# Ex 5.7, 12 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at March 11, 2021 by

Last updated at March 11, 2021 by

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Ex 5.7, 12 If y= γπππ γ^(β1) π₯ , Find π2π¦/ππ₯2 in terms of π¦ alone.Let y = γπππ γ^(β1) π₯ Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = (π(γπππ γ^(β1) π₯))/ππ₯ ππ¦/ππ₯ = (β1)/β(1 β π₯^2 ) Again Differentiating π€.π.π‘.π₯ π/ππ₯ (ππ¦/ππ₯) = π/ππ₯ ((β1)/β(1 β π₯^2 )) ("As" π(γπππ γ^(β1) π₯)/ππ₯=(β1)/β(1 β π₯^2 )) (π^2 π¦)/(ππ₯^2 ) = π/ππ₯ ((β1)/β(1 β π₯^2 )) (π^2 π¦)/(ππ₯^2 ) = βπ/ππ₯ (1βπ₯^2 )^((β1)/2) (π^2 π¦)/(ππ₯^2 ) = β((β1)/2) (1βπ₯^2 )^((β1)/2 β 1)Γ (1βπ₯^2 )^β² (π^2 π¦)/(ππ₯^2 ) = 1/2 (1βπ₯^2 )^((β3)/2) Γ (β2π₯) (π^2 π¦)/(ππ₯^2 ) = βπ₯(1βπ₯^2 )^((β3)/2) (π ^π π)/(π π^π ) = (β π)/( (π β π^π )^(π/π ) ) But we need to calculate (π^2 π¦)/(ππ₯^2 ) in terms of y . y = γπππ γ^(β1) π₯ cosβ‘π¦ = π₯ π₯ = cosβ‘π¦ Putting π₯ = cosβ‘π¦ in equation (π^2 π¦)/(ππ₯^2 ) = (β π₯)/( (1 β π₯^2 )^(3/2 ) ) (π^2 π¦)/(ππ₯^2 ) = (βcosβ‘π¦ " " )/( (1 β γ(cosβ‘π¦)γ^2 )^(3/2 ) ) = (β cosβ‘π¦ " " )/( γ(sin2β‘γπ¦) γγ^(3/2 ) ) (As 1βcosβ‘γ2 γ π = sinβ‘2 π) = (β cosβ‘π¦ " " )/( γ(sinβ‘γπ¦) γγ^(2 Γ 3/2 ) ) = (β πππ β‘π¦)/( γ(π ππβ‘γπ¦) γγ^3 ) = (β πππ β‘π¦)/sinβ‘π¦ Γ1/( γ(π ππβ‘π¦)γ^2 ) = βππ¨πβ‘π πππππ^π π

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.