Ex 5.5, 10
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Ex 5.5, 10 Differentiate the functions in, + 2+ 1 2 1 Calculating = Taking log both sides log = log log = . log Differentiating both sides . . . . ( log ) = . log ( log ) . = . log ( log ) . = . log 1 . = . log 1 . = . cos log + cos . log + ( log ) . 1 . = 1. cos . log + sin . log + 1 . 1 . = cos . log sin . log + 1 . = cos . log + cos . sin + log = u cos log +1 sin log = cos log +1 sin log Calculating = 2 + 1 2 1 Differentiating both sides . . . . = 2 + 1 2 1 . = 2 + 1 2 1 . = 2 + 1 2 1 1. = 2 + 1 2 1 = 2 + 1 2 1 = 2+ 1 . 2 1 2 1 . 2+ 1 2 1 2 = 2+ 1 + 1 2 1 2+ 1 + 1 2+ 1 2 1 2 = 2 + 0 2 1 2 0 2 + 1 2 1 2 = 2 2 1 2 2 + 1 2 1 2 = 2 2 1 2 1 2 1 2 = 2 2 2 1 2 = 4 2 1 2 Now = + Putting value of & = +
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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.