Discussion Forum

If   $\frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}}=0, x\neq y,$  then  $(1+x)^{2}\frac{dy}{dx}=$

Answer:

Explanation:

We have

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

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

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

$\Rightarrow$    $x^{2}(1+y)=y^{2}(1+x)$

$\Rightarrow$   $x^{2}+x^{2}y-y^{2}-xy^{2}=0$

$\Rightarrow$     $(x+y)(x-y)+xy(x-y)=0$

$\Rightarrow$     $(x-y)(x+y+xy)=0$

$\Rightarrow$     $y(1+x)=-x$    $[\because x-y\neq0]$

$\Rightarrow$    y=   $\frac{-x}{1+x}$

$\Rightarrow$    $\frac{dy}{dx}=-\left[\frac{(1+x)(1)-(x)}{(1+x)^{2}}\right]$

$\Rightarrow$    $(1+x)^{2}\frac{dy}{dx}=-1$