Finding second order derivatives- Implicit form

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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Ex 5.7, 17 (Method 1) If π¦= γ(γπ‘ππγ^(β1) π₯)γ^(2 ), show that γ(π₯^2+1)γ^(2 ) π¦2 + 2π₯ γ(π₯^2+1)γ^ π¦1 = 2 We have y = γ(γπ‘ππγ^(β1) π₯)γ^(2 ) Differentiating π€.π.π‘.π₯ yβ = 2 tanβ1 π₯ Γ 1/(1 + π₯^2 ) (1 + π₯^2) yβ = 2 tanβ1 π₯ Again differentiating π€.π.π‘.π₯ [π¦^β² (1+π₯^2 )]^β² = 2 Γ 1/(1 +γ π₯γ^2 ) [π¦^β² (1+π₯^2 )]^β² = 2/(1 + π₯^2 ) (tan^(β1)β‘π₯ )^β²=1/(1+π₯^2 ) Using product rule (1+π₯^2 )^β² + π¦^β²β² (1 + π₯^2) = 2/(1 + π₯^2 ) 2π₯ π¦^β²+π¦^β²β² (1 +π₯^2 ) = 2/(1 + π₯^2 ) 2π₯ π¦^β² (1 +π₯^2 )+π¦^β²β² (1 +π₯^2 )Γ(1 +π₯^2 ) = 2 2π₯(1 +π₯^2 ) π¦^β²+π¦^β²β² (1 +π₯^2 )^2 = 2 π^β²β² (π +π^π )^π+ππ(π +π^π ) π^β² = 2 Hence Proved Ex 5.7, 17 (Method 2) If π¦= γ(γπ‘ππγ^(β1) π₯)γ^(2 ), show that γ(π₯^2+1)γ^(2 ) π¦2 + 2π₯ γ(π₯^2+1)γ^ π¦1 = 2 We have y = γ(γπ‘ππγ^(β1) π₯)γ^(2 ) Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = (π (γ(γπ‘ππγ^(β1) π₯)γ^(2 )))/ππ₯ ππ¦/ππ₯ = 2 γπ‘ππγ^(β1) π₯ . π(γπ‘ππγ^(β1) π₯)/ππ₯ ππ¦/ππ₯ = 2 γπ‘ππγ^(β1) π₯ . 1/(1 +γ π₯γ^2 ) Hence, π¦1 = ππ¦/ππ₯ = (2 γπ‘ππγ^(β1) π₯)/(1 +γ π₯γ^2 ) Again Differentiating π€.π.π‘.π₯ π/ππ₯ (ππ¦/ππ₯) = π/ππ₯ ((2 γπ‘ππγ^(β1) π₯)/(1 +γ π₯γ^2 )) (π^2 π¦)/(ππ₯^2 ) = 2 π/ππ₯ ((γπ‘ππγ^(β1) π₯)/(1 +γ π₯γ^2 )) (π^2 π¦)/(ππ₯^2 ) = 2 [(π(γπ‘ππγ^(β1) π₯)/ππ₯ . (1 +γ π₯γ^2 ) β π(1 +γ π₯γ^2 )/ππ₯ .γ π‘ππγ^(β1) π₯)/(1 +γ π₯γ^2 )^2 ] (π^2 π¦)/(ππ₯^2 ) = 2 [(1/(1 +γ π₯γ^2 ) . (1 +γ π₯γ^2 ) β (π(1)/ππ₯ + π(π₯^2 )/ππ₯) . γ π‘ππγ^(β1) π₯)/(1 +γ π₯γ^2 )^2 ] Using quotient Rule As, (π’/π£)^β²= (π’βπ£ β π£βπ’)/π£^2 where u = tan-1 x & v = 1 + x2 (π^2 π¦)/(ππ₯^2 ) = 2 [( 1 β (0 + 2π₯) γ π‘ππγ^(β1) π₯)/(1 +γ π₯γ^2 )^2 ] (π^2 π¦)/(ππ₯^2 ) = 2 [( 1 β 2π₯ .γ π‘ππγ^(β1) π₯)/(1 +γ π₯γ^2 )^2 ] Thus, π2 = 2 [( π β ππ .γ πππγ^(βπ) π)/(π +γ πγ^π )^π ] We need to show γ(π₯^2+1)γ^(2 ) π¦2 + 2π₯ γ(π₯^2+1)γ^ π¦1 = 2 Solving LHS γ(π₯^2+1)γ^(2 ) π¦2 + 2π₯ γ(π₯^2+1)γ^ π¦1 = γ(π₯^2+1)γ^(2 ). 2 [( 1 β 2π₯ .γ π‘ππγ^(β1) π₯)/(1 +γ π₯γ^2 )^2 ] + 2π₯ (π₯2+1) . ((2 γπ‘ππγ^(β1) π₯)/(1 +γ π₯γ^2 )) = 2 (1β2π₯ γπ‘ππγ^(β1) π₯) + 4π₯ γπ‘ππγ^(β1) π₯ = 2 β 4x γπππγ^(βπ) x + 4x γπππγ^(βπ) x = 2 Hence proved .