Ex 5.2, 4 - Chapter 5 Class 12 Continuity and Differentiability

Transcript

Ex 5.2, 4 (Introduction) Differentiate the functions with respect to x sec (tan ( )) Finding derivative of sec x Let = sec = 1 cos Let = 1 & = cos So, y = (y) = = 2 Putting values = (1) cos ( cos ) 1 cos 2 y = 0 (cos ) ( sin ) 1 2 = 0 + sin 2 = sin cos . 1 cos = tan .sec Hence, = Ex 5.2, 4 Differentiate the functions with respect to x sec (tan ( )) Let = sec (tan ) We need to find Derivative of i.e. = ( ) = ( sec (tan )) = sec ( tan ) tan ( tan ) . ( tan ) = sec ( tan ) tan ( tan ) . (sec2 . ) = sec ( tan ) tan ( tan ) . sec2 . 1 2 = sec (tan ) . tan (tan ) . sec2 2 Hence = sec (tan ) . tan (tan ) . sec2