If $x\sqrt{1+y}+y\sqrt{1+x}=0, x≠ y $ then value of $\frac{d^2y}{dx^2}$ is :
Answer & explanation
Correct answer: option 1
$x\sqrt{1+y}+y\sqrt{1+x}=0,\;x\ne y.$
$x\sqrt{1+y}=-y\sqrt{1+x}.$
$x^2(1+y)=y^2(1+x).$
$x^2+x^2y=y^2+xy^2.$
$x^2-y^2=xy(y-x).$
$(x-y)(x+y)=-xy(x-y).$
$x+y=-xy.$
$y=-\frac{x}{1+x}.$
$\frac{dy}{dx}=-\frac{1}{(1+x)^2}.$
$\frac{d^2y}{dx^2}=\frac{2}{(1+x)^3}.$
$\displaystyle \frac{d^2y}{dx^2}=\frac{2}{(1+x)^3}.$