Finding derivative of Inverse trigonometric functions

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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### Transcript

Ex 5.3, 13 Find ππ¦/ππ₯ in, y = cosβ1 (2π₯/( 1+ π₯2 )) , β1 < x < 1 π¦ = cosβ1 (2π₯/( 1+ π₯2 )) Let π₯ = tanβ‘π π¦ = cosβ1 ((2 tanβ‘π)/( 1 + π‘ππ2π )) π¦ = cosβ1 (sin 2ΞΈ) π¦ ="cosβ1" (γcos γβ‘(π/2 β2π) ) π¦ = π/2 β 2π Putting value of ΞΈ = tanβ1 x π¦ = π/2 β 2 γπ‘ππγ^(β1) π₯ Since x = tan ΞΈ β΄ γπ‘ππγ^(β1) x = ΞΈ Differentiating both sides π€.π.π‘.π₯ (π(π¦))/ππ₯ = (π (" " π/2 " β " γ2π‘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 ) ((γπ‘ππγ^(β1) π₯") β = " 1/(1 + π₯^2 ))