Question

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

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

Answer 1

Answer: (C)

$x \sqrt{1+y}+y \sqrt{1+x}=0$
$=x \sqrt{1+y}=-y \sqrt{1+x}$
Squaring,
$x^{2}(1+y)=y^{2}(1+x)$
$=x^{2}+x^{2} y=y^{2}+x y^{2}$
$=x^{2}-y^{2}=x y^{2}-x^{2} y$
$=x^{2}-y^{2}=x y(y-x)$
$=(x +y)(x-y)=x y(y-x)$
$=(x +y)=-x y$
$=x+ x y+ y=0$
$=(1+x) y=-x$
$y=-x /(1+x)$
differentiating,
$\frac{d y}{d x}=\frac{(1+x)-x}{(1+x)^{2}}$
$=\frac{-1}{(1+x)^{2}}$