Question:
If $x y^{2}=1$, prove that $2 \frac{d y}{d x}+y^{3}=0$
Solution:
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, so by differentiating the equation on both sides with respect to $x$, we get,
$y^{2}(1)+2 x y \frac{d y}{d x}=0$
$\frac{d y}{d x}=\frac{-y}{2 x}$
Or we can further solve it by using the given equation,
$\frac{d y}{d x}=\frac{-y}{2 \frac{1}{y^{2}}}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{y}^{3}}{2}$
By putting this value in the L.H.S. of the equation, we get,
$2\left(\frac{-y^{2}}{2}\right)+y^{3}=0=$ R.H.S.