NCERT Class 12 · Ch 5 Miscellaneous Exercise · Q14
NCERT Class 12 Maths · Chapter 5 Continuity And Differentiability · Miscellaneous Exercise Question 14
If x√(1+y)+y√(1+x)=0, −1<x<1, prove y′=−1/(1+x)².
💡 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: y=−x/(1+x), hence y′=−1/(1+x)².
- The real equation has y≥−1; y=−1 would give a nonzero term −√(1+x), so y>−1. Divide by √[(1+x)(1+y)] to get x/√(1+x)+y/√(1+y)=0.
- Put u=√(1+x)>0, v=√(1+y)>0. This reads u−1/u+v−1/v=0, or (u+v)(uv−1)=0.
- Since u+v>0, uv=1. Thus (1+x)(1+y)=1 and y=−x/(1+x). Differentiation gives the claimed result.
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