Question

If \( x\sqrt{1+y} + y\sqrt{1+x} = 0 \), for \( -1<x<1, x \neq y \), then prove that \[ \frac{dy}{dx} = \frac{-1}{(1 + x)^2}. \]

Hint:

Use the product rule and chain rule carefully when differentiating implicit equations.

Answer 1

Differentiate implicitly. Given equation: \[ x\sqrt{1+y} + y\sqrt{1+x} = 0. \] Differentiate both sides using implicit differentiation: \[ \frac{d}{dx} \left( x\sqrt{1+y} \right) + \frac{d}{dx} \left( y\sqrt{1+x} \right) = 0. \] Using product rule: \[ \sqrt{1+y} \cdot \frac{dx}{dx} + x \cdot \frac{1}{2\sqrt{1+y}} \cdot \frac{dy}{dx} + \sqrt{1+x} \cdot \frac{dy}{dx} + y \cdot \frac{1}{2\sqrt{1+x}} = 0. \] Rearrange to solve for \( \frac{dy}{dx} \) and verify the given result.