Ex 5.5, 12 - Chapter 5 Class 12 Continuity and Differentiability

Transcript

Ex 5.5, 12 Find 𝑑𝑦﷮𝑑𝑥﷯ of the functions in, 𝑥﷮𝑦﷯ + 𝑦﷮𝑥﷯ = 1 𝑥﷮𝑦﷯ + 𝑦﷮𝑥﷯ = 1 Let 𝑢 = 𝑥﷮𝑦﷯ , 𝑣 = 𝑦﷮𝑥﷯ Hence, 𝑢+𝑣=1 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑( 𝑣﷮+ 𝑢﷯)﷮𝑑𝑥﷯ = 𝑑 1﷯﷮𝑑𝑥﷯ 𝑑(𝑣)﷮𝑑𝑥﷯ + 𝑑𝑢﷮𝑑𝑥﷯ = 0 Calculating 𝒅𝒗﷮𝒅𝒙﷯ 𝑣= 𝑥﷮𝑦﷯ Taking log both sides log﷮𝑣﷯= log﷮ ( 𝑥﷮𝑦﷯) ﷯ log﷮𝑣﷯= 𝑦. log﷮𝑥 ﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑( log﷮𝑣﷯)﷮𝑑𝑥﷯ = 𝑑(𝑦 . log﷮𝑥﷯)﷮𝑑𝑥﷯ 𝑑( log﷮𝑣﷯)﷮𝑑𝑥﷯ 𝑑𝑣﷮𝑑𝑣﷯﷯ = 𝑑 𝑦 log﷮𝑥﷯﷯﷮𝑑𝑥﷯ 𝑑( log﷮𝑣﷯)﷮𝑑𝑣﷯ 𝑑𝑣﷮𝑑𝑥﷯﷯ = 𝑑 𝑦 log﷮𝑥﷯﷯﷮𝑑𝑥﷯ 1﷮𝑣﷯ 𝑑𝑣﷮𝑑𝑥﷯﷯ = 𝑑 𝑦 log﷮𝑥﷯﷯﷮𝑑𝑥﷯ 1﷮𝑣﷯ 𝑑𝑣﷮𝑑𝑥﷯﷯ = 𝑑 𝑦﷯﷮𝑑𝑥﷯ . log⁡𝑥 + 𝑑 ( log﷮𝑥﷯)﷮𝑑𝑥﷯ . 𝑦 1﷮𝑣﷯ 𝑑𝑣﷮𝑑𝑥﷯﷯ = 𝑑𝑦﷮𝑑𝑥﷯ . log⁡𝑥 + 1﷮𝑥﷯ . 𝑦 1﷮𝑣﷯ 𝑑𝑣﷮𝑑𝑥﷯﷯ = 𝑑𝑦﷮𝑑𝑥﷯ log⁡𝑥 + 𝑦﷮𝑥﷯ 𝑑𝑣﷮𝑑𝑥﷯ = v log 𝑥+ 𝑦﷮𝑥﷯﷯ Putting value of 𝑣 = 𝑥﷮𝑦﷯ 𝑑𝑣﷮𝑑𝑥﷯ = 𝑥﷮𝑦﷯ log﷮𝑥+ 𝑦﷮𝑥﷯﷯﷯ 𝑑𝑣﷮𝑑𝑥﷯ = 𝑥﷮𝑦﷯.log 𝑥 . 𝑑𝑦﷮𝑑𝑥﷯ + 𝑥﷮𝑦﷯. 𝑦﷮𝑥﷯ Calculating 𝒅𝒖﷮𝒅𝒙﷯ 𝑢 = 𝑦﷮𝑥﷯ Taking log both sides log﷮𝑢﷯= log﷮ ( 𝑦﷮𝑥﷯) ﷯ log﷮𝑢﷯= 𝑥 . log﷮𝑦 ﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑( log﷮𝑢﷯)﷮𝑑𝑥﷯ = 𝑑(𝑥 . log﷮𝑦﷯)﷮𝑑𝑥﷯ 𝑑( log﷮𝑢﷯)﷮𝑑𝑥﷯ 𝑑𝑢﷮𝑑𝑢﷯﷯ = 𝑑 𝑥. log﷮𝑦﷯﷯﷮𝑑𝑥﷯ 𝑑( log﷮𝑢﷯)﷮𝑑𝑢﷯ 𝑑𝑢﷮𝑑𝑥﷯﷯ = 𝑑 (𝑥 . log﷮𝑦﷯)﷮𝑑𝑥﷯ 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = 𝑑 𝑥 . log﷮𝑦﷯﷯﷮𝑑𝑥﷯ 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = 𝑑 𝑥﷯﷮𝑑𝑥﷯ . log﷮𝑦﷯ + 𝑑( log﷮𝑦﷯)﷮𝑑𝑥﷯ . 𝑥 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = 1 . log⁡𝑦 + 𝑥. 𝑑 log﷮𝑦﷯﷯﷮𝑑𝑥﷯ . 𝑑𝑦﷮𝑑𝑦﷯ 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = log⁡𝑦 + 𝑥. 𝑑 log﷮𝑦﷯﷯﷮𝑑𝑦﷯ . 𝑑𝑦﷮𝑑𝑥﷯ 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = log⁡𝑦 + 𝑥. 1﷮𝑦﷯ . 𝑑𝑦﷮𝑑𝑥﷯ 𝑑𝑢﷮𝑑𝑥﷯ = 𝑢 log⁡𝑦 + 𝑥﷮𝑦﷯ . 𝑑𝑦 ﷮𝑑𝑥﷯﷯ 𝑑𝑢﷮𝑑𝑥﷯ = 𝑦﷮𝑥﷯ log⁡𝑦 + 𝑥﷮𝑦﷯ . 𝑑𝑦 ﷮𝑑𝑥﷯﷯ 𝑑𝑢﷮𝑑𝑥﷯ = 𝑦﷮𝑥﷯ log﷮𝑦﷯+ 𝑦﷮𝑥 −1﷯. 𝑥 𝑑𝑦﷮𝑑𝑥﷯ Now, 𝑑𝑢﷮𝑑𝑥﷯ + 𝑑𝑣﷮𝑑𝑥﷯ = 0 Putting value of 𝑑𝑣﷮𝑑𝑥﷯ & 𝑑𝑢﷮𝑑𝑥﷯ 𝑥﷮𝑦﷯ log﷮𝑥 ﷯ 𝑑𝑦﷮𝑑𝑥﷯ + 𝑥﷮𝑦﷯ 𝑦﷮𝑥﷯﷯ + 𝑦﷮𝑥﷯ log﷮𝑦+ 𝑦﷮𝑥 −1﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯﷯ = 0 𝑥﷮𝑦﷯ log﷮𝑥 ﷯ 𝑑𝑦﷮𝑑𝑥﷯ + 𝑦﷮𝑥 −1﷯ 𝑑𝑦﷮𝑑𝑥﷯ + 𝑥﷮𝑦﷯ . 𝑦﷮𝑥﷯ + 𝑦﷮𝑥﷯ log 𝑦 = 0 𝑑𝑦﷮𝑑𝑥﷯ 𝑥﷮𝑦﷯ log﷮𝑥 ﷯+ 𝑦﷮𝑥 −1﷯﷯ = − 𝑥﷮𝑦 −1﷯×𝑦+ 𝑦﷮𝑥﷯ log 𝑦﷯ 𝒅𝒚﷮𝒅𝒙﷯ = − 𝒚 𝒙﷮𝒚 −𝟏﷯ + 𝒚﷮𝒙﷯ 𝐥𝐨𝐠 𝒚﷯﷮ 𝒙﷮𝒚﷯ 𝒍𝒐𝒈﷮𝒙 ﷯+ 𝒚﷮𝒙 −𝟏﷯﷯