If $x\sqrt{1+y}+y\sqrt{1+x}=0, $ then $\frac{dy}{dx}=$
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{-1}{(1+x)^2}$
$x\sqrt{1+y}+y\sqrt{1+x}=0$
$\sqrt{1+y}+x\frac{1}{2\sqrt{1+y}}\frac{dy}{dx}+\sqrt{1+x}\frac{dy}{dx}+y\frac{1}{2\sqrt{1+x}}=0$
$\frac{dy}{dx}\left(\frac{x}{2\sqrt{1+y}}+\sqrt{1+x}\right)=-\left(\frac{y}{2\sqrt{1+x}+\sqrt{1+y}}\right)$
$⇒\frac{dy}{dx}=\frac{-1}{(1+x)^2}$