Chapter 5 Class 12 Continuity and Differentiability

Class 12
Important Questions for exams Class 12

### Transcript

Ex 5.3, 10 Find ππ¦/ππ₯ in, π¦ = tanβ1 ((3π₯β π₯^3)/( 1β 3π₯2 )) , β 1/β3 < π₯ < 1/β3 π¦ = tanβ1 ((3π₯β π₯^3)/( 1β 3π₯2 )) Putting x = tan ΞΈ y = γπ‘ππγ^(β1) ((3 tanβ‘γπ β tanβ‘3 πγ)/(1 β 3 tanβ‘2 π)) y = γπ‘ππγ^(β1) (tanβ‘γ3 πγ) π¦ = 3π Putting value of ΞΈ = γπ‘ππγ^(β1) π₯ π¦ = 3γ (π‘ππγ^(β1) π₯) (π‘ππβ‘3ΞΈ " = " (3 π‘ππβ‘γπβπ‘ππβ‘3 πγ)/(1β3 π‘ππβ‘2 π)) Since x = tan ΞΈ β΄ γπ‘ππγ^(β1) x = ΞΈ Differentiating both sides π€.π.π‘.π₯ . (π(π¦))/ππ₯ = (π 3γ(π‘ππγ^(β1) π₯") " )/ππ₯ ππ¦/ππ₯ =3 (πγ(π‘ππγ^(β1) π₯") " )/ππ₯ ππ¦/ππ₯ = 3 (1/(1 + π₯^2 )) ππ/ππ = π/(π +γ πγ^π ) ((γπ‘ππγ^(β1) π₯")β = " 1/(1 + π₯^2 ))

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.