Chapter 5 Class 12 Continuity and Differentiability

Class 12
Important Questions for exams Class 12

### Transcript

Ex 5.6, 11 If π₯=β(π^(γπ ππγ^(β1) π‘) ) , y=β(π^(γπππ γ^(β1) π‘) ), show that ππ¦/ππ₯ = β π¦/π₯Here ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) Calculating ππ/ππ π¦ = β(π^(γπππ γ^(β1) π‘) ) ππ¦/ππ‘ " "= π(β(π^(γπππ γ^(β1) π‘) ))/ππ‘ ππ¦/ππ‘ " "= 1/(2β(π^(γπππ γ^(β1) π‘) )) . π(π^(γπππ γ^(β1) π‘) )/ππ‘ ππ¦/ππ‘ " "= 1/(2β(π^(γπππ γ^(β1) π‘) )) . π^(γπππ γ^(β1) π‘). logβ‘π Γ π(γπππ γ^(β1) π‘)/ππ‘ ππ¦/ππ‘ " "= 1/(2β(π^(γπππ γ^(β1) π‘) )) . π^(γπππ γ^(β1) π‘). logβ‘π Γ (β1)/β(1 β π‘^2 ) (π΄π " " π(π^π₯ )/ππ=π^π₯.logβ‘π ) ππ¦/ππ‘ " "= (β π^(γπππ γ^(β1) π‘) ". " logβ‘π" " )/(2β(π^(γπππ γ^(β1) π‘) ) . β(1 β π‘^2 )) Calculating ππ/ππ π₯=β(π^(γπ ππγ^(β1) π‘) ) ππ₯/ππ‘ = π(β(π^(γπ ππγ^(β1) π‘) ))/ππ‘ ππ₯/ππ‘ = 1/(2β(π^(γπ ππγ^(β1) π‘) )) Γ π(π^(γπ ππγ^(β1) π‘) )/ππ‘ ππ₯/ππ‘ = 1/(2β(π^(γπ ππγ^(β1) π‘) )) Γ π^(γπ ππγ^(β1) π‘). logβ‘π . π(γπ ππγ^(β1) π‘)/ππ‘ ππ₯/ππ‘ = 1/(2β(π^(γπ ππγ^(β1) π‘) )) Γ π^(γπ ππγ^(β1) π‘). logβ‘π . 1/β(1 β π‘^2 ) ππ₯/ππ‘ = " " π^(γπ ππγ^(β1) π‘)/(2β(π^(γπ ππγ^(β1) π‘) )) Γ logβ‘π/β(1 β π‘^2 ) ππ₯/ππ‘ = " " (β(π^(γπππγ^(βπ) π) ) . πππβ‘π)/(πβ(π β π^π )) Finding ππ/ππ ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) ππ¦/ππ₯ = ((β β(π^(γπππ γ^(β1) π‘) ) . logβ‘π" " )/(2β(1 β π‘^2 )))/((β(π^(γπ ππγ^(β1) π‘) ) . logβ‘π)/(2β(1 β π‘^2 ))) ππ¦/ππ₯ = (β β(π^(γπππ γ^(β1) π‘) ) . β logβ‘π" " )/(2β(1 β π‘^2 )) Γ (2β(1 β π‘^2 ))/(β(π^(γπ ππγ^(β1) π‘) ) . logβ‘π ) ππ¦/ππ₯ = (ββ(π^(γπππ γ^(β1) π‘) )β‘)/β(π^(γπ ππγ^(β1) π‘) ) Γ (logβ‘π Γ 2β(1 β π‘^2 ))/(logβ‘π Γ 2β(1 β π‘^2 )) ππ¦/ππ₯ = (ββ(π^(γπππ γ^(β1) π‘) )β‘)/β(π^(γπ ππγ^(β1) π‘) ) ππ/ππ = (βπβ‘)/π Hence proved. π΄π  β(π^(γπππ γ^(β1) π‘) )=π¦ β(π^(γπ ππγ^(β1) π‘) )=π₯