Derivatives in parametric form

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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### Transcript

Ex 5.6, 3 If x and y are connected parametrically by the equations without eliminating the parameter, Find ππ¦/ππ₯, π₯=sinβ‘π‘, π¦=cosβ‘2π‘Here, ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) Calculating ππ/ππ ππ¦/ππ‘ " " = π(cosβ‘2π‘)/ππ‘ ππ¦/ππ‘ = βsinβ‘2π‘ . 2 ππ¦/ππ‘ = β2 sinβ‘2π‘ Calculating ππ/ππ ππ₯/ππ‘ " " = π(sinβ‘π‘ )/ππ‘ ππ₯/ππ‘ " " = cosβ‘π‘ Therefore ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) ππ¦/ππ₯ = (β2 sinβ‘2π‘" " )/cosβ‘π‘ ππ¦/ππ₯ = (β2(2 sinβ‘π‘ .γ cosγβ‘π‘ ))/cosβ‘π‘ ππ/ππ = βπ πππβ‘π (As sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ)

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.