Finding derivative of Inverse trigonometric functions

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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### Transcript

Ex 5.3, 10 Find ππ¦/ππ₯ in, π¦ = tanβ1 ((3π₯β π₯^3)/( 1β 3π₯2 )) , β 1/β3 < π₯ < 1/β3 π¦ = tanβ1 ((3π₯β π₯^3)/( 1β 3π₯2 )) Putting x = tan ΞΈ y = γπ‘ππγ^(β1) ((3 tanβ‘γπ β tanβ‘3 πγ)/(1 β 3 tanβ‘2 π)) y = γπ‘ππγ^(β1) (tanβ‘γ3 πγ) π¦ = 3π Putting value of ΞΈ = γπ‘ππγ^(β1) π₯ π¦ = 3γ (π‘ππγ^(β1) π₯) (π‘ππβ‘3ΞΈ " = " (3 π‘ππβ‘γπβπ‘ππβ‘3 πγ)/(1β3 π‘ππβ‘2 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 3γ(π‘ππγ^(β1) π₯") " )/ππ₯ ππ¦/ππ₯ =3 (πγ(π‘ππγ^(β1) π₯") " )/ππ₯ ππ¦/ππ₯ = 3 (1/(1 + π₯^2 )) ππ/ππ = π/(π +γ πγ^π ) ((γπ‘ππγ^(β1) π₯")β = " 1/(1 + π₯^2 ))