Ex 5.3, 10 - Find dy/dx in y=tan-1 (3x - x3/1 - 3x2) - Ex 5.3

Ex 5.3, 10 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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Ex 5.3, 10 Find 𝑑𝑦/𝑑π‘₯ in, 𝑦 = tan–1 ((3π‘₯βˆ’ π‘₯^3)/( 1βˆ’ 3π‘₯2 )) , βˆ’ 1/√3 < π‘₯ < 1/√3 𝑦 = tan–1 ((3π‘₯βˆ’ π‘₯^3)/( 1βˆ’ 3π‘₯2 )) Putting x = tan ΞΈ y = γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) ((3 tanβ‘γ€–πœƒ βˆ’ tan⁑3 πœƒγ€—)/(1 βˆ’ 3 tan⁑2 πœƒ)) y = γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (tan⁑〖3 πœƒγ€—) 𝑦 = 3πœƒ Putting value of ΞΈ = γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯ 𝑦 = 3γ€– (π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯) (π‘‘π‘Žπ‘›β‘3ΞΈ " = " (3 π‘‘π‘Žπ‘›β‘γ€–πœƒβˆ’π‘‘π‘Žπ‘›β‘3 πœƒγ€—)/(1βˆ’3 π‘‘π‘Žπ‘›β‘2 πœƒ)) Since x = tan ΞΈ ∴ γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) x = ΞΈ Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ . (𝑑(𝑦))/𝑑π‘₯ = (𝑑 3γ€–(π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯") " )/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ =3 (𝑑〖(π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯") " )/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 3 (1/(1 + π‘₯^2 )) π’…π’š/𝒅𝒙 = πŸ‘/(𝟏 +γ€– 𝒙〗^𝟐 ) ((γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯")β€˜ = " 1/(1 + π‘₯^2 ))

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.