Class 12 Continuity And Differentiability · Long Answer3 Marksimplicit_differentiation

Class 12 Maths · Continuity And Differentiability · Long Answer #73

If √(1−x²)+√(1−y²)=a(x−y) on a differentiable branch with |x|<1,|y|<1, show y′=√(1−y²)/√(1−x²).

✅ Step-by-Step Solution

Final Answer: √(1−y²)/√(1−x²)

Put A=√(1−x²)>0,B=√(1−y²)>0. The relation is A+B=a(x−y); it forces x≠y.
Differentiate: −x/A−(y/B)y′=a(1−y′). Rearranging gives (a−y/B)y′=a+x/A.
Substitute a=(A+B)/(x−y). Then a+x/A=(AB+1−xy)/[A(x−y)] and a−y/B=(AB+1−xy)/[B(x−y)]. The common numerator is positive on the open domain.
The ratio is B/A=√(1−y²)/√(1−x²), proving the result.

[0.5 marks] Put A=√(1−x²)>0,B=√(1−y²)>0. The relation is A+B=a(x−y); it forces x≠y.
[1 mark] Differentiate: −x/A−(y/B)y′=a(1−y′). Rearranging gives (a−y/B)y′=a+x/A.
[1 mark] Substitute a=(A+B)/(x−y). Then a+x/A=(AB+1−xy)/[A(x−y)] and a−y/B=(AB+1−xy)/[B(x−y)]. The common numerator is positive on the open domain.
[0.5 marks] The ratio is B/A=√(1−y²)/√(1−x²), proving the result.

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