Misc 7 - Differentiate (log x) log x - Chapter 5 Class 12 - Miscellaneous

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  1. Chapter 5 Class 12 Continuity and Differentiability
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Misc 7 Differentiate w.r.t. x the function, (log⁑π‘₯ ) log⁑π‘₯, π‘₯>1 Let y = (log⁑π‘₯ ) log⁑π‘₯ Taking log both sides log⁑𝑦 = log ((log⁑π‘₯ ) log⁑π‘₯ ) log⁑𝑦 = log⁑π‘₯. γ€– log〗⁑〖 (log⁑π‘₯ )γ€— Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑(log⁑𝑦 )/𝑑π‘₯ = 𝑑(log⁑π‘₯. γ€– log〗⁑〖 (log⁑π‘₯ )γ€— )/𝑑π‘₯ 𝑑(log⁑𝑦 )/𝑑π‘₯ (𝑑𝑦/𝑑𝑦) = 𝑑(log⁑π‘₯. γ€– log〗⁑〖 (log⁑π‘₯ )γ€— )/𝑑π‘₯ 𝑑(log⁑𝑦 )/𝑑𝑦 (𝑑𝑦/𝑑π‘₯) = 𝑑(log⁑π‘₯. γ€– log〗⁑〖 (log⁑π‘₯ )γ€— )/𝑑π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = 𝑑(log⁑π‘₯. γ€– log〗⁑〖 (log⁑π‘₯ )γ€— )/𝑑π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = 𝑑(log⁑π‘₯ )/𝑑π‘₯ . γ€– log〗⁑〖 (log⁑π‘₯ )γ€— + 𝑑(γ€– log〗⁑〖 (log⁑π‘₯ )γ€— )/𝑑π‘₯ .γ€– log〗⁑π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = 1/π‘₯ log⁑〖 (log⁑π‘₯ )γ€— + 1/log⁑π‘₯ . 𝑑(log⁑π‘₯ )/𝑑π‘₯ . log⁑π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = 1/π‘₯ log⁑〖 (log⁑π‘₯ )γ€— + (𝑑 (log⁑π‘₯ ))/𝑑π‘₯ . log⁑π‘₯/log⁑π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = 1/π‘₯ log⁑〖 (log⁑π‘₯ )γ€— + 1/π‘₯ 1/𝑦 . 𝑑𝑦/𝑑π‘₯ = 1/π‘₯ log⁑〖 (log⁑π‘₯ )γ€— + 1/π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑦 ( 1/π‘₯ + log⁑〖 (log⁑π‘₯ )γ€—/π‘₯) π’…π’š/𝒅𝒙 = (π₯𝐨𝐠⁑𝒙 )^π₯𝐨𝐠⁑𝒙 (𝟏/𝒙 + π’π’π’ˆβ‘γ€– (π’π’π’ˆβ‘π’™ )γ€—/𝒙) Hence, 𝑑𝑦/𝑑π‘₯ = (log⁑π‘₯ )^log⁑π‘₯ (1/π‘₯ + log⁑〖 (log⁑π‘₯ )γ€—/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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.