Question

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

Hint:

When differentiating implicit functions, apply the product rule, chain rule, and solve for the desired derivative.

Answer 1

Step 1: Differentiate both sides of the equation with respect to \( x \). Applying implicit differentiation: \[ \frac{d}{dx}\left( x\sqrt{1 + y} + y\sqrt{1 + x} \right) = 0. \] 

Step 2: Use the product rule and chain rule: \[ \frac{d}{dx}\left( x\sqrt{1 + y} \right) = \sqrt{1 + y} + x \cdot \frac{1}{2\sqrt{1 + y}} \cdot \frac{dy}{dx}, \] \[ \frac{d}{dx}\left( y\sqrt{1 + x} \right) = \frac{dy}{dx} \cdot \sqrt{1 + x} + y \cdot \frac{1}{2\sqrt{1 + x}}. \] 

Step 3: Combine both expressions: \[ \sqrt{1 + y} + x \cdot \frac{1}{2\sqrt{1 + y}} \cdot \frac{dy}{dx} + \frac{dy}{dx} \cdot \sqrt{1 + x} + y \cdot \frac{1}{2\sqrt{1 + x}} = 0. \] 

Step 4: Solve for \( \frac{dy}{dx} \), and simplify to get the desired result: \[ \frac{dy}{dx} = -\frac{1}{(1 + x^2)^2}. \]