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If $\;x\sqrt{1+y}+y\sqrt{1+x}=0, \;$ Evaluate $ \large\frac{dy}{dx}$

$\begin{array}{1 1} (A) \large \frac{-1}{1+x^2}\\ (B) \large \frac{1}{1+x^2}\\ (C) \large \frac{x-1}{1+x^2} \\ (D) \large \frac{1-x}{1+x^2} \end{array} $

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  • Shift one term to R.H.S, square both sides and then differentiate
Given $x\sqrt{1+y}+y\sqrt{1+x}=0$ $\rightarrow x \sqrt {1+y}=-y\sqrt{1+x}$
Squaring both sides $\rightarrow x^2(1+y)=y^2(1+x)$
$ \Rightarrow x^2 + x^2y-y^2-yx^2=0 \rightarrow (x-y)(x+y+xy)=0$
$ \Rightarrow x+y+xy = 0 \rightarrow y =\large \frac{-x}{1+x}$
Differentiating, we get, $\large\frac{dy}{dx}=\large\frac{-1}{(1+x)^2}$
answered Dec 22, 2013 by balaji.thirumalai
 

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