Ex 5.2, 4

Transcript

Ex 5.2, 4 (Introduction) Differentiate the functions with respect to x sec (tan ( ﷮𝑥﷯ )) Finding derivative of sec x Let 𝑦= sec﷮𝑥﷯ 𝑦 = 1﷮ cos﷮𝑥﷯﷯ Let 𝑢 = 1 & 𝑣 = cos⁡𝑥 So, y = 𝑢﷮𝑣﷯ (y)’ = 𝑢﷮𝑣﷯﷯﷮′﷯ 𝑦′ = 𝑢﷮′﷯𝑣 − 𝑣﷮′﷯𝑢﷮ 𝑣﷮2﷯﷯ Putting values 𝑦′ = (1)﷮′﷯ cos﷮𝑥﷯ − ( cos﷮𝑥 ﷯)﷮′﷯1﷮ cos﷮2﷯﷮𝑥﷯﷯ y’ = 0 (cos﷮𝑥)﷯ − ( −sin﷮𝑥) 1﷯﷮ 𝑐𝑜𝑠﷮2﷯𝑥﷯ = 0 + sin﷮𝑥﷯﷮ 𝑐𝑜𝑠﷮2﷯𝑥﷯ = sin﷮𝑥﷯﷮ cos﷮𝑥﷯﷯ . 1﷮ cos﷮𝑥﷯﷯ = tan﷮𝑥﷯.sec⁡𝑥 Hence, 𝒔𝒆𝒄﷮𝒙﷯﷯’ = 𝒔𝒆𝒄﷮𝒙﷯ 𝒕𝒂𝒏⁡𝒙 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’ = 𝑑𝑦﷮𝑑𝑥﷯ 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 ﷮𝑥﷯﷮𝑑𝑥﷯) = sec ﷮( tan﷮ ﷮𝑥﷯﷯)﷯ tan ﷮( tan﷮ ﷮𝑥﷯﷯)﷯. sec2 ﷮𝑥﷯ . 1﷮2 ﷮𝑥﷯﷯ = sec (tan ﷮𝑥﷯ ) . tan (tan ﷮𝑥﷯ ) . sec2 ﷮𝑥﷯﷮2 ﷮𝑥﷯ ﷯ Hence 𝒅𝒚﷮𝒅𝒙﷯ = sec (tan ﷮𝒙﷯ ) . tan (tan ﷮𝒙﷯ ) . sec2 ﷮𝒙﷯﷮𝟐 ﷮𝒙﷯ ﷯