If frac{x}{sqrt{1+x}}+frac{y}{sqrt{1+y}}=0 an | vedclass

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(A) Given the equation: $\frac{x}{\sqrt{1+x}} + \frac{y}{\sqrt{1+y}} = 0$ Rearranging the terms: $\frac{x}{\sqrt{1+x}} = -\frac{y}{\sqrt{1+y}}$ Squari

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$-\frac{1}{(1+x)^2}$

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(A) Given the equation: $\frac{x}{\sqrt{1+x}} + \frac{y}{\sqrt{1+y}} = 0$ Rearranging the terms: $\frac{x}{\sqrt{1+x}} = -\frac{y}{\sqrt{1+y}}$ Squaring both sides: $\frac{x^2}{1+x} = \frac{y^2}{1+y}$ Cross-multiplying: $x^2(1+y) = y^2(1+x)$ $x^2 + x^2y = y^2 + xy^2$ $x^2 - y^2 + x^2y - xy^2 = 0$ $(x-y)(x+y) + xy(x-y) = 0$ Since $x 
eq y$,we can divide by $(x-y)$: $x + y + xy = 0$ $y(1+x) = -x$ $y = -\frac{x}{1+x}$ Now,differentiate with respect to $x$ using the quotient rule: $\frac{dy}{dx} = -\frac{(1+x)(1) - x(1)}{(1+x)^2}$ $\frac{dy}{dx} = -\frac{1+x-x}{(1+x)^2}$ $\frac{dy}{dx} = -\frac{1}{(1+x)^2}$

If $\frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}}=0$ and $x \neq y$,then $\frac{dy}{dx}$ is equal to:

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