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Question 1 Important Deleted for CBSE Board 2025 Exams

Last updated at April 16, 2024 by Teachoo

Misc 7 Differentiate w.r.t. x the function, (logโก๐ฅ ) logโก๐ฅ, ๐ฅ>1 Let y = (logโก๐ฅ ) logโก๐ฅ Taking log both sides logโก๐ฆ = log ((logโก๐ฅ ) logโก๐ฅ ) logโก๐ฆ = logโก๐ฅ. ใ logใโกใ (logโก๐ฅ )ใ Differentiating both sides ๐ค.๐.๐ก.๐ฅ. ๐(logโก๐ฆ )/๐๐ฅ = ๐(logโก๐ฅ. ใ logใโกใ (logโก๐ฅ )ใ )/๐๐ฅ (As ๐๐๐โก(๐^๐) = ๐ ๐๐๐โก๐) ๐(logโก๐ฆ )/๐๐ฅ (๐๐ฆ/๐๐ฆ) = ๐(logโก๐ฅ. ใ logใโกใ (logโก๐ฅ )ใ )/๐๐ฅ ๐(logโก๐ฆ )/๐๐ฆ (๐๐ฆ/๐๐ฅ) = ๐(logโก๐ฅ. ใ logใโกใ (logโก๐ฅ )ใ )/๐๐ฅ 1/๐ฆ . ๐๐ฆ/๐๐ฅ = ๐(logโก๐ฅ. ใ logใโกใ (logโก๐ฅ )ใ )/๐๐ฅ 1/๐ฆ . ๐๐ฆ/๐๐ฅ = ๐(logโก๐ฅ )/๐๐ฅ . ใ logใโกใ (logโก๐ฅ )ใ + ๐(ใ logใโกใ (logโก๐ฅ )ใ )/๐๐ฅ .ใ logใโก๐ฅ 1/๐ฆ . ๐๐ฆ/๐๐ฅ = 1/๐ฅ logโกใ (logโก๐ฅ )ใ + 1/logโก๐ฅ . ๐(logโก๐ฅ )/๐๐ฅ . logโก๐ฅ 1/๐ฆ . ๐๐ฆ/๐๐ฅ = 1/๐ฅ logโกใ (logโก๐ฅ )ใ + (๐ (logโก๐ฅ ))/๐๐ฅ Using Product rule As (๐ข๐ฃ)โ = ๐ขโ๐ฃ + ๐ฃโ๐ข where u = log x & v = log (log x) 1/๐ฆ . ๐๐ฆ/๐๐ฅ = 1/๐ฅ logโกใ (logโก๐ฅ )ใ + 1/๐ฅ ๐๐ฆ/๐๐ฅ = ๐ฆ ( 1/๐ฅ + logโกใ (logโก๐ฅ )ใ/๐ฅ) ๐ ๐/๐ ๐ = (๐ฅ๐จ๐ โก๐ )^๐ฅ๐จ๐ โก๐ (๐/๐ + ๐๐๐โกใ (๐๐๐โก๐ )ใ/๐) Hence, ๐๐ฆ/๐๐ฅ = (logโก๐ฅ )^logโก๐ฅ (1/๐ฅ + logโกใ (logโก๐ฅ )ใ/๐ฅ)