Mathbee CBSE Mathbee · CBSE Class 12 Maths
NCERT Class 12 · Ch 5 Exercise 5.3 · Q3

NCERT Class 12 Maths · Chapter 5 Continuity And Differentiability · Exercise 5.3 Question 3

Find dy/dx for \(a x + b y^{2} - \cos{\left(y \right)} = 0\).

💡 Concept Hints

  1. Identify the domain and the inner and outer functions, or the relevant one-sided limits.
  2. Apply the appropriate differentiation or limit rule, keep all chain factors, and check exceptional points.

✅ Step-by-Step Solution

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

Master Chapter 5: Continuity And Differentiability

Practise interactive questions, flashcards, visual mind maps, and timed board tests for Continuity And Differentiability.
📘 Chapter Notes & All Exercises 🚀 Open Interactive Workspace →
← Previous Question
Find dy/dx for \(2 x + 3 y - \sin{\left(y \right)} = 0\).
Next Question →
Find dy/dx for \(x y + y^{2} - y - \tan{\left(x \right)} = 0\).

All Questions in Continuity And Differentiability — Exercise 5.3

Related Questions in Continuity And Differentiability