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Ex 5.5, 10 - Differentiate x^(x cos x) + (x^2 + 1)/(x^2 - 1)

Ex 5.5, 10 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.5, 10 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.5, 10 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Ex 5.5, 10 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Ex 5.5, 10 - Chapter 5 Class 12 Continuity and Differentiability - Part 6

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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 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.