# Ex 5.5, 4 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

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)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.