‘₯ 𝑑(𝑣)/𝑑𝑣 . 𝑑𝑣/𝑑π‘₯ = 𝑑((π‘₯2 + 1)/(π‘₯2 βˆ’ 1))/𝑑π‘₯ 1. 𝑑𝑣/𝑑π‘₯ = 𝑑((π‘₯2 + 1)/(π‘₯2 βˆ’ 1))/𝑑π‘₯ 𝑑𝑣/𝑑π‘₯ = (𝑑(π‘₯2+ 1)/𝑑π‘₯ . (π‘₯^2 βˆ’ 1) βˆ’ 𝑑(π‘₯^2 βˆ’ 1)/𝑑π‘₯ . (π‘₯2+ 1))/(π‘₯^2 βˆ’ 1)^2 𝑑𝑣/𝑑π‘₯ = ((2π‘₯ + 0) (π‘₯^2 βˆ’1) βˆ’ (2π‘₯ βˆ’ 0) (π‘₯^2 + 1))/(π‘₯^2 βˆ’1)^2 Using quotient rule (𝑒/𝑣)β€² = (𝑒^β€² 𝑣 βˆ’ 𝑣^β€² 𝑒)/𝑣^2 𝑑𝑣/𝑑π‘₯ = (2π‘₯ (π‘₯^2 βˆ’1) βˆ’ 2π‘₯ (π‘₯^2 + 1))/(π‘₯^2 βˆ’1)^2 𝑑𝑣/𝑑π‘₯ = (2π‘₯ (π‘₯^2 βˆ’1 βˆ’π‘₯^2 βˆ’1))/(π‘₯^2 βˆ’1)^2 𝑑𝑣/𝑑π‘₯ = (2π‘₯ (βˆ’ 2))/(π‘₯^2 βˆ’1)^2 𝑑𝑣/𝑑π‘₯ = (βˆ’4π‘₯)/(π‘₯^2 βˆ’1)^2 Now 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ Putting value of 𝑑𝑒/𝑑π‘₯ & 𝑑𝑣/𝑑π‘₯ π’…π’š/𝒅𝒙 = 𝒙^〖𝒙 πœπ¨π¬γ€—β‘π’™ (πœπ¨π¬β‘γ€–π’™ (𝟏+π₯𝐨𝐠⁑𝒙 ) βˆ’π’™ 𝐬𝐒𝐧⁑〖𝒙 π₯𝐨𝐠⁑𝒙 γ€— γ€— ) βˆ’ πŸ’π’™/(𝒙^𝟐 βˆ’πŸ)^𝟐

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.