If x√(1 + y) + y√(1 + x) = 0 and x ≠ y, then prove that dy/dx = -1/(x + 1)^2.

Question

If \(x\sqrt{1+y}+y\sqrt{x+1}\) = 0 and x ≠ y, then prove that \(\frac{dy}{dx}\) = \(-\frac{1}{(x+1)^2}\).

Answer 1

Given \(x\sqrt{1+y}+y\sqrt{x+1}\) = 0

\(x\sqrt{1+y}=-y\sqrt{1+x}\)

Squaring both sides we get 

⇒ x² (1 + y) = y² (1 + x)

⇒ x² + x² y = y² + y² x 

⇒ x² – y² + x² y – y² x = 0 

⇒ (x + y) (x – y) + xy(x – y) = 0 

⇒ (x – y) [(x + y) + xy] = 0 

∴ x – y = 0 (or) x + y + xy = 0 

x = y (or) x + y + xy = 0 

Given that x ≠ y 

x + y + xy = 0 

⇒ y + xy = -x 

⇒ y(1 + x) = -x

Hence proved.