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Question

If $$x\sqrt{1+y} +y\sqrt{1+x} =0$$, prove that $$\dfrac{dy}{dx}=-\dfrac{1}{(1+x)^{2}}$$


Solution

$$x\sqrt{1+y}+y\sqrt{1+x}=0$$

$$x\sqrt{1+y}=−y\sqrt{1+x}$$

Square on both sides
$$(x\sqrt{1+y})^2=(−y\sqrt{1+x})^2$$

$$x^2(1+y)=(−y)^2(1+x)$$

$$x^2(1+y)=y^2(1+x)$$

$$x^2+ x^2y=y^2+ y^2x$$

$$x^2-y^2=y^2x-x^2y $$

$$(x+y)(x−y)=−xy(x−y)$$

$$x+y=−xy$$

$$x=−xy−y$$

$$x=−y(x+1)$$

$$y=−\dfrac{x}{x+1}$$

Differentiate $$y$$ with respect to $$x$$ using Chain rule, we get ;

$$\dfrac{dy}{dx}=\dfrac{−1(x+1)−(−x)1}{(x+1)^2}$$

$$\dfrac{dy}{dx}=−\dfrac{1}{(x+1)^2}$$

Maths

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