Finding derivative of Inverse trigonometric functions

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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### Transcript

Ex 5.3, 15 Find ππ¦/ππ₯ in, y = secβ1 (1/( 2π₯2β1 )), 0 < x < 1/β2 y = secβ1 (1/( 2π₯^2 β 1 )) πππβ‘π = 1/(2π₯^2 β 1) π/ππ¨π¬β‘π = 1/(2π₯^2 β 1) cosβ‘π¦ = 2π₯2β1 y = cos β1 (2π₯2β1) Putting π₯ = cosβ‘ΞΈ π¦ = cos β1 (2πππ 2πβ1) π¦ = cos β1 (cosβ‘2 π) π¦ = 2π Putting value of ΞΈ = cosβ1 x π¦ = 2 γπππ γ^(β1) π₯ Differentiating both sides π€.π.π‘.π₯ . (π(π¦))/ππ₯ = (π (2γπππ γ^(β1) π₯" " ))/ππ₯ ππ¦/ππ₯ = 2 (π (γπππ γ^(β1) π₯" " ))/ππ₯ ππ¦/ππ₯ = 2 . ((β1)/β(1 β π₯^2 )) ππ/ππ = (βπ)/β(π β π^π ) (πππ β‘2π " = 2 " γπππ γ^2 πβ1) Since x = cos ΞΈ β΄ γπππ γ^(β1) x = ΞΈ ((γπππ γ^(β1) π₯")β = " (β1)/β(1 β π₯^2 ))

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#### Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.