# Ex 5.2, 4

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 5.2, 4 (Introduction) Differentiate the functions with respect to x sec (tan ( 𝑥 )) Finding derivative of sec x Let 𝑦= sec𝑥 𝑦 = 1 cos𝑥 Let 𝑢 = 1 & 𝑣 = cos𝑥 So, y = 𝑢𝑣 (y)’ = 𝑢𝑣′ 𝑦′ = 𝑢′𝑣 − 𝑣′𝑢 𝑣2 Putting values 𝑦′ = (1)′ cos𝑥 − ( cos𝑥 )′1 cos2𝑥 y’ = 0 (cos𝑥) − ( −sin𝑥) 1 𝑐𝑜𝑠2𝑥 = 0 + sin𝑥 𝑐𝑜𝑠2𝑥 = sin𝑥 cos𝑥 . 1 cos𝑥 = tan𝑥.sec𝑥 Hence, 𝒔𝒆𝒄𝒙’ = 𝒔𝒆𝒄𝒙 𝒕𝒂𝒏𝒙 Ex 5.2, 4 Differentiate the functions with respect to x sec (tan ( 𝑥 )) Let 𝑦 = sec (tan 𝑥 ) We need to find Derivative of 𝑦 i.e. 𝑦’ = 𝑑𝑦𝑑𝑥 𝑑(𝑦)𝑑𝑥 = 𝑑( sec (tan 𝑥)) 𝑑𝑥 = sec ( tan 𝑥) tan ( tan 𝑥). 𝑑( tan 𝑥)𝑑𝑥 = sec ( tan 𝑥) tan ( tan 𝑥). (sec2 𝑥 . 𝑑 𝑥𝑑𝑥) = sec ( tan 𝑥) tan ( tan 𝑥). sec2 𝑥 . 12 𝑥 = sec (tan 𝑥 ) . tan (tan 𝑥 ) . sec2 𝑥2 𝑥 Hence 𝒅𝒚𝒅𝒙 = sec (tan 𝒙 ) . tan (tan 𝒙 ) . sec2 𝒙𝟐 𝒙

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