# Example 47

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Example 47 For a positive constant a find 𝑑𝑦𝑑𝑥 , where 𝑦 = 𝑎𝑡+ 1𝑡 , and 𝑥 = 𝑡+ 1𝑡2 𝑦 = 𝑎𝑡+ 1𝑡 , and 𝑥 = 𝑡+ 1𝑡2 We need to find 𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑𝑥 × 𝑑𝑡𝑑𝑡 𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑𝑡 × 𝑑𝑡𝑑𝑥 𝒅𝒚𝒅𝒙 = 𝒅𝒚𝒅𝒕 𝒅𝒙𝒅𝒕 Calculating 𝒅𝒚𝒅𝒕 𝑦= 𝑎𝑡+ 1𝑡 Let 𝑡+ 1𝑡 = 𝑝 𝑦= 𝑎𝑝 Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑𝑦𝑑𝑡 = 𝑑 𝑎𝑝𝑑𝑡 𝑑𝑦𝑑𝑡 = 𝑑 𝑎𝑝𝑑𝑡 . 𝑑𝑝𝑑𝑝 𝑑𝑦𝑑𝑡 = 𝑑 𝑎𝑝𝑑𝑝 . 𝑑𝑝𝑑𝑡 𝑑𝑦𝑑𝑡 = 𝑎𝑝 . log𝑎. 𝑑 𝑝𝑑𝑡 Putting values of p = 𝑡+ 1𝑡 𝑑𝑦𝑑𝑡 = 𝑎 𝑡 + 1𝑡 . log𝑎. 𝑑 𝑡 + 1𝑡𝑑𝑡 𝑑𝑦𝑑𝑡 = 𝑎 𝑡 + 1𝑡 . log𝑎. 𝑑 𝑡𝑑𝑡 + 𝑑 1𝑡𝑑𝑡 𝑑𝑦𝑑𝑡 = 𝑎 𝑡 + 1𝑡 . log𝑎. 1+ 𝑑 𝑡−1𝑑𝑡 𝑑𝑦𝑑𝑡 = 𝑎 𝑡 + 1𝑡 . log𝑎. 1+ −1 𝑡−1−1 𝑑𝑦𝑑𝑡 = 𝑎 𝑡 + 1𝑡 . log𝑎. 1+ −1 𝑡−2 𝒅𝒚𝒅𝒕 = 𝒂 𝒕 + 𝟏𝒕 . 𝒍𝒐𝒈𝒂. 𝟏− 𝟏 𝒕𝟐 Calculating 𝒅𝒙𝒅𝒕 𝑥= 𝑡+ 1𝑡𝑎 Let 𝑡+ 1𝑡=𝑝 𝑥= 𝑝𝑎 Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑𝑥𝑑𝑡 = 𝑑 𝑝𝑎𝑑𝑡 𝑑𝑥𝑑𝑡 = 𝑑 𝑝𝑎𝑑𝑡 . 𝑑𝑝𝑑𝑝 𝑑𝑥𝑑𝑡 = 𝑑 𝑝𝑎𝑑𝑝 . 𝑑𝑝𝑑𝑡 𝑑𝑥𝑑𝑡 = a 𝑝𝑎 −1 . 𝑑 𝑝𝑑𝑡 Putting value of p = 𝑡+ 1𝑡 𝑑𝑥𝑑𝑡 = a 𝑡+ 1𝑡𝑎 −1 . 𝑑 𝑡 + 1𝑡𝑑𝑡 𝑑𝑥𝑑𝑡 = a 𝑡+ 1𝑡𝑎 −1 . 𝑑 𝑡𝑑𝑡 + 𝑑 1𝑡𝑑𝑡 𝑑𝑥𝑑𝑡 = a 𝑡+ 1𝑡𝑎 −1 . 1+ 𝑑 𝑡−1𝑑𝑡 𝑑𝑥𝑑𝑡 = a 𝑡+ 1𝑡𝑎 −1 . 1+ −1 𝑡−2 𝑑𝑥𝑑𝑡 = a 𝑡+ 1𝑡𝑎 −1 . 1− 1 𝑡2 Now, 𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑𝑡 𝑑𝑥𝑑𝑡 𝑑𝑦𝑑𝑥 = 𝑎𝑡 + 1𝑡 . log𝑎 1 − 1 𝑡2𝑎 𝑡 + 1𝑡𝑎 − 1 1 − 1 𝑡2. 𝒅𝒚𝒅𝒙 = 𝒂𝒕 + 𝟏𝒕 . 𝒍𝒐𝒈𝒂 𝒂 𝒕 + 𝟏𝒕𝒂 − 𝟏

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Example 41 Important

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Example 47 Important You are here

Misc 6 Important

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Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.