Question

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

• A. $\frac{1}{(1+x{)}^2}$
• B. $-\frac{1}{(1+x{)}^2}$
• C. $-\frac{1}{(1-x{)}^2}$
• D. $\frac{1}{(1-x{)}^2}$

Answer 1

Answer: (B)

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

Given, $x\sqrt{(1+y)}+y\sqrt{(1+x)}=0$
$\Rightarrow x\sqrt{(1+y)}=-y\sqrt{(1+x)}$
Squaring both side we get, $x^2(1+y)=y^2(1+x)$
$x+y+xy=0$ or $y=x$ (rejected)
$y=-\frac{x}{1+x}=\frac{1}{1+x}-1$
$\therefore \frac{dx}{dy}=-\frac{1}{(1+x{)}^2}$