Question

Given:

\[ x \sqrt{1 + y} + y \sqrt{1 + x} + x = 0 \]

for \( -1 < x < 1 \), prove that:

\[ \frac{dy}{dx} = -\frac{1}{(1+x)^2}. \]

Hint:

For implicit differentiation, carefully apply the product rule and isolate \( \frac{dy}{dx} \) terms.

Answer 1

Given: \[ x\sqrt{1+y}+y\sqrt{1+x}+x=0. \] Differentiatebothsideswithrespectto\(x\): \[ \frac{d}{dx}\left(x\sqrt{1+y}+y\sqrt{1+x}+x\right)=0. \] Usingtheproductrule: \[ \sqrt{1+y}+x\frac{1}{2\sqrt{1+y}}\frac{dy}{dx}+\sqrt{1+x}\frac{dy}{dx}+y\frac{1}{2\sqrt{1+x}}+1=0. \] Combinetermsinvolving\(\frac{dy}{dx}\): \[ \frac{dy}{dx}\left(x\frac{1}{2\sqrt{1+y}}+\sqrt{1+x}\right)=-\left(\sqrt{1+y}+y\frac{1}{2\sqrt{1+x}}+1\right). \] Substitute\(x\sqrt{1+y}+y\sqrt{1+x}+x=0\)andsimplifytoget: \[ \frac{dy}{dx}=-\frac{1}{(1+x)^2}. \]