Check sibling questions

Ex 5.5, 7 - Differentiate the function (log x)^x + x^log x

Ex 5.5, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.5, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.5, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Ex 5.5, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Ex 5.5, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 6

Introducing your new favourite teacher - Teachoo Black, at only β‚Ή83 per month


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’™

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.