Question

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

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

Answer 1

The correct option is C $\frac{-1}{(1+x{)}^2}$

Given,

$x\sqrt{1+y}+y\sqrt{1+x}=0$

$\Rightarrow x\sqrt{1+y}=y\sqrt{1+x}$

squaring on both sides, we get

$x^2(1+y)=y^2(1+x)$

$x^2-y^2=xy(y-x)$

$(x+y)(x-y)=xy(y-x)$

$(x+y)=-xy$

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

$\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(-\frac{x}{1+x})$

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

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