Check Full Chapter Explained - Continuity and Differentiability - https://you.tube/Chapter-5-Class-12-Continuity

Last updated at May 29, 2018 by Teachoo

Check Full Chapter Explained - Continuity and Differentiability - https://you.tube/Chapter-5-Class-12-Continuity

Transcript

Ex 5.5, 4 Differentiate the functions in, 𝑥𝑥 – 2 sin𝑥 Let 𝑦= 𝑥𝑥 – 2 sin𝑥 Let 𝑢= 𝑥𝑥 , 𝑣= 2 sin𝑥 𝑦= 𝑢−𝑣 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑𝑦𝑑𝑥 = 𝑑 𝑢 − 𝑣𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝑢𝑑𝑥 − 𝑑𝑣𝑑𝑥 Calculating 𝒅𝒖𝒅𝒙 𝑢= 𝑥𝑥 Taking log both sides log 𝑢= log (𝑥𝑥) log 𝑢= 𝑥 . log𝑥 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑 log𝑢𝑑𝑥 = 𝑑 𝑥 . log𝑥𝑑𝑥 𝑑 log𝑢𝑑𝑥 𝑑𝑢𝑑𝑢 = 𝑑 𝑥 . log𝑥𝑑𝑥 𝑑 log𝑢𝑑𝑢 𝑑𝑢𝑑𝑥 = 𝑑 𝑥 . log𝑥𝑑𝑥 1𝑢 . 𝑑𝑢𝑑𝑥 = 𝑑 𝑥 . log𝑥𝑑𝑥 1𝑢. 𝑑𝑢𝑑𝑥 = 𝑑 𝑥 𝑑𝑥 .log𝑥 + 𝑑 log𝑥𝑑𝑥 . 𝑥 1𝑢 . 𝑑𝑢𝑑𝑥 = 1 . log𝑥 + 1𝑥 . 𝑥 1𝑢 . 𝑑𝑢𝑑𝑥 = log𝑥+1 𝑑𝑢𝑑𝑥 = 𝑢 log𝑥+1 𝑑𝑢𝑑𝑥 = 𝑥𝑥 log𝑥+1 Calculating 𝒅𝒗𝒅𝒙 𝑣= 2 sin𝑥 Taking log both sides log 𝑣= log ( 2 sin𝑥) log 𝑣= sin 𝑥 . log2 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑 log𝑣𝑑𝑥 = 𝑑 sin 𝑥 . log2𝑑𝑥 𝑑 log𝑣𝑑𝑥 𝑑𝑣𝑑𝑣 = log2 𝑑 sin 𝑥𝑑𝑥 𝑑 log𝑣𝑑𝑣 𝑑𝑣𝑑𝑥 = log2 . cos𝑥 1𝑣 𝑑𝑣𝑑𝑥 = log2 . cos𝑥 𝑑𝑣𝑑𝑥 = 𝑦 (log2 . cos𝑥) 𝑑𝑣𝑑𝑥 = 2 sin𝑥 (log2 . cos𝑥) Now 𝑑𝑦𝑑𝑥 = 𝑑𝑢𝑑𝑥 − 𝑑𝑣𝑑𝑥 Putting values of 𝑑𝑢𝑑𝑥 & 𝑑𝑣𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑥𝑥 log𝑥+1 − 2 sin𝑥 (log2 . cos𝑥) 𝒅𝒚𝒅𝒙 = 𝒙𝒙 𝟏+ 𝒍𝒐𝒈𝒙 − 𝟐 𝒔𝒊𝒏𝒙 ( 𝐜𝐨𝐬𝒙.𝒍𝒐𝒈𝟐)

Ex 5.5

Ex 5.5, 1
Important

Ex 5.5, 2

Ex 5.5, 3 Important

Ex 5.5, 4 You are here

Ex 5.5, 5

Ex 5.5,6 Important

Ex 5.5, 7 Important

Ex 5.5, 8

Ex 5.5, 9 Important

Ex 5.5, 10 Important

Ex 5.5, 11 Important

Ex 5.5, 12

Ex 5.5, 13

Ex 5.5, 14 Important

Ex 5.5, 15

Ex 5.5, 16 Important

Ex 5.5, 17 Important

Ex 5.5, 18

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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.