Question

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

• A. $\frac{1}{(1+x{)}^2}$
• B. $-\frac{1}{(1+x{)}^2}$
• C. $-\frac{1}{(1+x)}+\frac{1}{(1+x{)}^2}$
• D. none of these

Answer 1

Answer: (B)

$x^2(1+y)=y^2(1+x)$
$\Rightarrow x^2-y^2+xy(x-y)=0$
$\Rightarrow (x-y)(x+y+xy)=0$
Now $x\ne y$ [does not satisfy the given equation]
$\therefore x+y+xy=0\Rightarrow y=\frac{-x}{1+x}$
$\therefore \frac{dy}{dx}=\frac{-(1+x)+x}{(1+x{)}^2}=-\frac{1}{(1+x{)}^2}$