Ex 5.3, 11 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at April 16, 2024 by Teachoo
Transcript
Ex 5.3, 11 Find ππ¦/ππ₯ in, π¦ = cosβ1 ((1β π₯^2)/( 1+ π₯2 )) , 0 < x < 1 π¦ = cosβ1 ((1β π₯^2)/( 1+ π₯2 )) Putting x = tan ΞΈ y = γπππ γ^(β1) ((1βtanβ‘2 π)/(1+ tanβ‘2 π)) y = cosβ1 (cos 2π) π¦ =2ΞΈ Putting value of ΞΈ = γπ‘ππγ^(β1) π₯ π¦=2 (γπ‘ππγ^(β1) π₯) (cosβ‘2ΞΈ " =" (1 β tanβ‘2 π)/(1+ tanβ‘2 π)) Since x = tan ΞΈ β΄ γπ‘ππγ^(β1) x = ΞΈ
