Ex 5.5, 10

Transcript

Ex 5.5, 10 Differentiate the functions in, 𝑥﷮𝑥 𝑐𝑜𝑠﷮𝑥﷯﷯ + 𝑥2+ 1﷮𝑥2− 1﷯ Calculating 𝒅𝒖﷮𝒅𝒙﷯ 𝑢 = 𝑥﷮𝑥 𝑐𝑜𝑠﷮𝑥﷯﷯ Taking log both sides log﷮𝑢﷯= log﷮ 𝑥﷮𝑥 𝑐𝑜𝑠﷮𝑥﷯﷯﷯ log﷮𝑢﷯= 𝑥﷮𝑥 𝑐𝑜𝑠﷮𝑥﷯﷯. log﷮ 𝑥﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑( log﷮𝑢﷯)﷮𝑑𝑥﷯ = 𝑑 𝑥﷮𝑥 𝑐𝑜𝑠﷮𝑥﷯﷯. log﷮ 𝑥﷯﷯﷮𝑑𝑥﷯ 𝑑( log﷮𝑢﷯)﷮𝑑𝑥﷯ . 𝑑𝑢﷮𝑑𝑢﷯ = 𝑑 𝑥﷮𝑥 𝑐𝑜𝑠﷮𝑥﷯﷯. log﷮ 𝑥﷯﷯﷮𝑑𝑥﷯ 𝑑( log﷮𝑢﷯)﷮𝑑𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = 𝑑 𝑥﷮𝑥 𝑐𝑜𝑠﷮𝑥﷯﷯. log﷮ 𝑥﷯﷯﷮𝑑𝑥﷯ 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = 𝑑 𝑥﷮𝑥 𝑐𝑜𝑠﷮𝑥﷯﷯. log﷮ 𝑥﷯﷯﷮𝑑𝑥﷯ 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = 𝑑 𝑥﷯﷮𝑑𝑥﷯ . cos 𝑥 log⁡𝑥 + 𝑑 cos﷮𝑥﷯﷯﷮𝑑𝑥﷯ . 𝑥 log 𝑥 + 𝑑( log﷮𝑥﷯)﷮𝑑𝑥﷯ . 𝑥 𝑐𝑜𝑠﷮𝑥﷯ 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = 1. cos﷮𝑥﷯ . log⁡𝑥 + − sin﷮𝑥﷯﷯ . 𝑥 log 𝑥 + 1﷮𝑥﷯ . 𝑥 𝑐𝑜𝑠﷮𝑥﷯ 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = cos﷮𝑥﷯ . log⁡𝑥 − sin﷮𝑥﷯ . 𝑥 log 𝑥 + 𝑥 𝑐𝑜𝑠﷮𝑥﷯ 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = cos﷮𝑥﷯ . log⁡𝑥 + cos﷮𝑥﷯ . 𝑥 sin 𝑥 + log﷮𝑥﷯ 𝑑𝑢﷮𝑑𝑥﷯ = u cos﷮𝑥﷯ log﷮𝑥﷯+1﷯−𝑥 sin﷮𝑥 log﷮𝑥﷯﷯﷯ 𝑑𝑢﷮𝑑𝑥﷯ = 𝑥﷮𝑥 𝑐𝑜𝑠﷮𝑥﷯﷯ cos﷮ log﷮𝑥+1﷯﷯−𝑥 sin﷮𝑥 log﷮𝑥﷯﷯﷯﷯ Calculating 𝒅𝒗﷮𝒅𝒙﷯ 𝑣= 𝑥2 + 1﷮𝑥2 − 1﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑 𝑣﷯﷮𝑑𝑥﷯ = 𝑑 𝑥2 + 1﷮𝑥2 − 1﷯﷯﷮𝑑𝑥﷯ 𝑑 𝑣﷯﷮𝑑𝑥﷯ . 𝑑𝑣﷮𝑑𝑣﷯ = 𝑑 𝑥2 + 1﷮𝑥2 − 1﷯﷯﷮𝑑𝑥﷯ 𝑑 𝑣﷯﷮𝑑𝑣﷯ . 𝑑𝑣﷮𝑑𝑥﷯ = 𝑑 𝑥2 + 1﷮𝑥2 − 1﷯﷯﷮𝑑𝑥﷯ 1. 𝑑𝑣﷮𝑑𝑥﷯ = 𝑑 𝑥2 + 1﷮𝑥2 − 1﷯﷯﷮𝑑𝑥﷯ 𝑑𝑣﷮𝑑𝑥﷯ = 𝑑 𝑥2 + 1﷮𝑥2 − 1﷯﷯﷮𝑑𝑥﷯ 𝑑𝑣﷮𝑑𝑥﷯ = 𝑑 𝑥2+ 1﷯﷮𝑑𝑥﷯ . 𝑥﷮2﷯ −1﷯ − 𝑑 𝑥﷮2﷯ −1﷯﷮𝑑𝑥﷯ . 𝑥2+ 1﷯﷮ 𝑥﷮2﷯ −1﷯﷮2﷯﷯ 𝑑𝑣﷮𝑑𝑥﷯ = 𝑑 𝑥2+ 1﷯﷮𝑑𝑥﷯ + 𝑑 1﷯﷮𝑑𝑥﷯﷯ 𝑥﷮2﷯ −1﷯ − 𝑑 𝑥2+ 1﷯﷮𝑑𝑥﷯ + 𝑑 1﷯﷮𝑑𝑥﷯﷯ 𝑥2+ 1﷯﷮ 𝑥﷮2﷯ −1﷯﷮2﷯﷯ 𝑑𝑣﷮𝑑𝑥﷯ = 2𝑥 + 0﷯ 𝑥﷮2﷯ −1﷯ − 2𝑥 − 0﷯ 𝑥﷮2﷯ + 1﷯﷮ 𝑥﷮2﷯ −1﷯﷮2﷯﷯ 𝑑𝑣﷮𝑑𝑥﷯ = 2𝑥 𝑥﷮2﷯ −1﷯ − 2𝑥 𝑥﷮2﷯ + 1﷯﷮ 𝑥﷮2﷯ −1﷯﷮2﷯﷯ 𝑑𝑣﷮𝑑𝑥﷯ = 2𝑥 𝑥﷮2﷯ −1 − 𝑥﷮2﷯ −1﷯﷮ 𝑥﷮2﷯ −1﷯﷮2﷯﷯ 𝑑𝑣﷮𝑑𝑥﷯ = 2𝑥 − 2﷯﷮ 𝑥﷮2﷯ −1﷯﷮2﷯﷯ 𝑑𝑣﷮𝑑𝑥﷯ = −4𝑥﷮ 𝑥﷮2﷯ −1﷯﷮2﷯﷯ Now 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑢﷮𝑑𝑥﷯ + 𝑑𝑣﷮𝑑𝑥﷯ Putting value of 𝑑𝑢﷮𝑑𝑥﷯ & 𝑑𝑣﷮𝑑𝑥﷯ 𝒅𝒚﷮𝒅𝒙﷯ = 𝒙﷮ 𝒙 𝐜𝐨𝐬﷮𝒙﷯﷯ 𝐜𝐨𝐬﷮𝒙 𝟏+ 𝐥𝐨𝐠﷮𝒙﷯﷯ −𝒙 𝐬𝐢𝐧﷮𝒙 𝐥𝐨𝐠﷮𝒙﷯﷯﷯﷯ − 𝟒𝒙﷮ 𝒙﷮𝟐﷯ −𝟏﷯﷮𝟐﷯﷯