# Ex 5.3, 9 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at April 16, 2024 by Teachoo

Last updated at April 16, 2024 by Teachoo

Ex 5.3, 9 Find ππ¦/ππ₯ in, y = sin^(β1) (2π₯/( 1 + 2π₯2 )) π¦ = sin^(β1) (2π₯/( 1 + 2π₯2 )) Putting x = tan ΞΈ π¦ = sin^(β1) (2π₯/( 1 + π₯2 )) π¦ = sin^(β1) ((2 π‘ππβ‘π)/(1 + γπ‘ππγ^2 π)) π¦ = sin^(β1) ( sinβ‘2ΞΈ) π¦ = 2π Putting value of ΞΈ = γπ‘ππγ^(β1) π₯ π¦ = 2 γπ‘ππγ^(β1) π₯ ("Since " π ππβ‘2π" = " (2 π‘ππβ‘π)/(1 + γπ‘ππγ^2 π)) Since x = tan ΞΈ β΄ γπ‘ππγ^(β1) x = ΞΈ Differentiating both sides π€.π.π‘.π₯ . (π(π¦))/ππ₯ = (π (2 γπ‘ππγ^(β1) π₯" ) " )/ππ₯ ππ¦/ππ₯ = 2 (π (γπ‘ππγ^(β1) π₯" ) " )/ππ₯ ππ¦/ππ₯ = 2 (1/(1+ π₯^2 )) π π/π π = π/(π+γ πγ^π ) ((γπ‘ππγ^(β1) π₯")β = " 1/(1 + π₯^2 ))