Question

$x\sqrt{1+y}+y\sqrt{1+x}=0.Then\frac{dy}{dx}=$

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

Answer 1

Answer: (B)

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

$x\sqrt{1+y}+y\sqrt{1+x}=0$
$\Rightarrow x\sqrt{1+y}=-y\sqrt{1+x}=0$
$\Rightarrow \frac{x^2}{1+x}=\frac{y^2}{1+y}=\frac{x^2-y^2}{x-y}$
$\Rightarrow \frac{y^2}{1+y}=x+y$
$\Rightarrow x+y+xy=0$
$\Rightarrow y=-\frac{x}{1+x}=\frac{1}{1+x}-1$
$\therefore \frac{dy}{dx}=-\frac{1}{(1+x{)}^2}$