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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