# Ex 5.3, 2 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at April 16, 2024 by Teachoo

Last updated at April 16, 2024 by Teachoo

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 )