Ex 5.3

Chapter 5 Class 12 Continuity and Differentiability
Serial order 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 ))

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.