Question

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

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

Answer 1

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

Solution :-

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

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

squaring both sides, we get

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

$\Rightarrow x^2+x^2.y=y^2+y^{2}x$

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

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

$\Rightarrow x+y=-xy$

$\Rightarrow x=-xy-y$

$\Rightarrow x=-y(x+1)$

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

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