Class 12 Continuity And Differentiability · Long Answer3 Marksimplicit_differentiation
Class 12 Maths · Continuity And Differentiability · Long Answer #129
If x√(1+y)+y√(1+x)=0, −1<x<1, y>−1 and x≠y, show y′=−1/(1+x)².
✅ Step-by-Step Solution
Final Answer: −1/(1+x)²
- Square the relation x√(1+y)=−y√(1+x) to get x²(1+y)=y²(1+x).
- Rearrange and factor: (x−y)(x+y+xy)=0. Since x≠y, x+y+xy=0.
- Because x>−1, solve y=−x/(1+x). Differentiating the quotient gives y′=−[(1+x)−x]/(1+x)².
- Therefore y′=−1/(1+x)². The branch y=−x/(1+x) satisfies the original unsquared equation.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] Square the relation x√(1+y)=−y√(1+x) to get x²(1+y)=y²(1+x).
- [1 mark] Rearrange and factor: (x−y)(x+y+xy)=0. Since x≠y, x+y+xy=0.
- [1 mark] Because x>−1, solve y=−x/(1+x). Differentiating the quotient gives y′=−[(1+x)−x]/(1+x)².
- [0.5 marks] Therefore y′=−1/(1+x)². The branch y=−x/(1+x) satisfies the original unsquared equation.
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