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Ex 5.5, 6 - Differentiate (x + 1/x)^x + x^(1 + 1/x) - Teachoo

Ex 5.5,6 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.5,6 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.5,6 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Ex 5.5,6 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Ex 5.5,6 - Chapter 5 Class 12 Continuity and Differentiability - Part 6
Ex 5.5,6 - Chapter 5 Class 12 Continuity and Differentiability - Part 7
Ex 5.5,6 - Chapter 5 Class 12 Continuity and Differentiability - Part 8

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Transcript

Ex 5.5, 6 Differentiate the functions in, (π‘₯+1/π‘₯)^π‘₯+ π‘₯^((1 + 1/π‘₯) ) Let 𝑦= (π‘₯+1/π‘₯)^π‘₯+ π‘₯^((1 + 1/π‘₯) ) Let 𝑒 = (π‘₯+1/π‘₯)^π‘₯ , 𝑣 = π‘₯^((1 + 1/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 = (π‘₯+1/π‘₯)^π‘₯ Taking log both sides log⁑𝑒 = log (π‘₯+1/π‘₯)^π‘₯ log⁑𝑒 = π‘₯ log (π‘₯+1/π‘₯) Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑(log⁑𝑒 )/𝑑π‘₯ = (𝑑 (π‘₯ log" " (π‘₯ + 1/π‘₯)))/𝑑π‘₯ 𝑑(log⁑𝑒 )/𝑑π‘₯ (𝑑𝑒/𝑑𝑒) = (𝑑 (π‘₯ log" " (π‘₯ + 1/π‘₯)))/𝑑π‘₯ (As π‘™π‘œπ‘”β‘(π‘Ž^𝑏 )=𝑏 . π‘™π‘œπ‘”β‘π‘Ž) 𝑑(log⁑𝑒 )/𝑑𝑒 (𝑑𝑒/𝑑π‘₯)" = " (𝑑 (π‘₯ log" " (π‘₯ + 1/π‘₯)))/𝑑π‘₯ 1/𝑒 (𝑑𝑒/𝑑π‘₯)" = " (𝑑 (π‘₯ log" " (π‘₯ + 1/π‘₯)))/𝑑π‘₯ 1/𝑒 (𝑑𝑒/𝑑π‘₯)" = " 𝑑(π‘₯)/𝑑π‘₯ . log (π‘₯ + 1/π‘₯) + 𝑑(log" " (π‘₯ + 1/π‘₯))/𝑑π‘₯ . π‘₯ 1/𝑒 (𝑑𝑒/𝑑π‘₯)" =" 1. log (π‘₯ + 1/π‘₯) + ((1/(π‘₯ + 1/π‘₯)).𝑑/𝑑π‘₯ (π‘₯ + 1/π‘₯)) . π‘₯ 1/𝑒 (𝑑𝑒/𝑑π‘₯)" =" log (π‘₯ + 1/π‘₯) + (1/(π‘₯ + 1/π‘₯) . (𝑑(π‘₯)/𝑑π‘₯+(𝑑 (1/π‘₯))/𝑑π‘₯)) . π‘₯ Using product rule in π‘₯ π‘™π‘œπ‘”" " (π‘₯ + 1/π‘₯) As (uv)’ = u’ v + v’ u 1/𝑒 (𝑑𝑒/𝑑π‘₯)" =" log (π‘₯+1/π‘₯) + (1/(π‘₯ + 1/π‘₯) . (1+(βˆ’1)/π‘₯^2 " " )) . π‘₯ 1/𝑒 (𝑑𝑒/𝑑π‘₯)" =" log (π‘₯+1/π‘₯) + (π‘₯/(π‘₯^2 + 1) . (1βˆ’1/π‘₯^2 " " )) . π‘₯ 1/𝑒 (𝑑𝑒/𝑑π‘₯)" =" log (π‘₯+1/π‘₯) + (π‘₯/(π‘₯^2 + 1) ((π‘₯^2 βˆ’ 1)/π‘₯^2 ).