Ex 5.6, 11 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at April 16, 2024 by Teachoo
Derivatives in parametric form
Derivatives in parametric form
Last updated at April 16, 2024 by Teachoo
Ex 5.6, 11 If π₯=β(π^(γπ ππγ^(β1) π‘) ) , y=β(π^(γπππ γ^(β1) π‘) ), show that ππ¦/ππ₯ = β π¦/π₯Here ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) Calculating π π/π π π¦ = β(π^(γπππ γ^(β1) π‘) ) ππ¦/ππ‘ " "= π(β(π^(γπππ γ^(β1) π‘) ))/ππ‘ ππ¦/ππ‘ " "= 1/(2β(π^(γπππ γ^(β1) π‘) )) . π(π^(γπππ γ^(β1) π‘) )/ππ‘ ππ¦/ππ‘ " "= 1/(2β(π^(γπππ γ^(β1) π‘) )) . π^(γπππ γ^(β1) π‘). logβ‘π Γ π(γπππ γ^(β1) π‘)/ππ‘ ππ¦/ππ‘ " "= 1/(2β(π^(γπππ γ^(β1) π‘) )) . π^(γπππ γ^(β1) π‘). logβ‘π Γ (β1)/β(1 β π‘^2 ) (π΄π " " π(π^π₯ )/ππ=π^π₯.logβ‘π ) ππ¦/ππ‘ " "= (β π^(γπππ γ^(β1) π‘) ". " logβ‘π" " )/(2β(π^(γπππ γ^(β1) π‘) ) . β(1 β π‘^2 )) Calculating π π/π π π₯=β(π^(γπ ππγ^(β1) π‘) ) ππ₯/ππ‘ = π(β(π^(γπ ππγ^(β1) π‘) ))/ππ‘ ππ₯/ππ‘ = 1/(2β(π^(γπ ππγ^(β1) π‘) )) Γ π(π^(γπ ππγ^(β1) π‘) )/ππ‘ ππ₯/ππ‘ = 1/(2β(π^(γπ ππγ^(β1) π‘) )) Γ π^(γπ ππγ^(β1) π‘). logβ‘π . π(γπ ππγ^(β1) π‘)/ππ‘ ππ₯/ππ‘ = 1/(2β(π^(γπ ππγ^(β1) π‘) )) Γ π^(γπ ππγ^(β1) π‘). logβ‘π . 1/β(1 β π‘^2 ) ππ₯/ππ‘ = " " π^(γπ ππγ^(β1) π‘)/(2β(π^(γπ ππγ^(β1) π‘) )) Γ logβ‘π/β(1 β π‘^2 ) ππ₯/ππ‘ = " " (β(π^(γπππγ^(βπ) π) ) . πππβ‘π)/(πβ(π β π^π )) Finding π π/π π ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) ππ¦/ππ₯ = ((β β(π^(γπππ γ^(β1) π‘) ) . logβ‘π" " )/(2β(1 β π‘^2 )))/((β(π^(γπ ππγ^(β1) π‘) ) . logβ‘π)/(2β(1 β π‘^2 ))) ππ¦/ππ₯ = (β β(π^(γπππ γ^(β1) π‘) ) . β logβ‘π" " )/(2β(1 β π‘^2 )) Γ (2β(1 β π‘^2 ))/(β(π^(γπ ππγ^(β1) π‘) ) . logβ‘π ) ππ¦/ππ₯ = (ββ(π^(γπππ γ^(β1) π‘) )β‘)/β(π^(γπ ππγ^(β1) π‘) ) Γ (logβ‘π Γ 2β(1 β π‘^2 ))/(logβ‘π Γ 2β(1 β π‘^2 )) ππ¦/ππ₯ = (ββ(π^(γπππ γ^(β1) π‘) )β‘)/β(π^(γπ ππγ^(β1) π‘) ) π π/π π = (βπβ‘)/π Hence proved. π΄π β(π^(γπππ γ^(β1) π‘) )=π¦ β(π^(γπ ππγ^(β1) π‘) )=π₯