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

Transcript

Example 37 (Method 1) Find 𝑑𝑦/𝑑π‘₯ , if π‘₯^(2/3) + 𝑦^(2/3) = π‘Ž^(2/3) . π‘₯^(2/3) + 𝑦^(2/3) = π‘Ž^(2/3) Differentiating w.r.t. x (𝑑(π‘₯^(2/3)))/𝑑π‘₯ + (𝑑(𝑦^(2/3)))/𝑑π‘₯ = (𝑑(π‘Ž^(2/3)))/𝑑π‘₯ 2/3 π‘₯^(2/3 βˆ’ 1) + (𝑑(𝑦^(2/3)))/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦 = 0 2/3 π‘₯^((βˆ’1)/3) + (𝑑(𝑦^(2/3)))/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯ = 0 2/3 π‘₯^((βˆ’1)/3) + 2/3 𝑦^(2/3 βˆ’ 1) Γ— 𝑑𝑦/𝑑π‘₯ = 0 2/3Γ—1/π‘₯^(1/3) + 2/3 y^((βˆ’1)/3) Γ— 𝑑𝑦/𝑑π‘₯ = 0 2/3Γ—1/π‘₯^(1/3) + 2/3 Γ— 1/𝑦^(1/3) Γ— 𝑑𝑦/𝑑π‘₯ = 0 2/3Γ—1/𝑦^(1/3) Γ— 𝑑𝑦/𝑑π‘₯ = 0 βˆ’ 2/3Γ— 1/π‘₯^(1/3) 2/3Γ—1/𝑦^(1/3) Γ— 𝑑𝑦/𝑑π‘₯ = (βˆ’2)/3Γ—1/π‘₯^(1/3) 1/𝑦^(1/3) Γ— 𝑑𝑦/𝑑π‘₯ = (βˆ’1)/π‘₯^(1/3) 𝑑𝑦/𝑑π‘₯ = βˆ’ 𝑦^(1/3)/π‘₯^(1/3) π’…π’š/𝒅𝒙 = βˆ’ (π’š/𝒙)^(𝟏/πŸ‘) Example 37 (Method 2) Find 𝑑𝑦/𝑑π‘₯ , if π‘₯^(2/3) + 𝑦^(2/3) = π‘Ž^(2/3) . π‘₯^(2/3) + 𝑦^(2/3) = π‘Ž^(2/3) π‘₯^(2/3) + 𝑦^(2/3) = π‘Ž^(2/3) . 1 π‘₯^(2/3) + 𝑦^(2/3) = π‘Ž^(2/3) . (sin^2β‘πœƒ+cos^2β‘πœƒ ) π‘₯^(2/3) + 𝑦^(2/3) = π‘Ž^(2/3) sin^2β‘πœƒ+π‘Ž^(2/3) cos^2β‘πœƒ Let 𝒙^(𝟐/πŸ‘) = 𝒂^(𝟐/πŸ‘) 〖𝒄𝒐𝒔〗^𝟐⁑𝜽 Cubing both sides π‘₯^(2/3 \ Γ— 3) = π‘Ž^(2/3 Γ— 3) (cos^2β‘πœƒ )^3 π‘₯^2 = π‘Ž^2 cos^6β‘πœƒ x = π‘Ž cos3 πœƒ Let π’š^(𝟐/πŸ‘) = 𝒂^(𝟐/πŸ‘) 〖𝐬𝐒𝐧〗^𝟐⁑𝜽 Cubing both sides 𝑦^(2/3 \ Γ— 3) = π‘Ž^(2/3 Γ— 3) (sin^2β‘πœƒ )^3 𝑦^2 = π‘Ž^2 sin^6β‘πœƒ y = a sin3 πœƒ Let π‘₯ = a cos^3β‘πœƒ & 𝑦 = a sin^3β‘πœƒ Putting values of x and y in L.H.S of equation Solving LHS π‘₯^(2/3) + 𝑦^(2/3) = (a cos^3β‘πœƒ )^(2/3) + (a sin^3β‘πœƒ )^(2/3) = a^(2/3) cos^(3 Γ— 2/3)β‘πœƒ + a^(2/3) sin^(3 Γ— 2/3)β‘πœƒ = a^(2/3) cos^2β‘πœƒ + a^(2/3) sin^2β‘πœƒ = a^(2/3) (cos^2β‘πœƒ+sin^2β‘πœƒ ) = a^(2/3) . 1 = a^(2/3) = R.H.S Hence it is proved that π‘₯ = a cos^3β‘πœƒ & 𝑦 = a sin^3β‘πœƒ Now, 𝑑𝑦/𝑑π‘₯ = 𝑑𝑦/𝑑π‘₯ . π‘‘πœƒ/π‘‘πœƒ = 𝑑𝑦/π‘‘πœƒ . π‘‘πœƒ/𝑑π‘₯ = (𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ). Calculating π’…π’š/π’…πœ½ 𝑦 = a sin^3β‘πœƒ 𝑑𝑦/π‘‘πœƒ = 𝑑(a sin^3β‘πœƒ )/𝑑π‘₯ 𝑑𝑦/π‘‘πœƒ = a 3 sin^(3 βˆ’ 2)β‘πœƒ . 𝑑(sinβ‘πœƒ )/𝑑π‘₯ 𝑑𝑦/π‘‘πœƒ = 3π‘Ž sin2 πœƒ cos πœƒ Calculating 𝒅𝒙/π’…πœ½ π‘₯ = a cos^3β‘πœƒ 𝑑π‘₯/π‘‘πœƒ = (𝑑 (γ€–a cosγ€—^3β‘γ€–πœƒ)γ€—)/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ = 3a cos^(3 βˆ’1)β‘πœƒ (𝒅(𝒄𝒐𝒔 𝜽) )/π’…πœ½ 𝑑π‘₯/π‘‘πœƒ = 3a cos^2β‘πœƒ . (βˆ’sinβ‘πœƒ ) 𝑑π‘₯/π‘‘πœƒ = βˆ’3a π‘ π‘–π‘›β‘πœƒ cos^2β‘πœƒ Therefore 𝑑𝑦/𝑑π‘₯ = 𝑑𝑦/π‘‘πœƒ Γ· 𝑑π‘₯/π‘‘πœƒ = (3a sin^2β‘πœƒ cosβ‘πœƒ)/(βˆ’3a cos^2β‘πœƒ sinβ‘πœƒ ) = (βˆ’sinβ‘πœƒ)/cosβ‘πœƒ = βˆ’tan πœƒ ∴ 𝑑𝑦/𝑑π‘₯ = βˆ’tan πœƒ Finding value of tan πœƒ 𝑦/π‘₯ = (π‘Ž sin^3β‘πœƒ)/(π‘Ž cos^3β‘πœƒ ) (π‘Ž sin^3β‘πœƒ)/(π‘Ž cos^3β‘πœƒ ) = 𝑦/π‘₯ sin^3β‘πœƒ/cos^3β‘πœƒ = 𝑦/π‘₯ (sinβ‘πœƒ/cosβ‘πœƒ )^3 =𝑦/π‘₯ sinβ‘πœƒ/cosβ‘πœƒ = (𝑦/π‘₯)^(1/3) tanβ‘πœƒ = βˆ›(𝑦/π‘₯) Thus, 𝑑𝑦/𝑑π‘₯ = βˆ’tan πœƒ π’…π’š/𝒅𝒙 = βˆ’βˆš(πŸ‘&π’š/𝒙)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.