Ex 5.5, 14 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Sept. 17, 2019 by Teachoo
Transcript
Ex 5.5, 14 Find of the functions in, ( cos ) = ( cos ) Given ( cos ) = ( cos ) Taking log both sides log ( cos ) = log ( cos ) . log ( cos )= . log ( cos ) Differentiating both sides . . . . ( . log ( cos )) = . log ( cos ) Finding ( . ( )) ( . ( )) = ( ) . log co + ( ( )) . = . log co + 1 . . = . log co + 1 . sin . = . log co + sin . = . log co tan . Finding . ( ) . ( ) = ( ) . log co + ( ( )) . = log co + 1 . . = log co + 1 . . . = log co + 1 . . . = log co + 1 . ( sin ) . . = log co + tan . . Now , ( . log ( cos )) = . log ( cos ) log co tan . = log co tan . . log co . tan = log co . tan . log co + tan = log co + tan log co + tan = log co + tan = + +
Ex 5.5
Ex 5.5, 1
Important
Ex 5.5, 2 Important
Ex 5.5, 3 Important
Ex 5.5, 4
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 You are here
Ex 5.5, 15
Ex 5.5, 16 Important
Ex 5.5, 17 Important
Ex 5.5, 18
About the Author
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.