Ex 5.5, 4 - Differentiate x2 - 2sin x - Chapter 5 Class 12

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

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

Transcript

Ex 5.5, 4 Differentiate the functions in, π‘₯^π‘₯ – 2^sin⁑π‘₯ Let 𝑦=π‘₯^π‘₯ – 2^sin⁑π‘₯ Let 𝑒=π‘₯^π‘₯ , 𝑣=2^sin⁑π‘₯ 𝑦= π‘’βˆ’π‘£ Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑦/𝑑π‘₯ = 𝑑(𝑒 βˆ’ 𝑣)/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ βˆ’ 𝑑𝑣/𝑑π‘₯ Calculating 𝒅𝒖/𝒅𝒙 𝑒=π‘₯^π‘₯ Taking log both sides log 𝑒=γ€–log⁑〖 (π‘₯γ€—γ€—^π‘₯) log 𝑒= π‘₯ . log⁑π‘₯ Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑(log⁑𝑒 )/𝑑π‘₯ = 𝑑(π‘₯ . log⁑π‘₯ )/𝑑π‘₯ 𝑑(log⁑𝑒 )/𝑑π‘₯ (𝑑𝑒/𝑑𝑒) = 𝑑(π‘₯ . log⁑π‘₯ )/𝑑π‘₯ 𝑑(log⁑𝑒 )/𝑑𝑒 (𝑑𝑒/𝑑π‘₯) = 𝑑(π‘₯ . log⁑π‘₯ )/𝑑π‘₯ v(As π‘™π‘œπ‘”β‘(π‘Ž^𝑏) = 𝑏 π‘™π‘œπ‘”β‘π‘Ž) 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = 𝑑(π‘₯ . log⁑π‘₯ )/𝑑π‘₯ 1/𝑒. 𝑑𝑒/𝑑π‘₯ = (𝑑 π‘₯ )/𝑑π‘₯ .log⁑π‘₯ + (𝑑 (log⁑π‘₯ ))/𝑑π‘₯ . π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = 1 . log⁑π‘₯ + 1/π‘₯ . π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = log⁑〖π‘₯+1γ€— 𝑑𝑒/𝑑π‘₯ = 𝑒 (log⁑〖π‘₯+1γ€— ) 𝑑𝑒/𝑑π‘₯ = π‘₯^π‘₯ (log⁑〖π‘₯+1γ€— ) Using product value in γ€–π‘₯ .γ€—β‘π‘™π‘œπ‘”β‘π‘₯ As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 Calculating 𝒅𝒗/𝒅𝒙 𝑣=2^sin⁑π‘₯ Taking log both sides log 𝑣=log⁑〖 (2^sin⁑π‘₯ ) " " γ€— log 𝑣= sin π‘₯ . log⁑2 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑(log⁑𝑣 )/𝑑π‘₯ = 𝑑(sin π‘₯" . " log⁑2 )/𝑑π‘₯ 𝑑(log⁑𝑣 )/𝑑π‘₯ (𝑑𝑣/𝑑𝑣) = log⁑2 𝑑(sin π‘₯)/𝑑π‘₯ 𝑑(log⁑𝑣 )/𝑑𝑣 (𝑑𝑣/𝑑π‘₯) = log⁑2 . cos⁑π‘₯ (As π‘™π‘œπ‘”β‘(π‘Ž^𝑏) = 𝑏 π‘™π‘œπ‘”β‘π‘Ž) 1/𝑣 (𝑑𝑣/𝑑π‘₯) = log⁑2 . cos⁑π‘₯ 𝑑𝑣/𝑑π‘₯ = 𝑦 (log⁑2 . cos⁑π‘₯) 𝑑𝑣/𝑑π‘₯ = 2^sin⁑π‘₯ (log⁑2 . cos⁑π‘₯) Now 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ βˆ’ 𝑑𝑣/𝑑π‘₯ Putting values of 𝑑𝑒/𝑑π‘₯ & 𝑑𝑣/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = π‘₯^π‘₯ (log⁑〖π‘₯+1γ€— ) βˆ’2^sin⁑π‘₯ (log⁑2" . " cos⁑π‘₯) π’…π’š/𝒅𝒙 = 𝒙^𝒙 (𝟏+π’π’π’ˆβ‘π’™ ) βˆ’πŸ^π’”π’Šπ’β‘π’™ (πœπ¨π¬β‘π’™.π’π’π’ˆβ‘πŸ)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.