Partial Differentiation | Multivariable Calculus | u = tan⁻¹(y/x), ∂²u/∂y∂x = ∂²u/∂x∂y

📘 Calculus & Multivariable Calculus | Partial Differentiation Problem Solution Lecture: Mixed Partial Derivatives 📝 Problem Statement: If u = tan⁻¹(y/x), then show that ∂²u/∂y∂x = ∂²u/∂x∂y. 🎯 Concepts Used: ✔️ Partial Differentiation ✔️ Mixed/Successive Partial Derivatives ✔️ Symmetry of Mixed Derivatives (Clairaut’s Theorem) ✔️ Multivariable Calculus ✨ This problem verifies that mixed partial derivatives are equal when the function is continuous and differentiable — a key concept in calculus and PDEs. ✅ Very useful for engineering mathematics, multivariable calculus, and VTU exam preparation. 💎 Support Us: Join this channel to support us & access exclusive perks 👇 🔗 / @officialmathematicstutor #Calculus #PartialDifferentiation #MultivariableCalculus #Laplacian #EngineeringMathematics #VTU #PDE #ChainRule #BMATS201 #BMATE201 #BMATC201 #BMATM201 #1BMATS101 #1BMATE101 #1BMATC101 #1BMATM101 #Module1 #Module2

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