Chapter 5 Class 12 Continuity and Differentiability

Class 12
Important Questions for exams Class 12

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Misc 23 If π¦=π^(γπ πππ γ^(β1) π₯) , β 1 β€ π₯ β€ 1, show that (1βπ₯^2 ) (π^2 π¦)/γππ₯γ^2 βπ₯ ππ¦/ππ₯ β π2 π¦ =0 . π¦=π^(γπ πππ γ^(β1) π₯) Differentiating π€.π.π‘.π₯. ππ¦/ππ₯ = π(π^(γπ πππ γ^(β1) π₯" " ) )/ππ₯ ππ¦/ππ₯ = π^(γπ πππ γ^(β1) π₯" " ) Γ π(γπ πππ γ^(β1) π₯)/ππ₯ ππ¦/ππ₯ = π^(γπ πππ γ^(β1) π₯" " ) Γ π ((β1)/β(1 β π₯^2 )) ππ¦/ππ₯ = (βπ π^(γπ πππ γ^(β1) π₯" " ))/β(1 β π₯^2 ) β(1 β π₯^2 ) ππ¦/ππ₯ = βππ^(γπ πππ γ^(β1) π₯" " ) β(1 β π₯^2 ) ππ¦/ππ₯ = βππ¦ Since we need to prove (1βπ₯^2 ) (π^2 π¦)/γππ₯γ^2 β π₯ ππ¦/ππ₯ βπ2 π¦ =0 Squaring (1) both sides (β(1 β π₯^2 ) ππ¦/ππ₯)^2 = (βππ¦)^2 (1βπ₯^2 ) (π¦^β² )^2 = π^2 π¦^2 Differentiating again w.r.t x π((1 β π₯^2 ) (π¦^β² )^2 )/ππ₯ = (d(π^2 π¦^2))/ππ₯ π((1 β π₯^2 ) (π¦^β² )^2 )/ππ₯ = π^2 (π(π¦^2))/ππ₯ π((1 β π₯^2 ) (π¦^β² )^2 )/ππ₯ = π^2 Γ 2π¦ Γππ¦/ππ₯ π(1 β π₯^2 )/ππ₯ (π¦^β² )^2+(1 β π₯^2 ) π((π^β² )^π )/ππ = π^2 Γ 2π¦π¦^β² (β2π₯)(π¦^β² )^2+(1 β π₯^2 )(ππ^β² Γ π(π^β² )/ππ) = π^2 Γ 2π¦π¦^β² (β2π₯)(π¦^β² )^2+(1 β π₯^2 )(ππ^β² Γ π^β²β² ) = π^2 Γ 2π¦π¦^β² Dividing both sides by ππ^β² βπ₯π¦^β²+(1 β π₯^2 ) π¦^β²β² = π^2 Γ π¦ βπ₯π¦^β²+(1 β π₯^2 ) π¦^β²β² = π^2 π¦ (π β π^π ) π^β²β²βππ^β²βπ^π π=π Hence proved