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)

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

$x\sqrt{(1+y)}+y\sqrt{(1+x)}=0$.

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

Squaring both sides, $x^2(1+y)=y^2(1+x)\Rightarrow x^2-y^2+xy(x-y)=0$

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

Now $x\ne y$ [does not satisfy the given equation]

$\therefore x+y+xy=0\Rightarrow y=\frac{-x}{1+x}$

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