Question

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

• A. True
• B. False

Answer 1

The correct option is B False

It is given that,

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

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

$\Rightarrow$ $x^2(1+y)=y^2(1+x)$ [ Squaring both the sides ]

$\Rightarrow$ $x^2+x^{2}y=y^2+x{y}^2$

$\Rightarrow$ $x^2-y^2=x{y}^2-x^{2}y$

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

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

$\Rightarrow$ $x+y=-xy$

$\Rightarrow$ $y+xy=-x$

$\Rightarrow$ $y(1+x)=-x$

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

Differentiating both sides w.r.t. $x,$ we get:

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

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

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

$\therefore$ The given statement is false.