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Miscellaneous
Misc 2
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Question 1 Important Deleted for CBSE Board 2024 Exams
Last updated at June 5, 2023 by Teachoo
Misc 7 Differentiate w.r.t. x the function, (log𝑥 ) log𝑥, 𝑥>1 Let y = (log𝑥 ) log𝑥 Taking log both sides log𝑦 = log ((log𝑥 ) log𝑥 ) log𝑦 = log𝑥. 〖 log〗〖 (log𝑥 )〗 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑(log𝑦 )/𝑑𝑥 = 𝑑(log𝑥. 〖 log〗〖 (log𝑥 )〗 )/𝑑𝑥 (As 𝑙𝑜𝑔(𝑎^𝑏) = 𝑏 𝑙𝑜𝑔𝑎) 𝑑(log𝑦 )/𝑑𝑥 (𝑑𝑦/𝑑𝑦) = 𝑑(log𝑥. 〖 log〗〖 (log𝑥 )〗 )/𝑑𝑥 𝑑(log𝑦 )/𝑑𝑦 (𝑑𝑦/𝑑𝑥) = 𝑑(log𝑥. 〖 log〗〖 (log𝑥 )〗 )/𝑑𝑥 1/𝑦 . 𝑑𝑦/𝑑𝑥 = 𝑑(log𝑥. 〖 log〗〖 (log𝑥 )〗 )/𝑑𝑥 1/𝑦 . 𝑑𝑦/𝑑𝑥 = 𝑑(log𝑥 )/𝑑𝑥 . 〖 log〗〖 (log𝑥 )〗 + 𝑑(〖 log〗〖 (log𝑥 )〗 )/𝑑𝑥 .〖 log〗𝑥 1/𝑦 . 𝑑𝑦/𝑑𝑥 = 1/𝑥 log〖 (log𝑥 )〗 + 1/log𝑥 . 𝑑(log𝑥 )/𝑑𝑥 . log𝑥 1/𝑦 . 𝑑𝑦/𝑑𝑥 = 1/𝑥 log〖 (log𝑥 )〗 + (𝑑 (log𝑥 ))/𝑑𝑥 Using Product rule As (𝑢𝑣)’ = 𝑢’𝑣 + 𝑣’𝑢 where u = log x & v = log (log x) 1/𝑦 . 𝑑𝑦/𝑑𝑥 = 1/𝑥 log〖 (log𝑥 )〗 + 1/𝑥 𝑑𝑦/𝑑𝑥 = 𝑦 ( 1/𝑥 + log〖 (log𝑥 )〗/𝑥) 𝒅𝒚/𝒅𝒙 = (𝐥𝐨𝐠𝒙 )^𝐥𝐨𝐠𝒙 (𝟏/𝒙 + 𝒍𝒐𝒈〖 (𝒍𝒐𝒈𝒙 )〗/𝒙) Hence, 𝑑𝑦/𝑑𝑥 = (log𝑥 )^log𝑥 (1/𝑥 + log〖 (log𝑥 )〗/𝑥)