If x\sqrt{1+y}+y\sqrt{1+x}=0 and x\neq y, prove that \frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{1}{(x+1)^2}.

 

 

 

 

 
 
 
 
 

Answers (1)

Given 

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

\Rightarrow x\sqrt{1+y}=-y\sqrt{1+x}

On squaring both sides, we get

\Rightarrow x^2(1+y)=y^2(1+x)

\Rightarrow x^2+x^2y=y^2+y^2x

\Rightarrow x^2-y^2+x^2y-y^2x=0\Rightarrow (x-y)(x+y)=xy(y-x)

\Rightarrow (x-y)\left [ x+y+xy \right ]=0\Rightarrow x+y+xy=0\left [ \because x\neq y \right ]

\Rightarrow y=-\frac{x}{1+x}\: \: -(i)

On differentiating equation (i) w.r.t x.

\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{(1+x)\frac{\mathrm{d} (-x)}{\mathrm{d} x}-(-x)\frac{\mathrm{d} (1+x)}{\mathrm{d} x}}{(1+x)^2}

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

\Rightarrow \frac{-1}{(1+x)^2}\: \: LHS=RHS

Hence proved 

 

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