Check Full Chapter Explained - Continuity and Differentiability - https://you.tube/Chapter-5-Class-12-Continuity

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

Transcript

Ex 5.5, 3 Differentiate the functions in, (log⁑π‘₯ )^cos⁑π‘₯ Let 𝑦=(log⁑π‘₯ )^cos⁑π‘₯ Taking log both sides log⁑𝑦 = log⁑〖〖 (log⁑π‘₯ )γ€—^cos⁑π‘₯ γ€— log⁑𝑦 = cos⁑〖π‘₯ .γ€–log 〗⁑(log⁑π‘₯ ) γ€— Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑(log⁑𝑦)/𝑑π‘₯ = 𝑑(cos⁑〖π‘₯ .γ€– log 〗⁑(log⁑π‘₯ ) γ€— )/𝑑π‘₯ 𝑑(log⁑𝑦 )/𝑑π‘₯ (𝑑𝑦/𝑑𝑦) = 𝑑(cos⁑〖π‘₯ .γ€– log 〗⁑(log⁑π‘₯ ) γ€— )/𝑑π‘₯ 𝑑(log⁑𝑦 )/𝑑π‘₯ (𝑑𝑦/𝑑π‘₯) = 𝑑(cos⁑〖π‘₯ .γ€– log 〗⁑(log⁑π‘₯ ) γ€— )/𝑑π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = 𝑑(cos⁑〖π‘₯ .γ€– log 〗⁑(log⁑π‘₯ ) γ€— )/𝑑π‘₯ 1/𝑦 𝑑𝑦/𝑑π‘₯ = 𝑑(cos⁑π‘₯ )/𝑑π‘₯ . γ€– log 〗⁑(log⁑π‘₯ ) + 𝑑(γ€– log 〗⁑(log⁑π‘₯ ) )/𝑑π‘₯ . cos⁑π‘₯ 1/𝑦 𝑑𝑦/𝑑π‘₯ = γ€–βˆ’sin〗⁑π‘₯ . γ€–log 〗⁑(log⁑π‘₯ ) + 1/log⁑π‘₯ . 𝑑(log⁑π‘₯ )/𝑑π‘₯ . cos⁑π‘₯ 1/𝑦 𝑑𝑦/𝑑π‘₯ = γ€–βˆ’sin〗⁑π‘₯ . γ€–log 〗⁑(log⁑π‘₯ ) + 1/log⁑π‘₯ Γ— 1/π‘₯ . cos⁑π‘₯ 1/𝑦 𝑑𝑦/𝑑π‘₯ = γ€–βˆ’sin〗⁑π‘₯ . γ€–log 〗⁑(log⁑π‘₯ ) + cos⁑π‘₯/(π‘₯ log⁑π‘₯ ) 𝑑𝑦/𝑑π‘₯ = 𝑦 (γ€–βˆ’sin〗⁑π‘₯ " . " γ€–log 〗⁑(log⁑π‘₯ )" + " cos⁑π‘₯/(π‘₯ log⁑π‘₯ )) Using product rule in π‘π‘œπ‘ β‘γ€–π‘₯ .γ€– π‘™π‘œπ‘” 〗⁑(π‘™π‘œπ‘”β‘π‘₯ ) γ€— (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 Putting values of 𝑦 𝑑𝑦/𝑑π‘₯ = (π‘™π‘œπ‘”β‘π‘₯ )^π‘π‘œπ‘ β‘π‘₯ (γ€–βˆ’π‘ π‘–π‘›γ€—β‘π‘₯ " . " γ€–π‘™π‘œπ‘” 〗⁑(π‘™π‘œπ‘”β‘π‘₯ )" + " π‘π‘œπ‘ β‘π‘₯/(π‘₯ π‘™π‘œπ‘”β‘π‘₯ )) π’…π’š/𝒅𝒙 = (π’π’π’ˆβ‘π’™ )^𝒄𝒐𝒔⁑𝒙 (𝒄𝒐𝒔⁑𝒙/(𝒙 π’π’π’ˆβ‘π’™ ) γ€– βˆ’ π’”π’Šπ’γ€—β‘π’™ " . " γ€–π₯𝐨𝐠 〗⁑(π’π’π’ˆβ‘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.