Ex 5.5, 4 - Differentiate x2 - 2sin x - Chapter 5 Class 12 - Ex 5.5

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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

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

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