NCERT Class 12 · Ch 5 Exercise 5.3 · Q2
NCERT Class 12 Maths · Chapter 5 Continuity And Differentiability · Exercise 5.3 Question 2
Find dy/dx for \(2 x + 3 y - \sin{\left(y \right)} = 0\).
💡 Concept Hints
- Identify the domain and the inner and outer functions, or the relevant one-sided limits.
- Apply the appropriate differentiation or limit rule, keep all chain factors, and check exceptional points.
✅ Step-by-Step Solution
Final Answer: \(\frac{2}{\cos{\left(y \right)} - 3}\) where \(3 - \cos{\left(y \right)}\)≠0.
- Differentiate the relation treating y as a function of x. The partial derivatives are Fₓ=\(2\) and Fᵧ=\(3 - \cos{\left(y \right)}\).
- The chain rule gives Fₓ+Fᵧy′=0. Thus \(3 - \cos{\left(y \right)}\)y′ = \(-2\).
- At points where Fᵧ≠0, divide to obtain y′=\(\frac{2}{\cos{\left(y \right)} - 3}\). Points with Fᵧ=0 require a separate local-curve check.
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