Finding derivative of Inverse trigonometric functions

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

### Transcript

Ex 5.3, 14 Find ππ¦/ππ₯ in, y = sinβ1 (2π₯ β(1βπ₯^2 )) , β 1/β2 < x < 1/β2 y = sinβ1 (2π₯ β(1βπ₯^2 )) Putting π₯ =π ππβ‘π π¦ = sinβ1 (2 sinβ‘π β(1βγπ ππγ^2 π)) π¦ = sinβ1 ( 2 sin ΞΈ β(γπππ γ^2 π)) π¦ ="sinβ1 " (γ"2 sin ΞΈ" γβ‘cosβ‘π ) π¦ = sinβ1 (sinβ‘γ2 π)γ π¦ = 2ΞΈ Putting value of ΞΈ = sinβ1 x π¦ = 2 γπ ππγ^(β1) π₯ Since x = sin ΞΈ β΄ γπ ππγ^(β1) x = ΞΈ Differentiating both sides π€.π.π‘.π₯ . (π(π¦))/ππ₯ = (π (γ2 sin^(β1)γβ‘π₯ ))/ππ₯ ππ¦/ππ₯ = 2 (πγ (π ππγ^(β1) π₯))/ππ₯ ππ¦/ππ₯ = 2 (1/β(1 βγ π₯γ^2 )) ππ/ππ = π/β(π β π^π ) ((sin^(β1)β‘π₯ )^β²= 1/β(1 β 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.