Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12

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

Transcript

Misc 3 Differentiate 𝑀.π‘Ÿ.𝑑. π‘₯ the function, (5π‘₯)^(3cos⁑2π‘₯) Let 𝑦" = " (5π‘₯)^(3cos⁑2π‘₯) Taking log on both sides log⁑𝑦 = log (5π‘₯)^(3cos⁑2π‘₯) π’π’π’ˆβ‘π’š = πŸ‘ 𝐜𝐨𝐬 πŸπ’™ . π’π’π’ˆ β‘πŸ“π’™ Differentiating both sides 𝑀.π‘Ÿ.𝑑. x 𝑑(log⁑𝑦 )/𝑑π‘₯ = 𝑑(3 cos 2π‘₯ . log ⁑5π‘₯)/𝑑π‘₯ 𝑑(log⁑𝑦 )/𝑑π‘₯ (𝑑𝑦/𝑑𝑦) = 𝑑(3 cos 2π‘₯ . log ⁑5π‘₯)/𝑑π‘₯ (As π‘™π‘œπ‘”β‘(π‘Ž^𝑏) = 𝑏 π‘™π‘œπ‘”β‘π‘Ž) 𝑑(log⁑𝑦 )/𝑑𝑦 (𝑑𝑦/𝑑π‘₯) = 𝑑(3 cos 2π‘₯ . log ⁑5π‘₯)/𝑑π‘₯ 𝟏/π’š . π’…π’š/𝒅𝒙 = 𝑑(3 cos 2π‘₯ . log ⁑5π‘₯)/𝑑π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = 𝑑(3 cos 2π‘₯ )/𝑑π‘₯ . log ⁑5π‘₯ + 𝑑(log ⁑5π‘₯)/𝑑π‘₯ .3 cos 2π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = 3 𝑑(cos 2π‘₯ )/𝑑π‘₯ . log ⁑5π‘₯ + 1/5π‘₯ . 𝑑(5π‘₯)/𝑑π‘₯ . 3 cos 2π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = 3(βˆ’sin⁑2π‘₯ ) Γ— 2 Γ— log ⁑5π‘₯ + 1/5π‘₯ Γ— 5 Γ— 3 cos 2π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = βˆ’6 sin 2π‘₯ . log⁑5π‘₯ + (3 cos⁑2π‘₯)/π‘₯ Using Product rule (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 where u = 3 π‘π‘œπ‘  2π‘₯ & 𝑣=π‘™π‘œπ‘” ⁑5π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑦 (βˆ’6 sin 2π‘₯ . log⁑5π‘₯ "+ " (3 cos⁑2π‘₯)/π‘₯) 𝑑𝑦/𝑑π‘₯ = (5π‘₯)^(3cos⁑2π‘₯) (βˆ’6 sin 2π‘₯ . log⁑5π‘₯ "+ " (3 cos⁑2π‘₯)/π‘₯) π’…π’š/𝒅𝒙 = (πŸ“π’™)^(πŸ‘π’„π’π’”β‘πŸπ’™) ((πŸ‘ π’„π’π’”β‘πŸπ’™)/π’™βˆ’πŸ” 𝐬𝐒𝐧 πŸπ’™ . π’π’π’ˆβ‘πŸ“π’™ )

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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.