Finding second order derivatives- Implicit form

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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Misc 15 If (π₯ β π)^2+ (π¦ β π)^2= π2, for some π > 0, prove that γ[1 + (ππ¦/ππ₯)^2 ]/((π^2 π¦)/γππ₯γ^2 )γ^(3/2)is a constant independent of a and b.First we will calculate ππ¦/ππ₯ (π₯ β π)^2+ (π¦ β π)^2= π2 Differentiating π€.π.π‘.π₯. π((π₯ β π)^2+ (π¦ β π)^2 )/ππ₯ = π(π^2 )/ππ₯ π((π₯ β π)^2 )/ππ₯ +" " π((π¦ β π)^2 )/ππ₯ = 0 2(π₯ β π). π(π₯ β π)/ππ₯ + 2 (π¦ β π). π(π¦ β π)/ππ₯ = 0 2 (π₯ β π) (1 β0) + 2(π¦ β π) . (ππ¦/ππ₯ β0) = 0 2 (π₯ β π) + 2(π¦ β π) . (ππ¦/ππ₯) = 0 2(π¦ β π) . ππ¦/ππ₯ = β2 (π₯ β π) ππ¦/ππ₯ = (β2 (π₯ β π))/2(π¦ β π) ππ/ππ = (β(π β π))/(π β π) Again Differentiating π€.π.π‘.π₯. π/ππ₯ (ππ¦/ππ₯) = π/ππ₯ ((β(π₯ β π))/(π¦ β π)) (π^2 π¦)/(ππ₯^2 ) = β π/ππ₯ ((π₯ β π)/(π¦ β π)) (π^2 π¦)/(ππ₯^2 )= β ((π(π₯ β π)/ππ₯ (π¦ β π) β π(π¦ β π)/ππ₯ . (π₯ β π))/(π¦ β π)^2 ) (π^2 π¦)/(ππ₯^2 ) = β (((1 β 0) (π¦ β π) β (ππ¦/ππ₯ β 0)(π₯ β π))/(π¦ β π)^2 ) Using Quotient rule As (π’/π£)β² = (π’^β² π£ β π£^β² π’)/π£^2 where u = x β π & v = y β b (π^2 π¦)/(ππ₯^2 ) = β (((π¦ β π) β (ππ¦/ππ₯)(π₯ β π))/(π¦ β π)^2 ) (π^2 π¦)/(ππ₯^2 ) = β (((π¦ β π) β (β (π₯ β π))/((π¦ β π) ) (π₯ β π))/(π¦ β π)^2 ) (π^2 π¦)/(ππ₯^2 )= β (((π¦ β π)^2 + (π₯ β π)^2)/((π¦ β π)^2 (π¦ β π) )) (π^π π)/(ππ^π )= (βπ^π)/(π β π)^π Now, finding value of γ[π+ (ππ/ππ)^π ]/((π^π π)/γππγ^π )γ^(π/π) (Given (π₯ β π)^2+ (π¦ β π)^2= π2) γ[π+ (ππ/ππ)^π ]/((π^π π)/γππγ^π )γ^(π/π) Putting values = γ[1+ ((β(π₯ β π))/(π¦ β π))^2 ]/((βπ^2)/(π¦ β π)^3 )γ^(3/2) = β γ[((π¦ β π)^2 + (π₯ β π)^2)/(π¦ β π)^2 ]/(π^2/(π¦ β π)^3 )γ^(3/2) = β γ[π^2/(π¦ β π)^2 ]/(π^2/(π¦ β π)^3 )γ^(3/2) = β [π^2/(π¦ β π)^2 ]^(3/2) Γ (π¦ β π)^3/π^2 = β (π/(π¦ β π))^(2 Γ 3/2) Γ (π¦ β π)^3/π^2 "= β" (π/(π¦ β π))^3 " Γ " (π¦ β π)^3/π^2 "= β" π^3/π^2 Γ (π¦ β π)^3/(π¦ β π)^3 = βπ = Constant Which is constant independent of a & b Hence proved