Example 47 - Chapter 5 Class 12 Continuity and Differentiability

Transcript

Example 47 For a positive constant a find 𝑑𝑦﷮𝑑𝑥﷯ , where 𝑦 = 𝑎﷮𝑡+ 1﷮𝑡﷯﷯ , and 𝑥 = 𝑡+ 1﷮𝑡﷯﷯﷮2﷯ 𝑦 = 𝑎﷮𝑡+ 1﷮𝑡﷯﷯ , and 𝑥 = 𝑡+ 1﷮𝑡﷯﷯﷮2﷯ We need to find 𝑑𝑦﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦﷮𝑑𝑥﷯ × 𝑑𝑡﷮𝑑𝑡﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦﷮𝑑𝑡﷯ × 𝑑𝑡﷮𝑑𝑥﷯ 𝒅𝒚﷮𝒅𝒙﷯ = 𝒅𝒚﷮𝒅𝒕﷯﷮ 𝒅𝒙﷮𝒅𝒕﷯﷯ Calculating 𝒅𝒚﷮𝒅𝒕﷯ 𝑦= 𝑎﷮𝑡+ 1﷮𝑡﷯﷯ Let 𝑡+ 1﷮𝑡﷯﷯ = 𝑝 𝑦= 𝑎﷮𝑝﷯ Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑𝑦﷮𝑑𝑡﷯ = 𝑑 𝑎﷮𝑝﷯﷯﷮𝑑𝑡﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑑 𝑎﷮𝑝﷯﷯﷮𝑑𝑡﷯ . 𝑑𝑝﷮𝑑𝑝﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑑 𝑎﷮𝑝﷯﷯﷮𝑑𝑝﷯ . 𝑑𝑝﷮𝑑𝑡﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑎﷮𝑝﷯ . log﷮𝑎﷯. 𝑑 𝑝﷯﷮𝑑𝑡﷯ Putting values of p = 𝑡+ 1﷮𝑡﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑎﷮ 𝑡 + 1﷮𝑡﷯﷯ ﷯ . log﷮𝑎﷯. 𝑑 𝑡 + 1﷮𝑡﷯﷯﷮𝑑𝑡﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑎﷮ 𝑡 + 1﷮𝑡﷯﷯ ﷯ . log﷮𝑎﷯. 𝑑 𝑡﷯﷮𝑑𝑡﷯ + 𝑑 1﷮𝑡﷯﷯﷮𝑑𝑡﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑎﷮ 𝑡 + 1﷮𝑡﷯﷯ ﷯ . log﷮𝑎﷯. 1+ 𝑑 𝑡﷮−1﷯﷯﷮𝑑𝑡﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑎﷮ 𝑡 + 1﷮𝑡﷯﷯ ﷯ . log﷮𝑎﷯. 1+ −1﷯ 𝑡﷮−1−1﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑎﷮ 𝑡 + 1﷮𝑡﷯﷯ ﷯ . log﷮𝑎﷯. 1+ −1﷯ 𝑡﷮−2﷯﷯ 𝒅𝒚﷮𝒅𝒕﷯ = 𝒂﷮ 𝒕 + 𝟏﷮𝒕﷯﷯ ﷯ . 𝒍𝒐𝒈﷮𝒂﷯. 𝟏− 𝟏﷮ 𝒕﷮𝟐﷯﷯﷯ Calculating 𝒅𝒙﷮𝒅𝒕﷯ 𝑥= 𝑡+ 1﷮𝑡﷯﷯﷮𝑎﷯ Let 𝑡+ 1﷮𝑡﷯﷯=𝑝 𝑥= 𝑝﷮𝑎﷯ Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑𝑥﷮𝑑𝑡﷯ = 𝑑 𝑝﷮𝑎﷯﷯﷮𝑑𝑡﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 𝑑 𝑝﷮𝑎﷯﷯﷮𝑑𝑡﷯ . 𝑑𝑝﷮𝑑𝑝﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 𝑑 𝑝﷮𝑎﷯﷯﷮𝑑𝑝﷯ . 𝑑𝑝﷮𝑑𝑡﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a 𝑝﷮𝑎 −1 ﷯ . 𝑑 𝑝﷯﷮𝑑𝑡﷯ Putting value of p = 𝑡+ 1﷮𝑡﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a 𝑡+ 1﷮𝑡﷯﷯﷮𝑎 −1 ﷯ . 𝑑 𝑡 + 1﷮𝑡﷯﷯﷮𝑑𝑡﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a 𝑡+ 1﷮𝑡﷯﷯﷮𝑎 −1 ﷯ . 𝑑 𝑡﷯﷮𝑑𝑡﷯ + 𝑑 1﷮𝑡﷯﷯﷮𝑑𝑡﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a 𝑡+ 1﷮𝑡﷯﷯﷮𝑎 −1 ﷯ . 1+ 𝑑 𝑡﷮−1﷯﷯﷮𝑑𝑡﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a 𝑡+ 1﷮𝑡﷯﷯﷮𝑎 −1 ﷯ . 1+ −1﷯ 𝑡﷮−2﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a 𝑡+ 1﷮𝑡﷯﷯﷮𝑎 −1 ﷯ . 1− 1﷮ 𝑡﷮2﷯﷯﷯ Now, 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦﷮𝑑𝑡﷯﷮ 𝑑𝑥﷮𝑑𝑡﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑎﷮𝑡 + 1﷮𝑡﷯﷯ . log﷮𝑎 ﷯ 1 − 1﷮ 𝑡﷮2﷯﷯﷯﷮𝑎 𝑡 + 1﷮𝑡﷯﷯﷮𝑎 − 1﷯ 1 − 1﷮ 𝑡﷮2﷯﷯﷯.﷯ 𝒅𝒚﷮𝒅𝒙﷯ = 𝒂﷮𝒕 + 𝟏﷮𝒕﷯﷯ . 𝒍𝒐𝒈﷮𝒂 ﷯﷮𝒂 𝒕 + 𝟏﷮𝒕﷯﷯﷮𝒂 − 𝟏﷯﷯