If $\sqrt{1+y}+y \sqrt{1+x}=0$, prove that $(1+x)^{2} \frac{d y}{d x}+1=0$
We are given with an equation $x y^{2}=1$, we have to prove that $2 \frac{d y}{d x}+y^{3}=0$ by using the given equation we
will first find the value of $\frac{d y}{d x}$ and we will put this in the equation we have to prove
But first we need to simplify this equation in accordance with our result, which is that in our result there is no square root and our derivative is only in the form of $x$.
$x \sqrt{1+y}+y \sqrt{1+x}=0$
$x \sqrt{1+y}=-y \sqrt{1+x}$
Squaring both sides,
$x^{2}(1+y)=y^{2}(1+x)$
$x^{2}+x^{2} y=y^{2}+x y^{2}$
$x^{2}-y^{2}=x y^{2}-x^{2} y$
$(x-y)(x+y)=x y(y-x)$
$x+y=-x y$
$y=\frac{-x}{1+x}$
So, now by differentiating the equation on both sides with respect to $x$, we get,
By using quotient rule, we get,
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{(1+\mathrm{x})(-1)-(-\mathrm{x})(1)}{(1+\mathrm{x})^{2}}$
$\frac{d y}{d x}=\frac{-1}{(1+x)^{2}}$