Ex 5.3, 10 Find ππ¦/ππ₯ in, π¦ = tanβ1 ((3π₯β π₯^3)/( 1β 3π₯2 )) , β 1/β3 < π₯ < 1/β3 π¦ = tanβ1 ((3π₯β π₯^3)/( 1β 3π₯2 ))
Putting x = tan ΞΈ
y = γπ‘ππγ^(β1) ((3 tanβ‘γπ β tanβ‘3 πγ)/(1 β 3 tanβ‘2 π))
y = γπ‘ππγ^(β1) (tanβ‘γ3 πγ)
π¦ = 3π
Putting value of ΞΈ = γπ‘ππγ^(β1) π₯
π¦ = 3γ (π‘ππγ^(β1) π₯)
(π‘ππβ‘3ΞΈ " = " (3 π‘ππβ‘γπβπ‘ππβ‘3 πγ)/(1β3 π‘ππβ‘2 π))
Since x = tan ΞΈ
β΄ γπ‘ππγ^(β1) x = ΞΈ
Differentiating both sides π€.π.π‘.π₯ .
(π(π¦))/ππ₯ = (π 3γ(π‘ππγ^(β1) π₯") " )/ππ₯
ππ¦/ππ₯ =3 (πγ(π‘ππγ^(β1) π₯") " )/ππ₯
ππ¦/ππ₯ = 3 (1/(1 + π₯^2 ))
π π/π π = π/(π +γ πγ^π )
((γπ‘ππγ^(β1) π₯")β = " 1/(1 + 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.