# Ex 5.3, 2 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at March 10, 2021 by Teachoo

Last updated at March 10, 2021 by Teachoo

Transcript

Ex 5.3, 2 Find ππ¦/ππ₯ in , 2π₯ + 3π¦ = sinβ‘π¦ . 2π₯ + 3π¦ = sinβ‘π¦ Differentiating both sides π€.π.π‘.π₯ π(2π₯ + 3π¦)/ππ₯ = (π (sinβ‘π¦ ))/ππ₯ π(2π₯)/ππ₯ +π(3π¦)/ππ₯= (π (sinβ‘π¦ ))/ππ₯ 2 ππ₯/ππ₯ +3π(π¦)/ππ₯= (π (sinβ‘π¦ ))/ππ¦ Γ ππ¦/ππ₯ 2 + 3 ππ¦/ππ₯ = cosβ‘π¦ "Γ " ππ¦/ππ₯ cosβ‘π¦ "Γ " ππ¦/ππ₯ β 3 ππ¦/ππ₯ = 2 (Derivative of π ππβ‘π₯ is πππ β‘π₯) ππ¦/ππ₯ (cos y β 3) = 2 π π/π π = π/((πππβ‘γπβπ)γ )

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.