NCERT Class 12 Maths · Chapter 5 Continuity And Differentiability · Exercise 5.3 Question 7

✅ Step-by-Step Solution

1. Differentiate the relation treating y as a function of x. The partial derivatives are Fₓ=\(- y \sin{\left(x y \right)}\) and Fᵧ=\(- x \sin{\left(x y \right)} + 2 \sin{\left(y \right)} \cos{\left(y \right)}\).
2. The chain rule gives Fₓ+Fᵧy′=0. Thus \(- x \sin{\left(x y \right)} + 2 \sin{\left(y \right)} \cos{\left(y \right)}\)y′ = \(y \sin{\left(x y \right)}\).
3. At points where Fᵧ≠0, divide to obtain y′=\(- \frac{y \sin{\left(x y \right)}}{x \sin{\left(x y \right)} - \sin{\left(2 y \right)}}\). Points with Fᵧ=0 require a separate local-curve check.