
Derivatives in parametric form
Derivatives in parametric form
Last updated at April 13, 2021 by Teachoo
Example 35 Find ππ¦/ππ₯ , if π₯ = ππ‘2, π¦ = 2ππ‘ . ππ¦/ππ₯ = ππ¦/ππ₯ Γ ππ‘/ππ‘ ππ¦/ππ₯ = ππ¦/ππ‘ Γ ππ‘/ππ₯ ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) Calculating π π/π π π¦ = 2ππ‘" " ππ¦/ππ‘ " " = π(2ππ‘)/ππ‘ ππ¦/ππ‘ = 2π . π(π‘)/ππ‘ ππ¦/ππ‘ = ππ Calculating π π/π π π₯=ππ‘2 ππ₯/ππ‘ " " = π(ππ‘2)/ππ‘ ππ₯/ππ‘ " " = π . π(π‘2)/ππ‘ " " ππ₯/ππ‘ " " = πππ Therefore ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) ππ¦/ππ₯ = 2π/2aπ‘ π π/π π = π/π