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Finding second order derivatives- Implicit form
Finding second order derivatives- Implicit form
Last updated at May 29, 2023 by Teachoo
Ex 5.7, 16 (Method 1) If π^π¦ (x+1)= 1, show that π2π¦/ππ₯2 = (ππ¦/ππ₯)^2 We need to show that π2π¦/ππ₯2 = (ππ¦/ππ₯)^2 π^π¦ (x+1)= 1 Differentiating π€.π.π‘.π₯ π(π^π¦ (x+1))/ππ₯ = (π(1))/ππ₯ π(π^π¦ (x + 1))/ππ₯ = 0 Using product rule in ey(x + 1) As (π’π£)β= π’βπ£ + π£βπ’ where u = ey & v = x + 1 (π(π^π¦))/ππ₯ . (x+1) + (π (x + 1))/ππ₯ . π^π¦ = 0 (π(π^π¦))/ππ₯ Γ ππ¦/ππ¦ (x+1) + ((π(π₯))/ππ₯ + (π(1))/ππ₯) . π^π¦ = 0 (π(π^π¦))/ππ¦ Γ ππ¦/ππ₯ (x+1) + (1+0) . π^π¦ = 0 π^π¦ Γ ππ¦/ππ₯ (x+1) + π^π¦ = 0 π^π¦ (ππ¦/ππ₯) (x+1) = β π^π¦ ππ¦/ππ₯ = ("β " π^π¦)/(π^π¦ (π₯ + 1)) ππ¦/ππ₯ = ("β " 1)/((π₯ + 1)) Again Differentiating π€.π.π‘.π₯ π/ππ₯ (ππ¦/ππ₯) = π/ππ₯ (("β " 1)/((π₯+1) )) (π^2 π¦)/(ππ₯^2 ) = β[((π(1))/ππ₯ . (π₯ + 1) β π(π₯ + 1)/ππ₯ . 1)/γ(π₯ + 1)γ^2 ] using Quotient Rule As, (π’/π£)^β²= (π’βπ£ β π£βπ’)/π£^2 where U = 1 & V = x + 1 = β[(0 . (π₯+1) β π(π₯+1)/ππ₯ . 1)/γ(π₯ + 1)γ^2 ] = β[(0 β (1 + 0) . 1)/γ(π₯ + 1)γ^2 ] = β[(β1)/γ(π₯ + 1)γ^2 ] = 1/γ(π₯ + 1)γ^2 Hence (π^2 π¦)/(ππ₯^2 ) = 1/γ(π₯ + 1)γ^2 = ((β1)/( π₯ + 1))^2 = (ππ¦/ππ₯)^2 Hence proved Ex 5.7, 16 (Method 2) If π¦= π^π¦ (x+1)= 1, show that π2π¦/ππ₯2 = (ππ¦/ππ₯)^2 If π¦= π^π¦ (x+1)= 1 We need to show that π2π¦/ππ₯2 = (ππ¦/ππ₯)^2 π^π¦ (π₯+1)= 1 Differentiating π€.π.π‘.π₯ π(π^π¦ (x + 1))/ππ₯ = (π(1))/ππ₯ π(π^π¦ (x + 1))/ππ₯ = 0 Using product rule in ey(x + 1) As (π’π£)β= π’βπ£ + π£βπ’ where u = ey & v = x + 1 (π(π^π¦))/ππ₯ . (x+1) + (π (x + 1))/ππ₯ . π^π¦ = 0 (π(π^π¦))/ππ₯ Γ ππ¦/ππ¦ (x+1) + ((π(π₯))/ππ₯ + (π(1))/ππ₯) . π^π¦ = 0 (π(π^π¦))/ππ¦ Γ ππ¦/ππ₯ (x+1) + (1+0) . π^π¦ = 0 π^π¦ Γ ππ¦/ππ₯ (x+1) + π^π¦ = 0 π^π¦ (ππ¦/ππ₯) (x+1) = β π^π¦ ππ¦/ππ₯ = ("β " π^π¦)/(π^π¦ (π₯ + 1)) ππ¦/ππ₯ = ("β " 1)/((π₯ + 1)) Given, π^π¦ (x + 1) = 1 π^π = π/(π + π) Putting (2) in (1) ππ¦/ππ₯ = ("β " 1)/((π₯ + 1)) ππ¦/ππ₯ = β π^π¦ β¦(1) β¦(2) Again Differentiating π€.π.π‘.π₯ π/ππ₯ (ππ¦/ππ₯) = (π("β" π^π¦))/ππ₯ (π^2 π¦)/(ππ₯^2 ) = β (π(π^π¦))/ππ₯ (π^2 π¦)/(ππ₯^2 ) = β (π(π^π¦))/ππ₯ Γ ππ¦/ππ¦ (π^2 π¦)/(ππ₯^2 ) = β (π(π^π¦))/ππ¦ Γ ππ¦/ππ₯ (π^2 π¦)/(ππ₯^2 ) = "β " π^πΓ ππ¦/ππ₯ (π^2 π¦)/(ππ₯^2 ) = π π/π π Γ ππ¦/ππ₯ (π^2 π¦)/(ππ₯^2 ) = ππ¦/ππ₯ Γ ππ¦/ππ₯ (From (1) "β " π^π¦ " = " ππ¦/ππ₯) (π^2 π¦)/(ππ₯^2 ) = (ππ¦/ππ₯)^2 Hence proved