π‘₯) 1/𝑒 (𝑑𝑒/𝑑π‘₯)" =" log (π‘₯+1/π‘₯) + (π‘₯/(π‘₯^2 + 1) ((π‘₯^2 βˆ’ 1)/π‘₯^2 ).π‘₯) 1/𝑒 (𝑑𝑒/𝑑π‘₯)" =" log (π‘₯+1/π‘₯) + (π‘₯^2/π‘₯^2 ((π‘₯^2 βˆ’ 1)/(π‘₯^2 + 1))) 1/𝑒 (𝑑𝑒/𝑑π‘₯)" =" log (π‘₯+1/π‘₯) + ((π‘₯^2 βˆ’ 1)/(π‘₯^2+ 1)) 𝑑𝑒/𝑑π‘₯ "= " 𝑒 (γ€–log 〗⁑(π‘₯+1/π‘₯)+((π‘₯^2 βˆ’ 1)/(π‘₯^2+ 1))) 𝑑𝑒/𝑑π‘₯ "=" (π‘₯+1/π‘₯)^π‘₯ (γ€–log 〗⁑(π‘₯+1/π‘₯)+((π‘₯^2 βˆ’ 1)/(π‘₯^2+ 1))) 𝒅𝒖/𝒅𝒙 "=" (𝒙+𝟏/𝒙)^𝒙 ((𝒙^𝟐 βˆ’ 𝟏)/(𝒙^𝟐+ 𝟏)⁑〖+ γ€–π’π’π’ˆ 〗⁑(𝒙+𝟏/𝒙) γ€— ) Calculating 𝒅𝒗/𝒅𝒙 𝑣 = π‘₯^(1 + 1/π‘₯)" " Taking log both sides log 𝑣 = log π‘₯^(1 + 1/π‘₯)" " log 𝑣 = (1 + 1/π‘₯)log π‘₯^" " Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑(log⁑𝑣 )/𝑑π‘₯ = (𝑑 ((1 + 1/π‘₯)" . " log π‘₯))/𝑑π‘₯ 𝑑(log⁑𝑣 )/𝑑π‘₯ (𝑑𝑣/𝑑𝑣) = (𝑑 ((1 + 1/π‘₯)" . " log π‘₯))/𝑑π‘₯ 𝑑(log⁑𝑣 )/𝑑𝑣 (𝑑𝑣/𝑑π‘₯) = (𝑑 ((1 + 1/π‘₯)" . " log π‘₯))/𝑑π‘₯ 1/𝑣 (𝑑𝑣/𝑑π‘₯) = (𝑑 ((1 + 1/π‘₯)" . " log π‘₯))/𝑑π‘₯ Using product rule in (π‘₯+ 1/π‘₯)" . " π‘™π‘œπ‘” π‘₯ 1/𝑣 (𝑑𝑣/𝑑π‘₯) = 𝑑(1 + 1/π‘₯)/𝑑π‘₯ . log⁑π‘₯ + 𝑑(log⁑π‘₯ )/𝑑π‘₯ . (1 + 1/π‘₯) 1/𝑣 (𝑑𝑣/𝑑π‘₯) = (𝑑(1)/𝑑π‘₯+𝑑(1/π‘₯)/𝑑π‘₯) . log⁑π‘₯ + 1/π‘₯ (1 + 1/π‘₯) 1/𝑣 (𝑑𝑣/𝑑π‘₯) = (0+((βˆ’1)/π‘₯^2 )) . log⁑π‘₯ + 1/π‘₯ (1 + 1/π‘₯) 1/𝑣 (𝑑𝑣/𝑑π‘₯) = (βˆ’1)/π‘₯^2 . log⁑π‘₯ + 1/π‘₯ (1 + 1/π‘₯) 1/𝑣 (𝑑𝑣/𝑑π‘₯) = (βˆ’log⁑π‘₯)/π‘₯^2 + 1/π‘₯ + 1/π‘₯^2 1/𝑣 (𝑑𝑣/𝑑π‘₯) = (βˆ’log⁑π‘₯)/π‘₯^2 + 1/π‘₯ + 1/π‘₯^2 1/𝑣 (𝑑𝑣/𝑑π‘₯) = ((βˆ’log⁑π‘₯ + π‘₯ + 1)/π‘₯^2 ) 𝑑𝑣/𝑑π‘₯ = 𝑣 ((βˆ’log⁑π‘₯ + π‘₯ + 1)/π‘₯^2 ) 𝑑𝑣/𝑑π‘₯ = π‘₯^((1 + 1/π‘₯) ) ((π‘₯ + 1 βˆ’ log⁑π‘₯ )/π‘₯^2 ) Now 𝑑𝑣/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ Putting values of 𝑑𝑒/𝑑π‘₯ & 𝑑𝑣/𝑑π‘₯ π’…π’š/𝒅𝒙 = (𝒙+𝟏/𝒙)^𝒙 ((𝒙^𝟐 βˆ’ 𝟏)/(𝒙^𝟐+ 𝟏)+π₯𝐨𝐠⁑(𝒙+ 𝟏/𝒙) ) + 𝒙^((𝟏 + 𝟏/𝒙) ) ((𝒙 + 𝟏 βˆ’ π’π’π’ˆβ‘π’™ )/𝒙^𝟐 )

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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.