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Given the equation $x\sqrt{1+y}+y\sqrt{1+x}=0$, we isolate one of the terms. Let's isolate $x\sqrt{1+y}$:
Rewrite the equation as: $x\sqrt{1+y} = -y\sqrt{1+x}$
Square both sides of the equation to eliminate the square roots: $(x\sqrt{1+y})^2 = (-y\sqrt{1+x})^2$ $x^2(1+y) = y^2(1+x)$
Expand both sides of the equation: $x^2 + x^2y = y^2 + y^2x$
Rearrange the terms to bring all terms to one side: $x^2 - y^2 + x^2y - y^2x = 0$
Factor the difference of squares and factor out $xy$: $(x-y)(x+y) + xy(x-y) = 0$
Factor out the common term $(x-y)$: $(x-y)(x+y+xy) = 0$
This gives us two possible solutions: 1. $x-y = 0 \Rightarrow x = y$ 2. $x+y+xy = 0$
If $x=y$, substituting into the original equation gives: $x\sqrt{1+x} + x\sqrt{1+x} = 0$ $2x\sqrt{1+x} = 0$ This implies $x=0$ or $x=-1$. Since $x > -1$, both are valid solutions. However, we are looking for a general expression for $\frac{dy}{dx}$.
Consider the second solution $x+y+xy = 0$. We want to find $\frac{dy}{dx}$. Rewrite the equation as $y(1+x) = -x$, so $y = \frac{-x}{1+x}$.
Differentiate $y$ with respect to $x$ using the quotient rule: $\frac{dy}{dx} = \frac{d}{dx} \left(\frac{-x}{1+x}\right) = \frac{(-1)(1+x) - (-x)(1)}{(1+x)^2} = \frac{-1-x+x}{(1+x)^2} = \frac{-1}{(1+x)^2}$
Therefore, $\frac{dy}{dx} = -\frac{1}{(1+x)^2}$.
Final Answer: $-\frac{1}{(1+x)^2}$
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