Finding derivative of Inverse trigonometric functions

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

### Transcript

Ex 5.3, 12 Find ππ¦/ππ₯ in, y = sinβ1 ((1β π₯^2)/( 1+ π₯2 )) , 0 < x < 1 y = sinβ1 ((1β π₯^2)/( 1+ π₯2 )) Putting x = tan ΞΈ y = γπ ππγ^(β1) (γ1 β tan^2γβ‘π/γ1 + tan^2γβ‘π ) y = γπ ππγ^(β1) (cosβ‘2π) π¦ =γπ ππγ^(β1) (γsin γβ‘(π/2 β2π) ) π¦ = π/2 β 2π "We know that" " " β(πππ β‘2ΞΈ " =" (1 β π‘ππβ‘2 π)/(1 + π‘ππβ‘2 π)) Putting value of ΞΈ = tanβ1 x π¦ = π/2 β 2 γπ‘ππγ^(β1) π₯ Differentiating both sides π€.π.π‘.π₯ . (π(π¦))/ππ₯ = (π (" " π/2 " β " 2γπ‘ππγ^(β1) π₯" " ))/ππ₯ ππ¦/ππ₯ = 0 β 2 (πγ (π‘ππγ^(β1) π₯))/ππ₯ ππ¦/ππ₯ = β2 (πγ (π‘ππγ^(β1) π₯))/ππ₯ ππ¦/ππ₯ = β2 (1/(1 + π₯^2 )) ππ/ππ = (βπ)/(π + π^π ) Since x = tan ΞΈ β΄ γπ‘ππγ^(β1) x = ΞΈ ((γπ‘ππγ^(β1) π₯") β = " 1/(1 + π₯^2 ))