Question

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

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

Answer 1

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

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

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

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

${\left(\frac{\sqrt{1+y}}{y}\right)}^2={(-\frac{\sqrt{1+x}}{x})}^2$

$\frac{1+y}{y^2}=\frac{1+x}{x^2}$

$x^2+x^{2}y=y^2+y^{2}x$

$x^2-y^2=y^{2}x-x^{2}y$

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

$\therefore x+y+xy=0\Rightarrow y=\frac{-x}{1+x}$ $(\because x\ne y)$

Differentiate on both sides w.r.t x

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

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

$=\frac{-(1-\frac{x}{1+x})}{1+x}$

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

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