NCERT Class 12 · Ch 5 Exercise 5.3 · Q4
NCERT Class 12 Maths · Chapter 5 Continuity And Differentiability · Exercise 5.3 Question 4
Find dy/dx for \(x y + y^{2} - y - \tan{\left(x \right)} = 0\).
💡 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: \(- \frac{y \cos^{2}{\left(x \right)} - 1}{\left(x + 2 y - 1\right) \cos^{2}{\left(x \right)}}\) where \(x + 2 y - 1\)≠0.
- Differentiate the relation treating y as a function of x. The partial derivatives are Fₓ=\(y - \tan^{2}{\left(x \right)} - 1\) and Fᵧ=\(x + 2 y - 1\).
- The chain rule gives Fₓ+Fᵧy′=0. Thus \(x + 2 y - 1\)y′ = \(- y + \tan^{2}{\left(x \right)} + 1\).
- At points where Fᵧ≠0, divide to obtain y′=\(- \frac{y \cos^{2}{\left(x \right)} - 1}{\left(x + 2 y - 1\right) \cos^{2}{\left(x \right)}}\). Points with Fᵧ=0 require a separate local-curve check.
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