Finding second order derivatives- Implicit form

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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### Transcript

Ex 5.7, 15 If π¦= γ500πγ^7π₯+ γ600πγ^(β7π₯), show that π2π¦/ππ₯2 = 49π¦ π¦= γ500πγ^7π₯+ γ600πγ^(β7π₯) Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = (π(γ500πγ^7π₯ " +" γ600πγ^(β7π₯) " " ))/ππ₯ ππ¦/ππ₯ = (π (γ500πγ^7π₯))/ππ₯ + (π (γ600πγ^(β7π₯)))/ππ₯ ππ¦/ππ₯ = 500 (π (π^7π₯))/ππ₯ + 600 (π (π^(β7π₯)))/ππ₯ ππ¦/ππ₯ = 500 . π^7π₯. (π (7π₯))/ππ₯ + 600 . π^(β7π₯) . (π (β7π₯))/ππ₯ ππ¦/ππ₯ = 500 . π^7π₯. 7 + 600 . π^(β7π₯) . (β7) ππ¦/ππ₯ = 500 . 7 . π^7π₯ β 600 . 7 . π^(β7π₯) Again Differentiating π€.π.π‘.π₯ π/ππ₯ (ππ¦/ππ₯) = π(500 . 7 . π^7π₯ " β" γ 600 . 7 . πγ^(β7π₯) )" " /ππ₯ (π^2 π¦)/(ππ₯^2 ) = π(500 Γ 7 . π^7π₯ )/ππ₯ β π(γ600 Γ 7πγ^(β7π₯) )" " /ππ₯ (π^2 π¦)/(ππ₯^2 ) = 500 Γ 7 π(π^7π₯ )/ππ₯ β 600 Γ 7 π(π^(β7π₯) )/ππ₯ = 500 Γ 7 Γ π^7π₯ π(7π₯)/ππ₯ β 600 Γ 7 Γ π^(β7π₯) π(β7π₯)/ππ₯ = 500 Γ 7 Γ π^7π₯. 7 β 600 Γ 7 Γ π^(β7π₯) (β7) = 500 Γ 7 Γ 7π^7π₯ + 600 Γ 7 Γ 7 Γ π^(β7π₯) = 7 Γ 7 (500γ πγ^7π₯+600γ πγ^(β7π₯) ) = 49 (500γ πγ^7π₯+600γ πγ^(β7π₯) ) = 49 π¦ Hence proved . (As π¦= γ500πγ^7π₯+ γ600πγ^(β7π₯))