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

Transcript

Ex 5.5, 7 Differentiate the functions in, γ€–(log⁑〖π‘₯)γ€—γ€—^π‘₯ + π‘₯^log⁑π‘₯ Let 𝑦 = γ€–(log⁑〖π‘₯)γ€—γ€—^π‘₯+ π‘₯^log⁑π‘₯ Let 𝑒 = γ€–(log⁑〖π‘₯)γ€—γ€—^π‘₯ , 𝑣 = π‘₯^log⁑π‘₯ 𝑦 = 𝑒+𝑣 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑦/𝑑π‘₯ = (𝑑 (𝑒 + 𝑣))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ Calculating 𝒅𝒖/𝒅𝒙 𝑒 = γ€–(log⁑〖π‘₯)γ€—γ€—^π‘₯ Taking log both sides log⁑𝑒 = log γ€–(log⁑〖π‘₯)γ€—γ€—^π‘₯ log⁑𝑒 = π‘₯ . log (log⁑〖π‘₯)γ€— Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑(log⁑𝑒 )/𝑑𝑒 . 𝑑𝑒/𝑑π‘₯ = (𝑑(π‘₯ . log (log⁑〖π‘₯)γ€— ) )/𝑑π‘₯ 1/𝑒 (𝑑𝑒/𝑑π‘₯)" =" 𝑑(π‘₯)/𝑑π‘₯ . log (log" " π‘₯) + 𝑑(log (log" " π‘₯))/𝑑π‘₯ Γ— π‘₯ Using product Rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 1/𝑒 (𝑑𝑒/𝑑π‘₯)" =" 1 . log (log" " π‘₯) + (1/(log" " π‘₯) .𝑑(log" " π‘₯)/𝑑π‘₯) Γ— π‘₯ 1/𝑒 (𝑑𝑒/𝑑π‘₯)" =" log (log" " π‘₯) + (1/(log" " π‘₯) . 1/π‘₯) Γ— π‘₯ 1/𝑒 (𝑑𝑒/𝑑π‘₯)" =" log (log" " π‘₯) + 1/log⁑π‘₯ Γ— π‘₯/π‘₯ 1/𝑒 (𝑑𝑒/𝑑π‘₯)" =" log (log" " π‘₯) + 1/log⁑π‘₯ 𝑑𝑒/𝑑π‘₯ " =" 𝑒 (log (log" " π‘₯)" + " 1/log⁑π‘₯ ) 𝑑𝑒/𝑑π‘₯ = (log⁑π‘₯ )^π‘₯ (log (log" " π‘₯)" + " 1/log⁑π‘₯ ) 𝑑𝑒/𝑑π‘₯ = (log⁑π‘₯ )^π‘₯ ((log⁑〖π‘₯ . γ€–log 〗⁑(log⁑π‘₯ ) +γ€— 1)/log⁑π‘₯ ) 𝑑𝑒/𝑑π‘₯ = (log⁑π‘₯ )^π‘₯/log⁑π‘₯ (log⁑〖π‘₯ . γ€–log 〗⁑(log⁑π‘₯ )+γ€— 1) 𝑑𝑒/𝑑π‘₯ = (log⁑π‘₯ )^(π‘₯ βˆ’1) (log⁑〖π‘₯ . γ€–log 〗⁑(log⁑π‘₯ )+γ€— 1) Calculating 𝒅𝒗/𝒅𝒙 𝑣 = π‘₯^log⁑π‘₯ Taking log both sides . log⁑𝑣=log⁑〖 (π‘₯^log⁑π‘₯ )γ€— log⁑𝑣 = log π‘₯ . log⁑π‘₯ log⁑𝑣 = (log⁑π‘₯ )^2 (As π‘™π‘œπ‘”β‘(π‘Ž^𝑏 )=𝑏 . π‘™π‘œπ‘”β‘π‘Ž) Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑(log⁑𝑣 )/𝑑π‘₯ = (𝑑(log⁑π‘₯ )^2)/𝑑π‘₯ 𝑑(log⁑𝑣 )/𝑑π‘₯ . 𝑑𝑣/𝑑𝑣 = (𝑑(log⁑π‘₯ )^2)/𝑑π‘₯ 𝑑(log⁑𝑣 )/𝑑𝑣 . 𝑑𝑣/𝑑π‘₯ = (𝑑(log⁑π‘₯ )^2)/𝑑π‘₯ 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = (𝑑(log⁑π‘₯ )^2)/𝑑π‘₯ 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = 2 log⁑π‘₯ . 𝑑(log⁑π‘₯ )/𝑑π‘₯ 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = 2 log⁑π‘₯ . 1/π‘₯ 𝑑𝑣/𝑑π‘₯ = 𝑣 ((2 log⁑π‘₯)/π‘₯) 𝑑𝑣/𝑑π‘₯ = π‘₯^log⁑π‘₯ ((2 log⁑π‘₯)/π‘₯) 𝑑𝑣/𝑑π‘₯ = π‘₯^log⁑π‘₯ /π‘₯ (2 log⁑π‘₯ ) 𝑑𝑣/𝑑π‘₯ = π‘₯^(log⁑π‘₯ βˆ’ 1) . 2 log⁑π‘₯ 𝑑𝑣/𝑑π‘₯ = γ€–2π‘₯γ€—^(log⁑π‘₯ βˆ’ 1) . log⁑π‘₯ Hence 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ Putting value of 𝑑𝑒/𝑑π‘₯ & 𝑑𝑣/𝑑π‘₯ π’…π’š/𝒅𝒙 = (π₯𝐨𝐠⁑𝒙 )^(𝒙 βˆ’πŸ) (𝟏+π₯𝐨𝐠⁑𝒙.π₯𝐨𝐠⁑(π₯𝐨𝐠⁑𝒙 ) ) + γ€–πŸπ’™γ€—^(π’π’π’ˆ 𝒙 βˆ’πŸ). 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.