Question

If $x \sqrt{(1+y)}+y \sqrt{(1+x)}=0$, then $\frac{d y}{d x}$ 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+x y(x-y)=0 $
$\Rightarrow (x-y)(x+y+x y)=0$
Now $x \neq y $ [does not satisfy the given equation]
$\therefore x+y+x y=0 \Rightarrow y=\frac{-x}{1+x} $
$\therefore \frac{d y}{d x}=\frac{-(1+x)+x}{(1+x)^2}=-\frac{1}{(1+x)^2}$