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Ex 5.5, 3 - Differentiate the function (log x)^cos x - Teachoo

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

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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‘₯ (As π‘™π‘œπ‘”β‘(π‘Ž^𝑏) = 𝑏 π‘™π‘œπ‘”β‘π‘Ž) 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.