# Ex 5.3, 14 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at March 11, 2021 by Teachoo

Last updated at March 11, 2021 by Teachoo

Transcript

Ex 5.3, 14 Find ππ¦/ππ₯ in, y = sinβ1 (2π₯ β(1βπ₯^2 )) , β 1/β2 < x < 1/β2 y = sinβ1 (2π₯ β(1βπ₯^2 )) Putting π₯ =π ππβ‘π π¦ = sinβ1 (2 sinβ‘π β(1βγπ ππγ^2 π)) π¦ = sinβ1 ( 2 sin ΞΈ β(γπππ γ^2 π)) π¦ ="sinβ1 " (γ"2 sin ΞΈ" γβ‘cosβ‘π ) π¦ = sinβ1 (sinβ‘γ2 π)γ π¦ = 2ΞΈ Putting value of ΞΈ = sinβ1 x π¦ = 2 γπ ππγ^(β1) π₯ Since x = sin ΞΈ β΄ γπ ππγ^(β1) x = ΞΈ Differentiating both sides π€.π.π‘.π₯ . (π(π¦))/ππ₯ = (π (γ2 sin^(β1)γβ‘π₯ ))/ππ₯ ππ¦/ππ₯ = 2 (πγ (π ππγ^(β1) π₯))/ππ₯ ππ¦/ππ₯ = 2 (1/β(1 βγ π₯γ^2 )) π π/π π = π/β(π β π^π ) ((sin^(β1)β‘π₯ )^β²= 1/β(1 β π₯^2 ))

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.